8.1. Visualize:
Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing on a rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contact force, and the earth, which exerts the long range force of gravity. Figure (ii) shows the barbell, the weightlifter, and the earth separated from one another. This separation helps to indicate the forces acting on each object. Figures (iii) and (iv) are free body diagrams for the barbell and the weightlifter, respectively. Altogether there are four interactions. There is the interaction between the barbell and the weightlifter, the weightlifter and the surface of the earth (contact force), the barbell and the earth, and the weightlifter and the earth. r r The interaction between the barbell and the weightlifter leads to two forces: F WL on BB and F BB on WL. These are an action/reaction pair. The interaction between the weightlifter and the surface of the earth is a contact interaction r r r leading to n S on WL and n WL onr S. These two forces also constitute an action/reaction pair. The weight force w E on BB r r has its action/reaction force w BB on E, and the weight force w E on WL has its action/reaction force w WL on E.
8.2. Visualize:
Solve: Figure (i) shows a softball player (P) with a ball (B) in her hand that has come forward beside her head and is just about ready to release the ball. We have represented the earth (E) and its rough surface (S). Contact forces exist between the surface and the softball player. Figure (ii) shows the ball, the player, and the earth separated. This separation helps to indicate the various forces on each object. We have altogether 6 interactions. The action/reaction pairs of all the forces have been indicated by dotted lines. Figure (iii) shows the free-body diagrams for the ball and the softball player.
8.3. Visualize:
Solve: Figure (i) shows the head-on collision of a bowling ball (BB) and a soccer ball (SB) on a hard surface (S) of the earth (E). There are 3 types of interactions in the present case: between the bowling ball and the soccer ball, between the earth and the balls (long-range force of gravity), and between the balls and the surface (contact force). Figure (ii) shows the bowling ball, the soccer ball, and the surface separated from one another. The force on the r r bowling ball due to the soccer ball FrSB on BB and the r force on the soccer ball due to the bowling ball F BB on SBr make an action/reaction pair. The weights r r w E on SB and r w SB on E are an action/reaction pair,r and so are rthe weights w E on BB and w BB on E. The contact forces n S on BB and n S on SB have action/reaction forces in n BB on S and n SB on S, respectively. All the action/reaction pairs are indicated by dotted lines. Figure (iii) shows the free-body diagrams for the bowling ball and the soccer ball.
8.4. Visualize:
Solve: Figure (i) shows the mountain climber (C), the rope (R), the bag of supplies (B), the surface of the mountain (S), and the earth (E). To indicate the forces on each of rthe various objects, we have separated C, R, B, S, r r and E. In the diagram, n denotes the normal (contact) forces, f denotes the frictional (contact) forces, and w denotes the weights. All action/reaction forces have also been identified in this figure with dotted lines. Figure (ii) shows the free-body diagrams on the bag, the rope, and the climber.
8.5. Visualize:
Solve: The top figure shows the pulley (P), block A, block B, the surface S of the incline, the rope (R), and the r earth (E). In indicating have denoted the normal (contact) forces by n , the kinetic frictional r the various forces, we r (contact) forces by f , and the weights by w . All the action/reaction forces have been identified with dotted lines. The bottom figure shows the free body diagrams of each block, the rope, and the pulley.
8.6. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-range magnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In addition, the table experiences a normal force due to the surface.
(b) Solve: Each object is in static equilibrium with (Fnet)y = 0. Start with the lower magnet. Because FU on L = 3wL = 6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N because these are an action/reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the table, action/reaction pairs are nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0. The table’s weight is wT = 20 N, leaving nS on T = 24.0 N in order for the table to be in equilibrium. Summarizing,
Upper magnet wU = 2.0 N nT on U = 8.0 N FL on U = 6.0 N
Table wT = 20.0 N nU on T = 8.0 N nL on T = 4.0 N n S on T = 24.0 N
Lower magnet wL = 2.0 N nT on L = 4.0 N FU on L = 6.0 N
Assess: The result nS on T = 24.0 N makes sense. The combined weight of the table and two magnets is 24.0 N. Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total weight of the table and magnets.
8.7.
Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denoted by A and C, respectively, and they are separate systems. The launch pad is a part of the environment. Visualize:
Solve:
(a) Newton’s second law for the astronaut is
∑(F )
on A y
= nC on A − wA = mA aA = 0 N ⇒ nC on A = wA = mA g
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA g = (80 kg)(9.8 m/s 2 ) = 784 N
(b) Newton’s second law for the astronaut is:
∑(F )
on A y
= nC on A − wA = mA aA ⇒ nC on A = wA + mA aA = mA ( g + aA )
By Newton’s third law, the astronaut’s force on the chair is
nA on C = nC on A = mA ( g + aA ) = (80 kg)(9.8 m/s 2 + 10 m/s 2 ) = 1580 N
Assess: This is a reasonable value because the astronaut’s acceleration is more than g.
8.8. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T, respectively. The ground will be denoted by the symbol G. Visualize:
Solve:
(a) The x-component of Newton’s second law for the car is
∑(F )
on C x
= FG on C − FT on C = mC aC
The x-component of Newton’s second law for the truck is
∑(F )
on T x
= FC on T = mT aT
Using aC = aT = a and FT on C = FC on T, we get
(F
C on G
1 − FC on T =a mC
)
( F ) m1 = a C on T
T
Combining these two equations,
(F
C on G
1 1 1 1 1 − FC on T + = FC on G = FC on T ⇒ FC on T mT mC m T mC mC
)
(
)
(
)
mT 2000 kg ⇒ FC on T = FC on G = ( 4500 N ) 1000 kg + 2000 kg = 3000 N mC + m T
(
)
(b) Due to Newton’s third law, FT on C = 3000 N.
8.9. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionless and along with the earth it is a part of the environment. The three blocks are our three systems of interest. Visualize:
The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a. Solve: Newton’s second law for the three blocks along the x-direction is
∑(F )
on 1 x
= FA on 1 − F2 on 1 = m1a
∑(F )
on 2 x
= F1 on 2 − F3 on 2 = m2 a
∑(F )
on 3 x
= F2 on 3 = m3 a
Adding these three equations and using Newton’s third law (F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3), we get
FA on 1 = ( m1 + m2 + m3 )a ⇒ (12 N ) = (1 kg + 2 kg + 3 kg)a ⇒ a = 2 m /s 2
Using this value of a, the force equation on block 3 gives
F2 on 3 = m3 a = (3 kg)(2 m/s 2 ) = 6 N
Substituting into the force equation on block 1,
12 N − F2 on 1 = (1 kg)(2 m/s 2 ) ⇒ F2 on 1 = 10 N
Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the 2 kg block exerts on the 1 kg block is reasonable.
