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Friction
P INTRODUCTION
We have so far dealt with smooth surfaces, which offer a single reaction force R. In this chapter w-e deal with rough surfaces which offer an additional reaction known as friction force. Friction force is developed whenever there is a motion or tendency of motion of one body with respect to the other body involving rubbing of the surfaces of contact. We will understand the concept of friction and also present the laws of friction. Application of friction to the problems of blocks, ladder, wedges, square threaded screw, and rope friction will be dealt with in this chapter.
FRICTIONAL FORCE Friction is of two types i) Dry Fiction ii) Fluid Fiction. Dry friction also known as Coulomb friction involves friction due to rubbing of rigid bodies, for example a block tending to move on table or a wheel rolling on the ground. Fluid friction is developed between layers of fluids as they move with different velocities inside a pipe, bodies moving over lubricated surface, etc. Our study would be limited to Dry Friction. Frictional force is generated whenever a body moves or tends to move over another surface. This is best illustrated by the following discussion. Consider a block of weight W resting on a rough surface. Let the normal reaction be N. The block is in equilibrium under the action of two forces. Fig. (a)
Now if an attempt is made to disturb the equilibrium by applying a force P as shown in Fig. (b), the rough surface generates a friction force F to maintain the equilibrium. The force F is equal to P and the block is maintained equilibrium. If P is now increased the friction force F also increases {refer shown in Fig. (c)}. However the surface can generate a maximum friction
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known as limiting fiction force Fmax. If P exceeds Fmax, the body would be set into motion and the block is said to be in a dynamic state. In dynamic state, the same surfaces develop a lower friction force known as kinetic frictional force Fk. Friction force generated is generally due to presence of hills and valleys on any surface and to smaller extent due to molecular attraction. Fig. Shows a macroscopic view of two contacting surfaces enlarged 1000 times.
When the body is in static condition, there is greater interlocking between the hills and the valleys of the two surfaces. As the body becomes dynamics, the interlocking reduces resulting in lowering the friction force to a new value Fk.
COEFFICIENT OF FRICTION It has been experimentally found that the ratio of the limiting frictional forces Fmax and the normal reaction N is a constant. This constant is referred to as coefficient of static friction, denoted as μ s .
μs =
Fmax N
…..4.2 Since Fk is less than Fmax , we have μ k < μ s Coefficient of friction μ s or μ k depend on the nature of surfaces of contact and have value less than 1. For example the approximate value of coefficient of static friction ( μ s ) between wood on glass ranges between 0.2 to 0.6. The corresponding coefficient of kinetic friction ( μ k ) is around 25 % lower.
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LAWS OF FRICTION 1. The frictional force is always tangential to the contact surface and acts opposite to the direction of impending motion. 2. The value of frictional force F increases as the applied disturbing forces increases till it reaches the limiting value Fmax. At this limiting stage the body is on the verge of motion. 3. The ratio of limiting frictional force (Fmax). and the normal reaction N is a constant and it is referred to as coefficient of static friction ( μ s ). 4. For bodies in motion, frictional force developed ( Fk ) is less than the limiting frictional force ( Fmax ). The ratio of Fk and the normal reaction N is a constant and is referred to as coefficient of kinetic friction ( μ k ). 5. The frictional force F generated between the two rubbing surfaces is independent of the area of contact.
ANGLE OF FRICTION, CONE OF FRICTION AND ANGLE OF REPOSE Angle of Friction: "It is the angle made by the resultant of the limiting frictional force Fmax and the normal reaction N with the normal reaction". Fig. shows a block of weight W under the action of the applied force P. Let N be the normal reaction. If the block is on the verge of impending motion, the frictional force Fmax . Would be developed as shown. Let R be the resultant of Fmax , and N, making an angle with the normal reaction. Here is known as the angle of fiction. Here
R = Fmax 2 + N 2 =
tan =
Also
μs N
2
+ N2
Fmax μ s N = N N
tan = μs
tan 1 μs
.....4.3
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CONE OF FRICTION Fig. shows a block of Weight W on the verge of motion acted upon by force P. Let R be the resultant reaction at the contact surface acting at an angle of friction . If the direction of force is changed by rotating it through 360° in a plane parallel to the contact surface, the force R also rotates and generates a right circular cone of semi-central angle equal to . This right circular cone is known as the cone of friction.
