CHAPTER 4 NEWTON’S LAW OF MOTION 4.1 4.2
Definition of force Types of forces 4.2.1 Gravitational force 4.2.2 Normal force 4.2.3 Frictional force 4.2.4 Tensional force 4.3 Newton’s Law of motion and its application 4.3.1 Newton’s First Law 4.3.2 Newton’s Second Law 4.3.3 Newton’s Third Law 4.4 Static equilibrium under concurrent force 1
4.1
Definition of force
First we need to define the word FORCE: • The cause of motion (what causes objects to move) • Two method in producing forces :
i) Pushes
ii) Pulls
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FORCE An object at rest needs a force to get it moving; a moving object needs a force to change its velocity. The magnitude of a force can be measured using a spring scale. SI Unit : Newton (N) @ kgm/s2. Quantity : Vector 3
Forces may cause the object balanced or unbalanced • Balanced forces – all forces acting on an object are equal • There is NO MOTION
• A net force is ZERO ( ∑F = 0 )
• Unbalanced forces – one or more forces acting on an object are stronger than others • There is MOTION
• A net force have value ( ∑F ≠ 0 )
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FORCE
• Arrows are used to represent forces.
The length of the
arrow is proportional to the magnitude of the force.
• Net force : The vector sum of all forces acting on that object. 15 N
5N
Individual Forces 4N
10 N
Net Force 6N 5
FORCE Individual Forces
Net Force
5N
64ď Ż
3N 4N
Example :
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4.2 Types of forces
Weight, W Weight is the force exerted on that object by gravity.
W = mg Mass is measured in Kilogram (Kg) Mass is not weight Direction: directly downward from the object.
W
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Normal Force, N Also called reaction force, R R object
floor
A reaction force exerted by surface Cannot exist on its own Direction: Always perpendicular to the surface
W 8
Frictional Force, f -Direction: Always oppose the motion direction -Exist at rough surface
F
f
f μR Coefficient of friction
Normal Force 9
Tension, T T1 T2
-Exist at rope, cable -Direction: always away from body -Tension at smooth pulley always similar, hence T1 = T2 10
4.3
Newton’s Law of motion and its application
NEWTON’S FIRST LAW • In the absence of external forces, an object at rest remains at rest and an object in motion continues in motion with constant speed. • When no force acts on an object, F = 0, a = 0
• Also called law of inertia • Inertia is the tendency for object to resist change in its state. 11
Mass & Inertia • Mass is the amount of matter in an object. • The more MASS an object has, the more INERTIA the object has. • Bigger objects are harder to start & stop Which vehicle has more inertia?
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Object at rest remain at rest & object at motion will remain at motion
A lot of Inertia!!!
Very little inertia..
Since the train is so huge, it is Since the baby carriage is so difficult to change its speed. In small, it is very easy to change fact, a large net force is its speed or direction. A small required to change its speed or net force is required to change direction. its speed or direction. 13
NEWTON’S SECOND LAW Newton’s second law is the relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass.
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Exercise 1 : Lets try this!! Determine the accelerations which result when a 12-N net force is applied to a 3-kg object and then to a 6-kg object.
Solution:
Answer : 4 m
s-2
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NEWTON’S THIRD LAW Newton’s third law: Whenever one object exerts a force on a second object, the
second exerts an equal force in the opposite direction on the first. (action reaction)
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Free Body Diagram 1. Draw a sketch. 2. For one object, draw a free-body diagram, showing all the forces acting on the object. Label each force. If there are multiple objects, draw a separate diagram for each one. 3. Resolve vectors into components.
4. Apply Newton’s second law to each component. 5. Solve. R
W 18
Example Draw Free Body Diagram for figure below
a)
b)
c)
d)
Inclined plane (smooth surface)
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Exercise 2 : Lets try this!! Suppose the coefficient of kinetic friction is 0.05 and the total mass is 40kg. What is the kinetic frictional force?
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Solution :
Answer : 19.6 N
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Exercise 3 : Lets try this!!
A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration. 30°
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Solution :
Answer : 1.12
m -2s-2
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Exercise 4 : Lets try this!! A box of mass m is placed on a smooth incline that makes an angle θ with the horizontal. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of θ = 30°.
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Solution:
Answer : R = Wsin(90°-θ) N , a = gcos(90°-θ) ms-2, R = 84.87 N , a = 4.9
ms-2
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Exercise 5 : Lets try this!!
Find the acceleration of the pulley system as in the figure. Calculate the tension in the string. Solution:
m1 2 kg
3 kg
Answer : a = 1.96
ms-2
, T = 23.52 N
m2
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Exercise 6 : Lets try this!!
A 15 kg block rests on a level frictionless surface and is attached by a light string to a 5.0 kg hanging mass where the string passes over a massless frictionless pulley. What is the tension in the connecting string? Solution:
Answer : T = 36.75 N
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4.4
Static Equilibrium Under Concurrent Force
CONCURRENT FORCES • Forces whose line of action are all passing through the same point of the object. • For a body, the concurrent forces pass through the centre of gravity. Centre of gravity - The point at which all the surrounding weight is equal. It is the point at which an object is in balance.
Exercise 7 : Lets try this!!
Figure shows a system of forces in static equilibrium. Determine force T and P. Solution :
Answer : Tx = P N , Ty = 19.6 N
T
30° P
2kg
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Exercise 8 : Lets try this!!
Find T1 and. T2 if the weight of the object is 30 N.
Solution :
60°
T1 T2
70° 30N
Answer : T1 = 43.48 N , T2 = 23.04 N
TORQUE measure the tendency of a force to course or change the rotational motion of a body • Torque is + (ve) for CCW and – (ve) for CW • Vector quantity • SI unit: N m
r F r sin F where : angle between r and F r : distance from axis of rotation to force r sin : torque arm 31
Equilibrium condition for rigid body : OBJECT
RIGID BODY
DEFINITION
an object with mass and has larger dimensions
1st CONDITION
F
= 0 (Total force is zero)
i.e : 2nd CONDITION
, =0
(Total torque is zero)
i.e : 32
Example:
N
Determine the net torque:
4m
500 N
Answer : -400Nm (counter-clockwise)
2m
800 N
Exercise 9 : Lets try this!!
A
W2
35 cm
O
75 cm
B
W1
A hanging flower basket having weight, W2 =23 N is hung out over the edge of a balcony railing on a uniform horizontal beam AB of length 110 cm that rests on the balcony railing. The basket is counterbalanced by a body of weight, W1 as shown in figure above. If the mass of the beam is 3.0 kg, calculate a. the weight, W1 needed, b. the force exerted on the beam at point O.
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Solution:
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Answer : W1 = 2.89 N, R = 55.29 N
Exercise 10 : Lets try this!!
The horizontal beam is uniform and has a mass of 250 kg. Calculate the reaction force of F1 and F2. F1
F2
4000 N 2m
3000 N 4m
3m
2000 N
1m
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Solution:
Answer : F1 = 5825 N, F2 = 5625 N
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Exercise 11 : Lets try this!!
A 20 kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole as shown in figure below. A cable at an angle of 30° with the beam helps to support the light. Find : a) The tension in the cable b) The horizontal and vertical forces exerted on the beam by the pole.
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Solution:
Answer : T = 196 N, Rx = 169.74 N, Ry = 98 N
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