8.10. Model: The astronaut and the satellite, the two systems of our interest, will be treated as particles. Visualize:
Solve: The astronaut and the satellite accelerate in opposite directions for 0.5 s. The force on the satellite and the force on the astronaut are an action/reaction pair, so both are 100 N. Newton’s second law for the satellite along the x-direction is
∑(F )
on S x
= FA on S = mS aS ⇒ aS =
FA on S
−(100 N ) = −0.156 m /s 2 640 kg
=
mS
Newton’s second law for the astronaut along the x-direction is
∑(F )
= FS on A = mA aA ⇒ aA =
on A x
FS on A mA
FA on S
=
mA
=
100 N = 1.25 m /s 2 80 kg
Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.5s under the accelerations aA and aS:
(1.25 m/s )(0.5 s − 0 s) = 0.156 m + v (t − t ) + a (t − t ) = 0 m + 0 m + ( −0.156 m/s )(0.5 s − 0 s) = −0.02 m v = v + a (t − t ) = 0 m/s + (1.25 m/s )(0.5 s − 0 s) = 0.625 m/s v = v + a (t − t ) = 0 m/s + ( −0.156 m/s )(0.5 s − 0 s) = −0.078 m/s
x1A = x 0A + v0A (t1 − t 0 ) + 12 aA (t1 − t 0 ) = 0 m + 0 m + 2
x1S = x 0S
0S
1
2
0
1 2
S
0A
A
1
1
0
1 2
1 2
2
2
2
2
2
1A
0
2
1S
0S
S
1
0
With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain the new positions at (t2 − t1 ) = 59.5 s:
x 2A = x1A + v1A (t2 − t1 ) = 0.156 m + (0.625 m /s)(59.5 s) = 37.34 m x 2S = x1S + v1S (t2 − t1 ) = − 0.02 m + ( −0.078 m /s)(59.5 s) = −4.66 m Thus the astronaut and the satellite are x 2A − x 2S = (37.34 m ) − ( −4.66 m ) = 42.0 m apart.
8.11. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume the pulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley. The two systems are the man and the block. Visualize:
Solve: Clearly the entire system remains in equilibrium since mB > mM. The block would move downward but it is already on the ground. From the free-body diagrams, we can write down Newton’s second law in the vertical direction as
∑(F )
on M y
(
)
= TR on M − wM = 0 N ⇒ TR on M = wM = (60 kg) 9.8 m/s 2 = 588 N
Since the tension is the same on both sides, TB on R = TM on R = T = 588 N .
8.12.
Model: The carp (C) and the trout (T) are two systems that will be represented through the particle model. The fishing rod line (R) is assumed to be massless. Visualize:
Solve: Jimmy’s pull T2 is larger than the total weight of the fish, so they accelerate upward. They are tied together, so each fish has the same acceleration a. Newton’s second law along the y-direction for the carp and the trout is
∑(F )
on C y
= T2 − T1 − wC = mC a
∑(F )
on T y
= T1 − wT = mT a
Adding these two equations gives
a=
2 2 T2 − wC − wT 60 N − (1.5 kg)( 9.8 m /s ) − (3 kg)(9.8 m /s ) = = 3.533 m/s2 1.5 kg + 3.0 kg ( mC + m T )
Substituting this value of acceleration back into the force equation for the trout, we find that
T1 = mT ( a + g) = (3 kg)(3.533 m/s 2 + 9.8 m/s 2 ) = 40 N wT = mT g = (3 kg)(9.8 m/s 2 ) = 29.4 N Thus, T2 > T1 > wT > wC .
wC = mC g = (1.5 kg)(9.8 m/s 2 ) = 14.7 N
8.13. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We must therefore consider both the block of ice and the rope as systems. Visualize:
r Solve: The force Fext acts only on the rope. Since the rope and the ice block move together, they have the same acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts on the rear end of the rope. Newton’s second law along the x-axis for the ice block and the rope is
∑(F )
= FR on I = mI a = (10 kg)(2.0 m /s 2 ) = 20 N
∑(F )
= Fext − FI on R = mR a ⇒ Fext − FR on I = mR a
on I x
on R x
⇒ Fext = FR on I + mR a = 20 N + (0.5 kg)(2.0 m/s 2 ) = 21 N
8.14. Model: The hanging block and the rail car are separate systems. Visualize:
Solve: The mass of the rope is very small in comparison to the 2000 kg block, so we will assume a massless rope. r r In that case, the forces T1 and T1′ act as if they are an action/reaction pair. The hanging block is in static r equilibrium, with Fnet = 0 N , so T1′ = mblock g = 19,600 N . The rail car with the pulley is also in static equilibrium:
T2 + T3 − T1 = 0 N Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now, T1 = T1′ = 19, 600 N by Newton’s third law. Also, the cable tension is T2 = T3 = T . Thus, T = 12 T1′ = 9800 N .
8.15.
Model: The two hanging blocks, which can be modeled as particles, form two systems. The other two systems are the two knots where rope 1 meets with rope 2 and rope 2 meets with rope 3. All the four systems are in static equilibrium. The ropes are assumed to be massless. Visualize:
Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are in static equilibrium r r with Fnet = 0 N. Note that there are three action/reaction pairs. For Block 1 and Block 2, Fnet = 0 N and we have
T4′ = w1 = m1 g
T5′ = w2 = m2 g
T4 = T4′ = m1 g
T5 = T5′ = m2 g
Then, by Newton’s third law: The knots are also in equilibrium. Newton’s law applied to the left knot is
( Fnet ) x = T2 − T1 cosθ1 = 0 N
( Fnet ) y = T1 sin θ1 − T4 = T1 sin θ1 − m1g = 0 N
The y-equation gives T1 = m1 g sin θ1 . Substitute this into the x-equation to find
T2 =
m1 g cosθ1 mg = 1 sin θ1 tan θ1
Newton’s law applied to the right knot is
( Fnet ) x = T3 cosθ 3 − T2′ = 0 N
( Fnet ) y = T3 sin θ 3 − T5 = T3 sin θ 3 − m2 g = 0 N
These can be combined just like the equations for the left knot to give
T2′ =
m2 g cosθ 3 mg = 2 sin θ 3 tan θ 3
r r But the forces T2 and T2′ are an action/reaction pair, so T2 = T2′ . Therefore, m1 g mg m = 2 ⇒ tan θ 3 = 2 tan θ1 ⇒ θ 3 = tan −1 (2 tan 20°) = 36° tan θ1 tan θ 3 m1 We can now use the y-equation for the right knot to find T3 = m2 g sin θ 3 = 66.6 N .
8.16. Model: The block (B) and the steel cable (C), the two systems of interest to us, are treated like particles. The motion of these systems is governed by the constant-acceleration kinematic equations. Visualize:
Solve:
Using v12x = v02x + 2 a x ( x1 − x 0 ) ,
(4.0 m/s)2 = 0 m 2 /s 2 + 2ax (2.0 m) ⇒ ax
= 4.0 m /s 2
From the free-body diagram on the block:
∑(F )
on B x
= FC on B = mB a x ⇒ FC on B = (20 kg)( 4.0 m/s 2 ) = 80 N
Also, according to Newton’s third law FB on C = FC on B = 80 N. Newton’s second law on the cable is:
∑(F )
on C x
= Fext − FB on C = mC a x ⇒ 100 N − 80 N = mC ( 4.0 m/s 2 ) ⇒ mC = 5 kg
8.17.