The significance of cone of friction is that for static body the resultant reaction force at the rough contact surface lies within the cone, while for a static body on the verge of motion, the reaction force lies on the surface of the cone of friction. ANGLE OF RESPONSE It is defined as the minimum angle of inclination of a plane with the horizontal for which a body kept on it will just slide down on it without application of any external force. Consider a block of weight W resting on a rough horizontal plane. The plane is slowly tilted till the block is just on the verge of sliding down the plane, Fig. The angle of inclination of the plane at this position is known as the angle of repose. It is denoted by letter α. Angle of repose is independent of the weight of the body and depends only on the coefficient of static friction. Let us
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derive the relation between angle of repose α and coefficient of static friction μ s . Applying COE
F 0 x
+ ve
Fmax -Wsinα = 0
----(1)
μsN- W sin α = 0
F 0 y
+ ve
N- W cos = 0 N = Wcos α Substituting (2) in (1) μ s (W cos α) – W sin α = 0 tan α = μ α = tan-1 μ s
or
----(2)
…..(3)
but we have seen that tan 1 μ s
i.e. Angle of Response = Angle of Friction Though magnitude wise both angle of repose and angle of friction have the same value, their meaning and application are different, as we have seen. PROBLEM ON BLOCKS We will encounter dimensionless blocks acted upon by forces tending to cause motion. The blocks can be modelled as a particle forming a concurrent system of forces. The following steps are adopted in the solution of problems on blocks. Step 1: Draw the FBD of the block showing the weight W, the normal reaction N, the friction force F and other applied forces. The direction of F is always directed opposite to the direction of impending motion. When the block is on verge of motion, F = μ s N should be taken. Step 2: Since we have a concurrent system of forces in equilibrium we apply two conditions of equilibrium viz.
F F
x
=0
y
=0
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EXERCISE 1 1. For the following cases find force P needed to just impend the motion of the block. Take weight of block to be 100 N and coefficient of static friction at the contact Surface to be 0.4.
2. A block weighing 800 N has to rest on an incline of 30°. If the angle of limiting friction is 18°, Find the least and greatest force that need to be applied on the block, parallel to the plane so as to keep the block in equilibrium.
3. A support block is acted upon by two forces as shown. Knowing μ s = 0.35 and μ k = 0.25, determine the force P required. a) to start the block moving up the incline. b) to keep it moving up. c) to prevent it from sliding down. 4. A block of weight 200 N rests on a horizontal surface. The co-efficient of friction between the block and the horizontal surface is 0.4. Find the frictional force acting on the block if a horizontal force of 40 N is applied to the block.
5. A horizontal force P is applied to the block A of mass 50 kg kept on an inclined plane as shown. If P = 200 N, find whether the block is in equilibrium. Also find the magnitude and direction of frictional force. Take μ s = 0.25.
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6. A car of 1000 kg mass is to be parked on the same 10° incline year around. The static coefficient of friction between the tires and the road varies between the extremes of 0.05 and 0.9. Is it always possible to park the car at this place? Assume that the car can be modelled as a particle. 7. Block A of weight 2000 N is kept on a plane inclined at 35°. It is connected to weight B by an in extensible string passing over a smooth pulley. Determine weight of B so that B just moves down. (a) = 0 (b) 0 = 20°
Takeμ = 0.2
8. Two blocks A and B of weight 500 N and 750 N respectively are connected by a cord that passes over a frictionless pulley as shown. μ between the block A and plane is 0.4 and between the block B and plane is 0.3. Determine the force P to be applied to block B to produce the impending motion of block B down the plane.
9. Determine the force P to cause motion to impend. Take masses A and B as 9 kg and 4 kg respectively and coefficient of static friction as 0.25. The force P and rope are parallel to the inclined plane. Assume smooth pulley. b) Solve taking mA = 8 kg, mB 4 kg, μ s 0.3.
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10. What is the minimum value of mass of block B required to maintain the equilibrium? The rope connection A and B passes over a frictionless pulley.
11. Determine the least and greatest value of W for the equilibrium of the whole system.
12. Three blocks are placed on the surface one above the other as shown. The coefficient of friction between the surfaces is shown. Determine the maximum value of P that can be applied before any slipping takes place. Hint: check for three possibilities. 13. Block A weighing 1000 N rests over a block B weighing 2000 N as shown. Block A is tied to a wall with a horizontal string. If p = o.25 between blocks A and B and μ = l/3 between block B and floor, determine the force P needed to move the block if (a) P is horizontal, P acts at 30° upwards to horizontal. 14. Block A of mass 27 kg rests on block B of mass 36kg. Block A restrained from moving by a horizontal rope to the wall at C. What force P, parallel to the plane inclined at 30° with the horizontal is necessary to start B down the plane? Assume p for all surfaces = 0.33.
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15. What should be the value of ' ' so that the motion of block A impends down the plane? Take mA = 40 kg and mB = 13.5 kg. μ = 1 / 3 for all surfaces. 16. Block A has a mass of 25 kg and block B has a mass of 15 kg. Knowing μ s = 0.2 for all surfaces, determine value of 0 for which motion impends. Assume frictionless pulley.
17. A block A of mass 10 kg rests on a rough inclined plane as shown. This block is connected to another block B of mass 30 kg resting' on a rough horizontal plane by a rigid bar inclined at an angle of 30°. Find the horizontal P required to be applied to block B to just move block A in the upward direction. Take μ = 0.25 for all contact surfaces.
18. Find the maximum value of WB for the rod AB to remain horizontal. Also find the corresponding axial force in the rod. Take μ = O.2 for all contact surfaces.
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19. Two blocks connected by a horizontal link AB are supported on two rough planes, Ο between block A, and horizontal surface is 0.4. The limiting angle of friction between block B and inclined plane is 20°. What is the smallest weight W of the block A for which equilibrium of the system can exist, if the weight of block B is 5 kN?
