# Static Friction

## Trending Questions

**Q.**Give some methods to increase friction.

**Q.**

A block of mass 2 kg is pushed normally against a rough vertical wall with a force of 40 N, co-efficient of static friction being 0.5. Another horizontal force of 15 N is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction and with what acceleration? If no, find the frictional force exerted by the wall on the block.

No Frictional force = 20N

Yes

Direction = 53° with horizontal

Acceleration = 5 m/s

^{2}Yes

Direction = 37° with vertical

Acceleration = 2.5 m/s

^{2}No

Frictional force = 20N

**Q.**A block of weight 5 N is pushed against a vertical wall by a force 12 N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

- 12 N
- 5 N
- 7.2 N
- 13 N

**Q.**A block of mass m is placed on a surface with a vertical cross-section represented by the equation y=x36. If the coefficient of static friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is

- 13 m
- 12 m
- 23 m
- 16 m

**Q.**If a man is walking on a rough horizontal ground without slipping, then identify the correct statement(s) from the options given below:

- Frictional force on the man will act in backward direction.
- Frictional force on the man will act in forward direction.
- Work done by frictional force on the man will be negative.
- Work done by frictional force on the man will be zero.

**Q.**A block of mass 4 kg is placed on a rough horizontal plane. A time dependent force F=kt2 acts on the block. where k=2 N/s2. Coefficient of friction μ=0.8 . Force of friction between block and the plane at t=2 s is

- 8 N
- 4 N
- 2 N
- 32 N

**Q.**Figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The coefficients of static and kinetic friction are 0.28 and 0.25 respectively at each of the surfaces. Find the minimum and maximum values of m for which the system remains at rest. (Both in kg)

- 98, 329
- 38, 3210
- 38, 329
- 98, 3210

**Q.**A uniform chain of length L has one of its ends attached to the wall at point A, while 3L4 length of the chain is lying on the table as shown in figure. Then, minimum co-efficient of static friction between table and chain, so that the chain remains in equilibrium is

- 13
- 15
- 14
- 34

**Q.**

Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms−2. What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

**Q.**The ratio of the limiting force of friction (F) to the normal reaction (R) is known as

- angle of friction
- coefficient of static friction
- coefficient of kinetic friction
- none of these

**Q.**A block of mass 3 kg is placed on a rough horizontal surface (μs=0.4). A force of 8.7 N is applied on the block. If g=10 ms−2, then the force of friction between the block and floor is

- 8.7 N
- Zero
- 12 N
- 10 N

**Q.**Carbon, silicon, and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)Si, (Eg)cand (Eg)Ge. Which of the following statements is true?

- (Eg)Si<(Eg)Ge<(Eg)C
- (Eg)C=(Eg)Si=(Eg)Ge
- (Eg)C>(Eg)Si>(Eg)Ge
- (Eg)C<(Eg)Ge>(Eg)Si

**Q.**A horse pulls forward on a carriage with a given force. By Newton's Third Law, the carriage must be pulling on the horse backward with an equal and opposite force. Given this, what explains why the horse and carriage can move forward?

- The cart's force is only in reaction to the horse's force so it does not define direction of movement
- The forward and backward forces are equal, so it actually can't move forward
- The net force of friction by the ground on the (carriage+horse) system acts in the forward direction thus enabling a forward motion.
- The forward force of the horse is just big enough to overcome the backward force of the cart and start the cart forward

**Q.**A block of mass 1 kg is projected from the lowest point up along the inclined plane, if g=10 m/s2, the retardation experienced by the block is

- 15√2 m/s2
- 5√2 m/s2
- 10√2 m/s2
- zero

**Q.**A block of mass m is at rest on a rough wedge of angle θ as shown. What is the force exerted by the wedge on the block?

- mgcosθ
- mgsinθ
- mgtanθ
- mg

**Q.**A lift is moving upwards with a uniform velocity v in which a block of mass m is lying. The frictional force offered by the block, when coefficient of friction is μ=0.5, will be

- Zero
- mg2
- mg
- 2mg

**Q.**Which of the following statement is correct, when a man walks on a rough surface:

- The frictional force exerted by the surface keeps him moving
- The force which the man exerts on the floor keeps him moving
- None of the above
- The reaction of the force which the man exerts on floor keeps him moving

**Q.**

Why is it difficult to move a bike with its brakes on?

**Q.**The upper portion of an inclined plane of inclination α is smooth and the lower portion is rough. A particle slides down from rest from the top and just comes to rest at the foot. If the ratio of smooth length to rough length is m : n, find the coefficient of friction

**Q.**Choose the incorrect statement from statements given below:

- Friction is an electromagnetic force.
- Coefficient of static friction is greater than coefficient of kinetic friction.
- None of these
- Value of static friction is always greater than kinetic friction.

**Q.**The tension T in the string in figure is

- (√3−1)50 N
- Zero
- 50 N
- 35√3 N

**Q.**A block of mass 1 kg is placed on a rough inclined plane at an angle θ=60∘ with the horizontal and the block is connected with a string as shown in figure. If coefficient of friction between the block and inclined surface, μs=3/4, find the tension in the string. [Take g=10 m/s2]

- 0.66 N
- 4.9 N
- 5 N
- 0 N

**Q.**A man standing stationary with respect to a horizontal conveyer belt which is accelerating with 1 m/s2 as shown in figure. If the coefficient of static friction between man's shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man =65 kg). Take g=10 m/s2.

- 1 m/s2
- 2.5 m/s2
- 3 m/s2
- 2 m/s2

**Q.**In the arrangement shown in the figure, the wall is smooth and the coefficient of static friction between the blocks is μ=0.1. A horizontal force F=1000 N is applied on the 2 kg block. Identify the wrong statement from the options given below :

- The normal reaction force between the blocks is 1000 N.
- The friction force between the blocks is zero.
- Both the blocks accelerate downwards with acceleration g.
- Friction force acting on 2 kg block is 2g N.

**Q.**

A block is placed on a rough floor and a horizontal external force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. (taking f on Y and F on X axis)

(a)The graph is a straight line of slope

The graph is a straight line parallel to the F-axis

The graph is a straight line of slope for small F and a straight line parallel to the F-axis for large F

There is a small kink on the graph

**Q.**A block of mass 5 kg is kept on a horizontal floor having coefficient of friction 0.09. Two mutually perpendicular horizontal forces of 3 N and 4 N act on this block. The acceleration of the block is: (g=10 m/s2)

- Zero
- 0.1 m/s2
- 0.2 m/s2
- 0.3 m/s2

**Q.**

The coefficient of static friction between the block of 2 kg and the table shown in figure is μs = 0.2. what should be the maximum value of m so that the blocks do not move? Take g = 10 m/s2. The string and the pulley are light and smooth.

2 kg

1 kg

0.4 kg

0.5 kg

**Q.**

Blocks A and B in the fig. Are connected by a bar of negligible weight. Mass of each block is 170 kg and μA = 0.2 and μB = 0.4, where μA and μB are the coefficients of limiting friction between blocks and plane, calculate the force developed in the bar (g=10 ms−2)

75 N

150 N

200 N

250 N

**Q.**Angle θ shown in graph between friction and applied force on a body is :

- 60∘
- 45∘
- 30∘
- 25∘

**Q.**Mass of block A is 100 kg and that of block B is 200 kg. As shown in figure, the block A is attached to a string tied to the wall. The coefficient of static friction between blocks A and B is (μs)A=0.2 and the coefficient of static friction between block B and floor is (μs)B=0.3. Then calculate the minimum value of force F required to move the block B. (g=10 m/s2)

- 980 N
- 1000 N
- 1100 N
- 1200 N