Model: The block (B) and the steel cable (C), the two systems of our interest, are considered particles, and their motion is determined by the constant-acceleration kinematic equations. Visualize:
Solve:
Using v1 x = v0 x + a x (t1 − t 0 ) ,
4.0 m /s = 0 m /s + a x (2.0 s − 0 s) ⇒ a x = 2.0 m /s 2 Newton’s second law along the x-direction for the block is
∑(F )
on B x
= FC on B = mB a x = (20 kg)(2.0 m/s 2 ) = 40 N
Fext acts on the right end of the cable and FB on C acts on the left end. According to Newton’s third law, FB on C = FC on B = 40 N. The difference in tension between the two ends of the cable is thus Fext − FB on C = 100 N – 40 N = 60 N
8.18.
Model: The block (B) and the steel cable (C), the two systems of our interest, are treated as particles, and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of the cable. Visualize:
Solve:
Using x1 = x 0 + v0 x (t1 − t 0 ) + 12 a x (t1 − t 0 ) , 2
4.0 m = 0 m + 0 m + 12 a x (2.0 s − 0 s) ⇒ a x = 2.0 m /s 2 Newton’s second law along the x-direction for the block is 2
∑(F )
onB x
= FC on B = mB a x = (20 kg)(2.0 m/s 2 ) = 40 N
Therefore, the change in tension (T) in the cable from one end to the other is Fext − FB on C = 100 N − 40 N = 60 N. The tension in the cable as a function of x is (60 N) T ( x ) = 40 N + x = ( 40 + 60x ) N 1m with x in meters. x = 0 m is where the cable attaches to the box and x = 1.0 m is at the right end of the cable.
8.19.
Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weight of this segment of rope is a downward force. It is balanced by the tension force at height y.
Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is a fraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope segment M m= y L This segment of rope is in static equilibrium, so the tension force pulling up on it is Mg (2.2 kg)(9.8 m /s 2 ) T = w = mg = y= y = 7.19 y N L 3.0 m where y is in m. The tension increases linearly from 0 N at the bottom (y = 0 m) to 21.6 N at the top (y = 3 m). This is shown in the graph.
8.20. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction. Visualize:
Solve:
The acceleration constraint is ( aA ) x = ( aB ) x = a x . Newton’s second law on sled A is r r ∑ Fon A = nA − wA = 0 N ⇒ nA = wA ∑ Fon A = T1 on A − fA = mA ax
( )
( )
y
Using fA = µknA, the x-equation yields
(
x
)
T 1 on A − µ k nA = mA a x ⇒ 150 N − (0.1)(100 kg) 9.8 m/s 2 = (100 kg)a x ⇒ a x = 0.52 m /s 2 On sled B:
r
∑(F
on B
)
y
= nB − w B = 0 N ⇒ nB = w B
r
∑(F
on B
)
x
= T2 − T1 on B − fB = mB a x
T 1 on B and T 1 on A act as if they are an action/reaction pair, so T 1 on B = 150 N. Using fB = µ k nB = (0.10)(80 kg)
(9.8 m/s ) = 78.4 N , we get 2
Thus the tension T2 = 270 N.
T2 − 150 N − 78.4 N = (80 kg)(0.52 m/s 2 ) ⇒ T2 = 270 N
8.21. Model: The coffee mug (M) is our system of interest, and it will be treated as a particle. The model of friction and the constant-acceleration kinematic equations will also be used. Visualize:
Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the same. Using v12x = v02x + 2 a x ( x1 − x 0 ) , we get
0 m 2 /s 2 = (20 m/s) + 2 a x (50 m ) ⇒ a x = −4.0 m / s 2 2
The static force needed to stop the mug is
( Fnet ) x = − fs = max = (0.5 kg)(−4.0 m/s 2 ) = −2.0 N ⇒ fs = 2.0 N
The maximum force of static friction is
fs max = µ s n = µ s w = µ s mg = (0.50)(0.50 kg)(9.8 m/s 2 ) = 2.45 N
Since fs < fs max, the mug does not slide.
8.22.
Visualize:
The car and the ground are denoted by C and G, respectively. r Solve: (a) As the drive wheels turn they push backward against the ground. This is a static friction force FC on G r because the wheels don’t slip against the ground. By Newton’s third law, the ground exerts a reaction force FG on C . r This reaction force is opposite in direction to FC on G , hence, is in the forward direction. This is the force that accelerates the car. r (b) The car has an internal source of energy – fuel – that allows it to turn the wheels and exert the force FC on G . r This is an active push by the car, generating the force FG on C in response. Houses do not have an internal source of energy that allows them to push sideways against the ground. (c) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although that is not critical) and the nondrive wheels. For this car, two-thirds of the weight rests on the front wheels. Physically, r force FG on C is a static friction force. The maximum acceleration of the car on the ground (or concrete surface) occurs when the static friction reaches its maximum possible value.
FG on C = ( fs ) max = µ s nF = µ s wF = (1.00)( 23 )(1500 kg)(9.8 m/s 2 ) = 9800 N ⇒ amax =
FG on C m
=
9800 N = 6.53 m /s 2 1500 kg
8.23. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:
r (a) The tractor beam is some kind of long-range force FM on m . Regardless of what kind of force it is, by r Newton’s third law there must be a reaction force Fm on M on the starship. As a result, both the shuttlecraft and the starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However, the very different masses of the two craft means that the distances they each move will also be very different. The pictorial representation shows that they meet at time t1 when xM1 = xm1 . There’s only one force on each craft, so Newton’s second law is very simple. Furthermore, because the forces are an action/reaction pair, Solve:
FM on m = Fm on M = Ftractor beam = 4 × 104 N The accelerations of the two craft are
aM =
Fm on M M
4 × 10 4 N = = 0.020 m /s 2 2 × 10 6 kg
am =
r FM on m m
=
−4 × 10 4 N = −2.0 m /s 2 2 × 10 4 kg
Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. The accelerations are very different even though the forces are the same. Now we have a constant-acceleration problem in kinematics. At a later time t1 the positions of the crafts are
x M1 = x M 0 + v M 0 (t1 − t 0 ) + 12 a M (t1 − t 0 ) = 12 a M t12 2
x m1 = x m 0 + vm 0 (t1 − t 0 ) + 12 am (t1 − t 0 ) = x m 0 + 12 am t12 2
The craft meet when xM1 = xm1, so 1 2
a M t12 = x m 0 + 12 am t12 ⇒ t1 =
2 xm0 = a M − am
2 xm0 = a M + am
2(10,000 m ) = 99.5 s 2.02 m /s 2
Knowing t1, we can now find the starship’s position as it meets the shuttlecraft:
x M1 = 12 a M t12 = 99.0 m The starship moves 99.0 m as it pulls in the shuttlecraft from 10 km away.