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WEDGES Wedges are tapering shaped devices which are used for lifting or shifting to little extent heavy blocks, machinery, pre-cast beams and columns etc. Usually two or more wedges are used in combination. Fig. shows two arrangements of wedges. In Fig. (a) Two wedges are used for imparting a small vertical movement to heavy machinery of weight W. Effort P is applied to the wedge B which is driven inside causing the vertical movement. In Fig. (b) Two wedges are used in combination to cause a small horizontal movement to the heavy block of weight W. Effort P is applied to wedge B, driving it down, thereby causing horizontal movement of the heavy block.
For solving problems on wedges, we require to isolate the wedges and then apply COE to each wedge separately. Since the dimensions are neglected, wedges are treated as articles forming a concurrent force system. Only two COE viz. Fx = 0 and
F
y
= 0 are therefore useful.
Solving equations (1) and (2), we get N1 = 537.25 N and N3 = 587.8 N Applying COE to Wedge A
F 0 y
N2 - 0.2 N3 sin 75- N3 cos 75-800 = 0
Substituting N3 = 587.8 N, we get
N2 = 1065.7 N
F 0 x
P +0.3N2 + 0.2 N3 cos 75- N3 sin 75 = 0
Substituting N2 = 1065.7 N and N3 = 587.8 N, we get P = 217.6 N
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EXERCISE 2 1. The horizontal position of the 1000 kg block is adjusted by 6° wedges. If coefficient of friction for all surfaces is 0.6, determine the least value of force P required to move the block.
2. A block weighing 1000 N is raised against a surface inclined at 60° to the horizontal by mean of a 15° wedge as shown in figure. Find the horizontal force P which will just start the block to move if the coefficient of friction between all the surfaces of contact be 0.2 Assume the wedge to be of negligible weight.
3. The vertical position of the 200 kg mass I section is being adjusted by two 15° Wedges as shown. Find force P to just raise the mass. Take μ = 0.2 for all surfaces.
4. Wedge A supports a load of W = 5000 N which is to be raised by forcing the wedge B under it. The angle of friction for all surfaces is 15°. Determine the necessary force P to initiate upward motion of the load Neglect the weight of the wedges. Hint: Wedge A comes in contact with the right wall, as the load gets lifted.
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5. Two blocks A and B are resting against the wall and floor as shown in figure. Find minimum value of P that will hold the system in equilibrium. μ = 0.25 at the floor, μ = 0.3 at the wall and μ = 0.2 between the blocks.
PROBLEMS ON LADDERS We have all made use of ladders at one time or the other to reach for things kept high. But every time while climbing the ladder we fear that the ladder might slip. All this is because the friction force at the floor and at the wall is responsible for preventing the slip. If these surfaces do not produce sufficient frictional force the ladder would slip. Invariably we adjust the slope of the ladder and even ask someone to hold the ladder as we use it. The mechanics involving ladders is dealt in this section Forces acting on ladder viz. the normal reactions and friction-force at the floor and wall contact points, the weight of the ladder, the weight of the person climbing the ladder and any other force, form a non-concurrent force system. It will therefore require all the three COE viz.
F = 0, F 0 x
y
and
M = 0 for analyzing the equilibrium of a ladder. EXERCISE 3 1. A person of weight P = 600 N ascends the 5 m ladder of weight 4OO N as shown. How far up the ladder may the person climb before sliding motion of ladder takes place. 2. A 1OO N uniform ladder AB is held in equilibrium as shown. If μ = 0.15 at A and B, calculate the range of values of P for which equilibrium is maintained.
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3. A 150 N uniform ladder 4 m long supports a 500 N weight person at its top. Assuming the wall to be smooth, find the minimum coefficient of friction which is required at the bottom rough surface to prevent the ladder from slipping.
4. The ladder shown is 6m long and is supported by a horizontal floor and a vertical wall. Îź Between floor and ladder is 0.4 and between wall and ladder is 0.25. The weight of ladder is 200 N. The ladder also supports a vertical load of 900 N at C. Determine the greatest value of 0 for which the ladder may be placed without slipping.
5. A 6.5m ladder AB of mass 10kg leans against a wall as shown. Assuming that the coefficient of static friction Îź is the same at both surface of contact, determine the smallest value of p for which equilibrium can be maintained.
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6. A ladder shown is 4 m long and is supported by a horizontal floor and vertical wall. The coefficient of friction at the wall is 0.25 and that at the floor is 0.5. The weight of ladder is 30 N. It also supports a vertical load of 150 N at 'C'. Determine the least value of 'Îą' at which the ladder may be placed without slipping to the left.
7. A uniform ladder of length 2.6m and weight 24O N is placed against a smooth vertical wall at A, with its lower end 1 m from the wall on the floor at B. The coefficient of friction between the ladder and the floor is 0.3. Find the frictional force acting on the ladder at the point of contact between the ladder and the floor. Will the ladder remain in equilibrium in this position? Give reason.
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