8.24. Model: The rock (R) and Bob (B) are two systems of our interest, and will be treated as particles. We will also use the constant-acceleration kinematic equations. Visualize:
r Solve: (a) Bob exerts a forward force F B on R on the rock to accelerate it forward. The rock’s acceleration is calculated as follows: 2 (30 m /s) = 450 m /s 2 v1R = 2 ∆x 2(1 m ) 2
2 2 v1R = v0R + 2 a0R ∆x ⇒ aR =
The force is calculated from Newton’s second law:
FB on R = mR aR = (.5 kg)( 450 m/s 2 ) = 225 N
r r r (b) Because Bob pushes on the rock, the rock pushes back on Bob with a force F R on B. Forces F R on B and F B on R are an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with an acceleration equal to aB =
(F
)
net on B x
mB
=−
FR on B mB
=−
225 N = −3.0 m /s 2 75 kg
This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time interval by returning to the kinematics of the rock:
v1R = v0R + aR ∆t = aR ∆t ⇒ ∆t =
v1R = 0.0667 s aR
At the end of this interval, Bob’s velocity is v1B = v0B + aB ∆t = aB ∆t = −0.200 m /s Thus his recoil speed is 0.200 m/s.
8.25. Model: Assume package A and package B are particles. Use the model of kinetic friction and the constant-acceleration kinematic equations. Visualize:
Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it. Package A will push back on B. The acceleration constraint is ( aA ) x = ( aB ) x = a. Newton’s second law for each package is
∑(F )
on A x
= FB on A + wA sin θ − fkA = mA a
⇒ FB on A + mA g sin θ − µ kA ( mA g cosθ ) = mA a
∑(F )
on B x
= − FA on B − fkB + wB sin θ = mB a
⇒ − FA on B − µ kB ( mB g cosθ ) + mB g sin θ = mB a where we have used nA = mA cosθ g and nB = mB cosθ g. Adding the two force equations, and using FA on B = FB on A because they are an action/reaction pair, we get
a = g sin θ −
( µ kA mA + µ kB mB )(g cosθ ) = 1.817 m /s 2 mA + mB
Finally, using x1 = x 0 + v0x (t1 − t 0 ) + 12 a(t1 − t 0 ) , 2
2m =0m+0m+
1 2
(1.817 m/s )(t 2
1
− 0 s) ⇒ t1 = 1.48 s 2
8.26. Model: The two ropes and the two blocks (A and B) will be treated as particles. Visualize:
Solve: (a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined system is accelerating upward at a = 3.0 m/s2 under the influence of a force F and the weight −Mgjˆ . Newton’s second law applied to the combined system is
( Fnet ) y = F − Mg = Ma ⇒ F = M(a + g) = 32.0 N (b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s second law to each system, starting at the top. Each has an acceleration a = 3.0 m/s2. For block A:
(F
)
net on A y
= F − mA g − T1 on A = mA a ⇒ T1 on A = F − mA ( a + g) = 19.2 N
(c) Applying Newton’s second law to rope 1:
(F
)
net on 1 y
= TA on 1 − m1 g − TB on 1 = m1a
r r r T A on 1 and T 1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces T A on 1 and r T B on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is TB on 1 = TA on 1 − m1 ( a + g) = 16.0 N (d) We can continue to repeat this procedure, noting from Newton’s third law that
T1 on B = TB on 1 and T2 on B = TB on 2 Newton’s second law applied to block B is
(F
)
net on B y
= T1 on B − mB g − T2 on B = mB a ⇒ T2 on B = T1 on B − mB ( a + g) = 3.2 N
8.27. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes are assumed to be massless, and the model of kinetic friction will be used. Visualize:
Solve: (a) The separate free body diagrams for the two blocks show that there are two action/reaction pairs. r r Notice how block 1 both pushes down on block 2 (force n1′ ) and exerts a retarding friction force f 2 top on the top surface of block 2. Block 1 is in static equilibrium (a1 = 0 m/s2) but block 2 is accelerating. Newton’s second law for block 1 is
(F
)
net on 1 x
(F
= f1 − Trope = 0 N ⇒ Trope = f1
)
net on 1 y
= n1 − m1 g = 0 N ⇒ n1 = m1 g
Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2. The friction model means f1 = µ k n1 = µ k m1 g . Substitute this result into the x-equation to get the tension in the rope:
Trope = f1 = µ k m1 g = 3.92 N (b) Newton’s second law for block 2 is
ax = a =
(F
)
net on 2 x
m2
=
Tpull − f2 top − f2 bot m2
a y = 0 m /s 2 =
(F
)
net on 2 y
m2
=
n2 − n1′ − m2 g m2
r r Forces n1 and n1′ are an action/reaction pair, so n1′ = n1 = m1 g . Substituting into the y-equation gives n2 = ( m1 + m2 )g. This is not surprising because the combined weight of both objects presses down on the surface. The kinetic friction on the bottom surface of block 2 is then
f2 bot = µ k n2 = µ k ( m1 + m2 )g r r The forces f1 and f2 top are an action/reaction pair, so f2 bot = f1 = µ k m1 g . Inserting these friction results into the x-equation gives a=
(F
)
net on 2 x
m2
=
Tpull − µ k m1 g − µ k ( m1 + m2 )g m2
= 2.16 m /s 2
8.28. Model: Blocks 1 and 2 are our systems of interest and will be treated as particles. Assume a frictionless rope and massless pulley. Visualize:
Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraint is a2 = a = −a1, where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Make sure you r understand why the friction forces point in the directions shown in the free-body diagrams, especially force f1′ exerted on block 2 by block 1. We have quite a few pieces of information to include. First, Newton’s second law for blocks 1 and 2: r Fnet on 1 = n1 − m1 g = 0 N ⇒ n1 = m1 g Fnet on 1 = f1 − T1 = µ k n1 − T1 = m1a1 = − m1a
(
)
(
x
(F
)
net on 2 x
)
y
= Tpull − f1′− f2 − T2 = Tpull − f1′− µ k n2 − T2 = m2 a2 = m2 a
(F
)
net on 2 y
= n2 − n1′ − m2 g = 0 N ⇒ n2 = n1′ + m2 g
We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:
n1′ = n1 = m1 g
f1′ = f1 = µ k n1 = µ k m1 g
T1 = T2 = T
Knowing n1′ , we can now use the y-equation of block 2 to find n2. Substitute all these pieces into the two x-equations, and we end up with two equations in two unknowns:
µ k m1 g − T = − m1a
Tpull − T − µ k m1 g − µ k ( m1 + m2 )g = m2 a
Subtract the first equation from the second to get
Tpull − µ k (3m1 + m2 )g = ( m1 + m2 )a ⇒ a =
Tpull − µ k (3m1 + m2 )g m1 + m2
= 1.77 m /s 2
8.29. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will be used. Visualize:
In the sled’s physical representation nS is the normal (contact) force on the sled due to the snow. Similarly fkS is the force of kinetic friction on the sled due to snow. Solve: Newton’s second law on the box in the y-direction is
nS on B − wB cos 20° = 0 N ⇒ nS on B = (10 kg)(9.8 m/s 2 ) cos 20° = 92.09 N r The static friction force fS on B accelerates the box. The maximum acceleration occurs when static friction reaches its maximum possible value. fs max = µSnS on B = (0.50)(92.09 N ) = 46.05 N Newton’s second law along the x-direction thus gives the maximum acceleration
(
)
fS on B − wB sin 20° = mB a ⇒ 46.05 N − (10 kg) 9.8 m/s 2 sin 20° = (10 kg)a ⇒ a = 1.25 m /s 2 Newton’s second law for the sled along the y-direction is nS − nB on S − wS cos20° = 0 N
(
)
⇒ nS = nB on S + mS g cos20° = (92.09 N ) + (20 kg) 9.8 m/s 2 cos20° = 276.27 N Therefore, the force of friction on the sled by the snow is fkS = (µk)nS = (0.12)(276.27 N) = 33.15 N Newton’s second law along the x-direction is Tpull – wSsin20° − fkS – fB on S = mSa The friction force fB on S = fS on B because these are an action/reaction pair. We’re using the maximum acceleration, so the maximum tension is
Tmax − (20 kg)(9.8 m/s 2 ) sin 20° − 33.15 N − 46.05 N = (20 kg)(1.25 m/s 2 ) ⇒ Tmax = 171.2 N Assess: Tmax of 171.2 N corresponds to a mass of 171.2 N/9.8 m/s2 = 17.5 kg. For the masses and the angle in this problem, a maximum tension of 171.2 N is physically understandable.
8.30. Model: Masses m1 and m2 are considered particles. The string is assumed to be massless. Visualize:
Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it r r acts like a pulley. Thus tension forces T1 and T2 act as if they were an action/reaction pair. Mass m1 is in circular motion of radius r, so Newton’s second law for m1 is
∑F
r
= T1 =
m1v 2 r
Mass m2 is at rest, so the y-equation of Newton’s second law is
∑F
y
= T2 − m2 g = 0 N ⇒ T2 = m2 g
Newton’s third law tells us that T1 = T2 . Equating the two expressions for these quantities:
m1v 2 = m2 g ⇒ v = r
m2 rg m1
8.31. Model: The boy (B) and the crate (C) are the two systems of our interest, and they will be treated in the particle model. We will also use the static and kinetic friction models. Visualize:
Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must exist between the boy’s feet and the sidewalk. That is, fsB = µsB nB. Also fkC = µkC nC. Newton’s second law for the crate is
∑(F )
on C y
= nC − w C = 0 N ⇒ nC = m C g
∑(F )
= FB on C − fkC = 0 N ⇒ FB on C = fkC = µ kC nC = µ kC mC g
∑(F )
= fsB − FC on B = 0 N ⇒ FC on B = fsB = µ sB nB = µ sB mB g
on C x
Newton’s second law for the boy is
∑(F )
on B y
= nB − w B = 0 N ⇒ nB = m B g
r r FC on B and FB on C are an action/reaction pair, so
on B x
FC on B = FB on C ⇒ µ sB mB g = µ kC mC g ⇒ mC =
µ sB mB (0.8)(50 kg) = = 200 kg µ kC (0.2)
8.32. Model: The 3 kg and 4 kg blocks are to be treated as particles. The models of kinetic and static friction and the constant-acceleration kinematic equations will be used. Visualize:
Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newton’s second law for the 4-kg block is
∑(F )
on 4 y
= n3 on 4 − w4 = 0 N ⇒ n3 on 4 = w4 = m4 g = ( 4.0 kg)(9.8 m/s 2 ) = 39.2 N ⇒ fs4 = fs max = µ s n3 on 4 = (0.60)(39.2 N ) = 23.52 N
Newton’s second law for the 3-kg block is
∑(F )
on 3 y
(
)
= n3 − n4 on 3 − w3 = 0 N ⇒ n3 = n4 on 3 + w3 = 39.2 N + (3.0 kg) 9.8 m/s 2 = 68.6 N
Friction forces f and fs4 are an action/reaction pair. Thus
∑(F )
on 3 x
= fs3 − fk3 = m3 a3 ⇒ fs4 − µ k n3 = m3 a3 ⇒ 23.52 N − (0.20)(68.6 N ) = (3 kg)a3 ⇒ a3 = 3.267 m /s 2
Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows:
x1 − x 0 + v0 x (t1 − t 0 ) + 12 a(t1 − t 0 ) ⇒ 5.0 m = 0 m + 0 m + 2
1 2
(3.267 m/s )(t 2
1
− 0 s) ⇒ t1 = 1.75 s 2
8.33. Model: The masses m and M are to be treated in the particle model. We will also assume a massless rope and frictionless pulley, and use the constant-acceleration kinematic equations for m and M. Visualize:
Solve:
Using y1 = y0 + v0 y (t1 − t 0 ) + 12 aM (t1 − t 0 ) , 2
(−1 m) = 0 m + 0 m + 12 aM (6.0 s − 0 s)2 ⇒ aM = −0.0556 m /s 2 Newton’s second law for m and M is:
∑(F )
on m y
= TR on m − wm = mam
∑(F )
on M y
= TR on M − w M = Ma M
The acceleration constraint is am = − a M . Also, the tensions are an action/reaction pair, thus TR on m = TR on M . With these, the second law equations are TR on M − Mg = Ma M
TR on M − mg = − ma M Subtracting the second from the first gives
g + aM − Mg + mg = Ma M + ma M ⇒ m = M g − aM 9.8 − 0.556 = 98.9 kg = (100 kg) 9.8 + 0.556 Assess: Note that am = − aM = .0556 m/s2. For such a small acceleration, a mass of 98.9 kg for m compared to M = 100 kg is understandable.
8.34. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the r rope is massless and the pulleys are frictionless. The block is kept in place by an applied force F . Visualize:
Solve:
Since there is no friction on the pulleys, T2 = T3 and T2 = T5. Newton’s second law for mass M is
T1 − w = 0 N ⇒ T1 = Mg = (10.2 kg)(9.8 m/s 2 ) = 100 N Newton’s second law for the small pulley is
T2 + T3 − T1 = 0 N ⇒ T2 = T3 =
T1 = 50 N = T5 = F 2
Newton’s second law for the large pulley is T4 − T2 − T3 − T5 = 0 N ⇒ T4 = T2 + T3 + T5 = 150 N
8.35. Model: Assume the particle model for m1, m2, and m3, and the model of kinetic friction. Assume the ropes to be massless, and the pulleys to be frictionless and massless. Visualize:
Solve:
Newton’s second law for m1 is T1 – w1 = m1a1. Newton’s second law for m2 is
∑(F )
y
= n2 − w2 = 0 N ⇒ n2 = (2 kg)(9.8 m /s 2 ) = 19.6 N
∑(F )
= T2 − fk2 − T1 = m2 a2 ⇒ T2 − µ k n2 − T1 = (2 kg)a2
on m2
on m2
Newton’s second law for m3 is
x
∑(F
on m3
)
y
= T2 − w3 = m3 a3
Since m1, m2, and m3 move together, a1 = a2 = − a3 = a. The equations for the three masses thus become T1 − w1 = m1a = (1 kg)a T2 − µ k n2 − T1 = m2 a = (2 kg)a T2 − w3 = − m3 a = −(3 kg)a Subtracting the third equation from the sum of the first two equations yields: − w1 − µ k n2 + w3 = (6 kg)a
⇒ −(1 kg)(9.8 m/s 2 ) − (0.300)(19.6 N ) + (3 kg)(9.8 m/s 2 ) = (6 kg)a ⇒ a = 2.29 m /s 2
8.36. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction. Visualize:
Solve: (a) If the mass m is too small, the hanging 2 kg mass will pull it up the slope. We want to find the smallest mass that will stick because of friction. The smallest mass will be the one for which the force of static friction is at its maximum possible value: fs = ( fs ) max = µ s n . As long as the mass m is stuck, both blocks are at rest with r Fnet = 0 N . Newton’s second law for the hanging mass M is
( Fnet ) y = TM − Mg = 0 N ⇒ TM = Mg = 19.6 N For the smaller mass m,
( Fnet ) x = Tm − fs − mg sin θ = 0 N
( Fnet ) y = n − mg cosθ ⇒ n = mg cosθ r r For a massless string and frictionless pulley, forces Tm and TM act as if they are an action/reaction pair. Thus Tm = TM . Mass m is a minimum when ( fs ) max = µ s n = µ s mg cosθ . Substituting these expressions into the x-equation, TM − µ s mg cosθ − mg sin θ = 0 N ⇒ m =
TM
( µs cosθ + sin θ )g
= 1.83 kg
(b) Because µk < µS, the 1.83 kg block will begin to slide up the ramp, and the 2 kg mass will begin to fall, if the block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pictorial representation, we chose different coordinate systems for the two masses. This gives block M an acceleration with only a y-component and block m an acceleration with only an x-component. The magnitudes of the accelerations r are the same because the blocks are tied together. But block M has a negative acceleration component ay (vector a points down) whereas block m has a positive ax. Thus the acceleration constraint is (am)x = − (aM)y = a, where a will have a positive value. Newton’s second law for block M is
( Fnet ) y = T − Mg = M(aM ) y = − Ma For block m we have
( Fnet ) x = T − fk − mg sin θ = T − µ k mg cosθ − mg sin θ = m(am ) x = ma In writing these equations, we used Newton’s third law to write Tm = TM = T . Also, we noticed that the y-equation and the friction model for block m don’t change, except for µs becoming µk, so we already know fk from part (a). Notice that the tension in the string is not the weight Mg. We have two equations in the two unknowns T and a:
T − Mg = − Ma
T − ( µ k cosθ + sin θ )mg = ma
Subtracting the second equation from the first to eliminate T,
− Mg + ( µ k cosθ + sin θ )mg = − Ma − ma = −( M + m)a ⇒a=
M − ( µ k cosθ + sin θ )m M+m
g = 1.32 m /s 2
8.37. Model: Assume the particle model for the two blocks. Visualize:
Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks will r move one way or the other. As long as m is held, the blocks are in static equilibrium with Fnet = 0 N . Newton’s second law for the hanging block M is
(F
)
net on M y
= TM − Mg = 0 N ⇒ TM = Mg = 19.6 N
By Newton’s third law, TM = Tm = T = 19.6 N is the tension in the string. (b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends on whether the acceleration a is positive or negative. The acceleration constraint is (am)x = a = −(aM)y. Newton’s second law for each system gives
(F
)
net on m x
= T − mg sin35° = m( am ) x = ma
(F
)
net on M y
= T − Mg = M ( a M ) y = − Ma
We have two equations in two unknowns. Subtract the second from the first to eliminate T: M − m sin35° − mg sin35° + Mg = ( m + M )a ⇒ a = g = −0.481 m /s 2 M+m Since a < 0 m/s2, the box accelerates down the slope. (c) It is now straightforward to compute T = Mg − Ma = 20.6 N . Notice how the tension is larger than when the blocks were motionless.
8.38. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic and static friction, and the constant-acceleration kinematic equations. Visualize:
Solve:
(a) Using v12x = v02x + 2 a( x1 − x 0 ) ,
0 m 2 /s 2 = (3.0 m/s) + 2 a( x1 ) ⇒ ax1 = −4.5 m 2 /s 2 2
To find x1, we must first find a. Newton’s second law for the book and the coffee cup is
∑(F )
on B y
= nB − wB cos 20° = 0 N ⇒ nB = (1.0 kg)(9.8 m /s 2 ) cos 20° = 9.21 N
∑(F )
on B x
∑(F )
= − T − fk − wB sin 20° = mB aB
on C y
= T − w C = mC aC
The last two equations can be rewritten, using aC = aB = a, as
− T − µ k nB − mB g sin 20° = mB a
T − mC g = mC a
Adding the two equations,
a( mC + mB ) = − g( mC + mB sin 20°) − µ k (9.21 N )
[
]
⇒ (1.5 kg)a = −(9.8 m/s 2 ) 0.5 kg + (1.0 kg) sin 20° − (0.20)(9.21 N ) ⇒ a = −6.73 m /s 2 Using this value for a, we can now find x1 as follows:
x1 =
−4.5 m 2 /s 2 −4.5 m 2 /s 2 = = 0.669 m a −6.73 m /s 2
(b) The maximum static friction force is fs max = µ s nB = (0.50)(9.21 N ) = 4.60 N. We’ll see if the force fs needed to keep the book in place is larger or smaller than fs max . When the cup is at rest, the string tension is T = mC g. Newton’s first law for the book is
∑(F )
on B x
= fs − T − wB sin 20° = fs − mC g − mB g sin 20° = 0 ⇒ fs = ( MC + MB sin 20°)g = 8.25 N
Because fs > fs max , the book slides back down.
8.39. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable. Visualize:
Solve: (a) Notice the separate coordinate systems for the cable car (system 1) and the counterweight (system 2). r r r r Forces T1 and T2 act as if they are an action/reaction pair. The braking force FB works with the cable tension T1 to r allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so Fnet = 0 N for both systems. Newton’s second law for the cable car is
(F
)
net on 1 x
= T1 + FB − m1 g sin θ1 = 0 N
(F
)
= n1 − m1 g cosθ1 = 0 N
(F
)
= n2 − m2 g cosθ 2 = 0 N
net on 1 y
Newton’s second law for the counterweight is
(F
)
net on 2 x
= m2 g sin θ 2 − T2 = 0 N
net on 2 y
From the x-equation for the counterweight, T2 = m2 g sin θ 2 . By Newton’s third law, T1 = T2 . Thus the x-equation for the cable car becomes
FB = m1 g sin θ1 − T1 = m1 g sin θ1 − m2 g sin θ 2 = 3770 N (b) If the brakes fail, then FB = 0 N . The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is a1x = a2x = a, where a will have a negative value. Using T1 = T2 = T , the two x-equations are
( Fnet on 1 ) x = T − m1g sin θ1 = m1a1x = m1a
(F
)
net on 2 x
= m2 g sin θ 2 − T = m2 a2 x = m2 a
Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T:
− m1 g sin θ1 + m2 g sin θ = ( m1 + m2 )a ⇒ a = −
m1 sin θ1 − m2 sin θ 2 g = −0.991 m /s 2 m1 + m2
Now we have a problem in kinematics. The speed at the bottom is calculated as follows:
v12 = v02 + 2 a( x1 − x 0 ) = 2 ax1 ⇒ v1 = 2 ax1 = 2( −0.991 m /s 2 )( −400 m ) = 28.2 m /s Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slope of 30° is reasonable.
8.40. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionless pulleys. Visualize:
Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to be lengthened by one-half meter. Thus the acceleration constraint is a2 = − 12 a1 . Solve: Newton’s second law for block 1 is T = m1a1 . Newton’s second law for block 2 is 2T – w 2 = m2a2. Combining these two equations gives
2( m1a1 ) − m2 g = m2 ( − 12 a1 ) ⇒ a1 [ 4 m1 + m2 ] = 2 m2 g ⇒ a1 =
2 m2 g 4 m1 + m2
where we have used a2 = − 12 a1 . Assess: If m1 = 0 kg, then a2 = –g. This is what is expected for a freely falling object.
8.41. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys. Visualize:
For every one meter that the 1-kg block goes down, each rope on the 2-kg block will be shortened by one-half meter. Thus the acceleration constraint is a1 = –2a2. Solve: Newton’s second law for the two blocks is 2T = m2a2
T − w1 = m1a1
Since a1 = −2 a2 , the above equations become 2T = m2a2
⇒ m2
T − m1g = m1(− 2a2)
2(1 kg)(9.8 m /s 2 ) 2 m1 g a2 = = 3.27 m/s2 + m1 (2 a2 ) = m1 g ⇒ a2 = m2 + 4 m1 2 (2 kg + 4 kg)
Assess: If m1 = 0 kg, then a2 = 0 m/s2, which is expected.
8.42. Model: The painter and the chair are treated as a single system and represented as a particle. We assume that the rope is massless and that the pulley is massless and frictionless. Visualize:
Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on the painter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tension force F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) is
2F − w = ( mP + mC )a
)[(mP + mC )a + (mP + mC )g] = 12 (mP + mC )(a + g) = ( 12 )(70 kg + 10 kg)(0.20 m/s 2 + 9.8 m/s 2 ) = 400 N
⇒F=(
1 2
Assess: A force of 400 N, which is approximately one-half the total weight, is reasonable since the upward acceleration is small.
8.43. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless, so the tension in the rope is uniform. Visualize:
Solve:
Newton’s second law for the tightrope walker is
FR on W − w = ma ⇒ FR on W = m( a + g) = (70 kg)(8.0 m/s 2 + 9.8 m/s 2 ) = 1246 N Now, Newton’s second law for the rope gives
∑(F )
on R y
= T sin θ + T sin θ − FW on R = 0 N ⇒ T =
We used FW on R = FR on W
FW on R
2 sin 10° because they are an action/reaction pair.
=
FR on W 2 sin 10°
=
1246 = 3590 N 2 sin 10°
8.44. Model: Use the particle model for the wedge and the block. Visualize:
The block will not slip relative to the wedge if they both have the same acceleration a. Solve: The y-component of Newton’s second law for block m2 is
∑(F )
on 2 y
= n2 cosθ − w2 = 0 N ⇒ n2 =
m2 g cosθ
Combining this equation with the x-component of Newton’s second law yields: n sin θ ∑ Fon 2 x = n2 sinθ = m2 a ⇒ a = 2m = g tanθ 2
( )
Now, Newton’s second law for the wedge is
∑(F )
on 1 x
= F − n2 sin θ = m1a
⇒ F = m1a + n2 sin θ = m1a + m2 a = (m1 + m2) a = (m1 + m2) g tanθ
8.45. A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg block across a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide? Visualize:
Solve: The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if fs < fs max. It slides if the force that would be needed to keep it in place exceeds fs max. Begin by assuming the blocks move together with common acceleration a. Newton’s second is
∑ ( Fon 1 ) x = fs = m1a Bottom block: ∑ ( Fon 2 ) x = Tpull − fs = m2 a Top block:
Adding these two equations gives Tpull = (m1 + m2)a, or a = (21.0 N)/(1.0 kg + 3.0 kg) = 7.0 m/s2. The static friction force needed to accelerate the top block at 7.0 m/s2 is fs = m1a = (1.0 kg)(7.0 m/s2) = 7.0 N To find the maximum possible static friction force fs max = µsn1, the y-equation of Newton’s second law for the top block shows that n1 = m1g. Thus fs max = µsm1g = (0.50)(1.0 kg)(9.80 m/s2) = 4.9 N Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides.
8.46. A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients of kinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushed forward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping? Visualize:
Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rear of the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. There’s no vertical acceleration, so n1 = m1g and n2 = m2g, leading to f1 = µ1m1g and f2 = µ2m2g. Newton’s second law along the x-axis is
1 kg block: 2 kg block:
∑(F
) = − F2 on 1 − f1 = − F2 on 1 − µ1 m1 g = m1a
on 1 x
∑ ( Fon 2 ) x = F1 on 2 − f2 = F2 on 1 − µ 2 m2 g = m2 a
where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an action/reaction pair. Adding these two equations gives
–( µ1 m1 + µ 2 m2 )g = ( m1 + m2 )a a=−
µ1 m1 + µ 2 m2 (0.3)(1.0 kg) + (0.5)(2.0 kg) g=− × 9.80 m/s 2 = −4.25 m/s 2 m1 + m2 1.0 kg + 2.0 kg
We can now use constant-acceleration kinematics to find
v12x = 0 = v02x + 2 a( x1 − x 0 ) ⇒ x1 = −
v02x (2.0 m /s) 2 =− = 0.47 m 2a 2( −4.25 m /s 2 )
8.47. Model: Visualize:
Treat the ball of clay and the block as particles.
→
→
Solve: (a) Forces F C on B and F B on C are an action/reaction pair, so FB on C = FC on B. Note that aB ≠ aC because the clay is decelerating while the block is accelerating. Newton’s second law in the x-direction is
∑ ( Fon C ) x = − FB on C = mC aC Block: ∑ ( Fon B ) x = FC on B = FB on C = mB aB
Clay:
Equating the two expressions for FB on C gives
aC = −
mB aB mC
Turning to kinematics, the velocity of each after ∆t is
(vC )1 = (vC ) 0 + aC ∆t (vB )1 = (vB ) 0 + aB ∆t = aB ∆t But (vC)1 = (vB)1 because the clay and the block are moving together after ∆t has elapsed. Equating these two expressions gives (vC)0 + aC∆t = aB∆t, from which we find
aC = a B −
( vC ) 0 ∆t
We can now equate the two expressions for aC:
−
(v ) (vC ) 0 / ∆t mB (10 m /s) / (0.01 s) = = 100 m /s 2 aB = aB − C 0 ⇒ aB = ∆t 1 + mB / mC 1 + (900 g)/ (100 g) mC
Then aC = –9aB = –900 m/s2. With the acceleration now known, we can use either kinematic equation to find (vC)1 = (vB)1 = (100 m/s2)(0.010 s) = 1.0 m/s (b) FC on B = mBaB = (0.90 kg)(100 m/s2) = 90 N. (c) FB on C = mCaC = (0.10 kg)(900 m/s2) = 90 N. Assess: The two forces are of equal magnitude, as expected from Newton’s third law.
8.48. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations. Visualize:
Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightens r his legs. He exerts a force FP on F on the floor. The player experiences a weight force wP as well as a normal force r r by the floor n F on P. The only force that the floor experiences is the one exerted by the player F P on F. r r r (b) The player standing atr rest exerts a force F P on F on the floor. The normal force n F on P is the reaction force to F P on F. But nF on P = FP on F , so F net = 0 N. When the basketball player accelerates upwards by straightening his legs, his speed has to increase from zero to v1y with which he leaves the floor. Thus, according to Newton’s second law, there must be a net upward force on him during this time. This can be true only if nF on P > wP . In other words, the r player presses on the floor with a force FP on F larger than his weight. The reaction force n F on P then exceeds his weight and accelerates him upward until his feet leave the floor. (c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once his feet are off the floor, he is simply in free fall, with a1 = − g. From kinematics,
v22y = v12y + 2 a1 ( y2 − y1 ) ⇒ 0 m 2 /s 2 = v12y + 2( − g)(0.80 m ) ⇒ v1y = 2 g(0.80 m ) = 3.96 m/s (d) The basketball player reached v1y = 3.96 m/s by accelerating from rest through a distance of 0.60 m. Assuming a0 to be constant during the jump, we find
v12y = v02y + 2 a0 ( y1 − y0 ) = 0 m 2 /s 2 + 2 a0 ( y1 − 0 m ) ⇒ a0 =
v12y 2 y1
=
(3.96 m /s)2 2(0.60 m )
= 13.1 m /s 2
(e) The scale reads the value of nF on P, the force exerted by the scale on the player. Before jumping,
nF on P − wP = 0 N ⇒ nF on P = wP = (100 kg)(9.8 m/s 2 ) = 980 N While accelerating up,
g 9.8 nF on P − wP = ma0 ⇒ nF on P = ma0 + mg = m1 + = (980 N )1 + = 1710 N a 13 .1 0 After leaving the scale, nF on P = 0 N because there is no contact with the scale.
8.49. Model: The hamster of mass m and the wedge with mass M will be treated as systems 1 and 2, respectively. They will be treated as particles. Visualize:
The scale is denoted by the letter s. r Solve: (a) The reading of the scale is the magnitude of the force n2 that the scale exerts upward. There are two r action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with Fnet = 0 N. r Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force f s on 2 to prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is
(F
)
net on 1 x
= mg sin θ − f2 on 1 = 0 N ⇒ f2 on 1 = mg sin θ
(F
)
net on 1 y
= n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ
For the wedge, we see from Newton’s third law that n1′ = n1 = mg cosθ and that f2 on 1 = f1 on 2 = mg sin θ . Using these equations, Newton’s second law for the wedge is
( Fnet on 2 ) x =
(F
f1 on 2 cosθ + fs on 2 − n1′ sin θ = mg sin θ cosθ + fs on 2 − mg cosθ sin θ = 0 N ⇒ fs on 2 = 0 N
)
net on 2 y
= n2 − n1′ cosθ − f1 on 2 sin θ − Mg = n2 − mg cos 2 θ − mg sin 2 θ − Mg = 0 N
⇒ n2 = mg(cos 2 θ + sin 2 θ ) + Mg = ( M + m)g = (0.8 kg + 0.2 kg)(9.8 m/s 2 ) = 9.8 N First we find that fs on 2 = 0 N , so no horizontal static friction is needed to prevent motion. More interesting, the scale reading is ( M + m)g which is the total weight resting on the scale. This is the expected result. (b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the reading still (M + m)g? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now has an acceleration. However, the acceleration is along the hamster’s x-axis, so a1y = 0 m/s2. The hamster’s y-equation is still
( Fnet on 1 ) y = n1 − mg cosθ = 0 N ⇒ n1 = mg cosθ We still have n1′ = n1 = mg cosθ , so the y-equation for block 2 (with a2y = 0 m/s2) is
(F
)
net on 2 y
= n2 − n1′ cosθ − Mg = n2 − mg cos 2 θ − Mg = 0 N
⇒ n2 = mg cos 2 θ + Mg = ( M + m cos 2 θ )g = 8.99 N Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit θ → 90°, in which case cos θ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and can have no effect on the scale (at least until impact!) So for θ = 90° we expect the scale to record Mg only, and that is indeed what the expression for n2 gives.
8.50. Model: The hanging masses m1, m2, and m3 are modeled as particles. Pulleys A and B are massless and frictionless. The strings are massless. Visualize:
Solve:
(a) The length of the string over pulley B is constant. Therefore,
( yB − y3 ) + ( yB − yA ) = LB ⇒ yA = 2 yB − y3 − LB The length of the string over pulley A is constant. Thus,
( yA − y2 ) + ( yA − y1 ) = LA = 2 yA − y1 − y2 ⇒ 2(2 yB − y3 − LB ) − y1 − y2 = LA ⇒ 2 y3 + y2 + y1 = constant This constraint implies that
2
dy3 dy2 dy1 + + = 0 m /s = 2v3 y + v2 y + v1y dt dt dt
Also by differentiation, 2 a3 y + a2 y + a1y = 0 m /s 2 . (b) Newton’s second law for the masses m3, m2, m1, and pulley A is TB − m3g = m3a3y
TA – m2g = m2a2y
TA – m1g = m1a1y
TB − 2TA = 0 N The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for TA, multiply the m3 equation by 2, substitute 2TB = 4TA, then divide each of the mass equations by the mass. This gives the three equations
4TA / m3 − 2 g = 2 a3 y TA / m 2 − g = a2 y TA / m 1 − g = a1y
If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus
( 4 / m3 + 1/ m2 + 1/ m2 )TA − 4 g = 0 ⇒ TA =
4g ( 4 / m3 + 1 / m 2 + 1 / m 2 )
(c) Using numerical values, we find TA = 18.97 N. Then
a1y = TA / m1 − g = −2.21 m /s 2 a2 y = TA / m 2 − g = 2.85 m /s 2 a3 y = 2 TA / m 3 − g = −0.315 m /s 2 (d) m3 = m1 + m2, so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need to equal m3g. However, TB is not (m1 + m2)g because masses m1 and m2 are accelerating rather than hanging at rest. Consequently, tension TB is not able to balance the weight of m3.