Question 1

For most of the surfaces used in daily life, the friction coefficient is less than 1. Is it always necessary that the friction coefficient is less than 1?

Accepted Answer

It is not necessary that the friction coefficient is always less than 1. When the friction is stronger than the normal reaction force, the coefficient of friction is greater than 1. For example, silicon rubber has the coefficient of friction greater than 1.

Question 2

Why is it easier to push a heavy block from behind than to press it on the top and push?

Accepted Answer

It is easier to push a heavy block from behind than from the top because when we try to push a heavy block from the top, we increase the normal reaction force, which, in turn, increases the friction between the object and the ground.

Question 3

What is the average friction force when a person has a usual 1 km walk?

Accepted Answer

The person started with zero initial velocity, covered a 1 km distance and ended with zero velocity, so the acceleration is zero. Hence, the average friction force is zero.

Question 4

Why is it difficult to walk on solid ice?

Accepted Answer

It is difficult to walk on solid ice because the coefficient of friction between our foot and ice is very less; hence, a person trying to walk on solid ice may slip.

Question 5

Can you accelerate a car on a frictionless horizontal road by putting more petrol in the engine? Can you stop a car going on a frictionless horizontal road by applying brakes?

Accepted Answer

No, we cannot accelerate or stop a car on a frictionless horizontal road. The car will not move on a frictionless surface because rolling is not possible without friction.

Question 6

Spring fitted doors close by themselves when released. You want to keep the door open for a long time, say for an hour. If you put a half kg stone in front of the open door, it does not help. The stone slides with the door and the door gets closed. However, if you sandwitch a 20 g piece of wood in the small gap between the door and the floor, the door stays open. Explain why a much lighter piece of wood is able to keep the door open while the heavy stone fails.

Accepted Answer

In the first case, the normal reaction force is equal to the weight of the stone, hence the stone slides easily because the friction force is very less. However, in the second case, a small piece of wood is sandwiched, which increases the normal reaction force on the wood due to the weight of the door. Hence, greater the normal reaction force on the wood, the greater will be the frictional force between wood and the floor.

Question 7

A classroom demonstration of Newton's first law is as follows : A glass is covered with a plastic card and a coin is placed on the card. The card is given a quick strike and the coin falls in the glass. (a) Should the friction coefficient between the card and the coin be small or large? (b) Should the coin be light or heavy? (c) Why does the experiment fail if the card is gently pushed?

Accepted Answer

(a) The coefficient of friction between the card and the coin should be small.(b) The coin should be heavy.(c) If the card is pushed gently, the experiment fails because the frictional force gets more to time to act and it may gain some velocity and move with the card.

Question 8

Can a tug of war be ever won on a frictionless surface?

Accepted Answer

No, a tug of war cannot be won on a frictionless surface because the tension in the rope on both the sides of both the teams will be same. So, to win, one of the teams must apply some greater force, which is the force of friction.

Question 9

Why do tyres have a better grip of the road while going on a level road than while going on an incline?

Accepted Answer

The normal reaction force on a level road is mg, whereas on an inclined plane it is mg cos θ, which means that on an incline road the friction force between the tyre and the road is less. Hence, tyres have less grip on an incline plane and better grip on a level road.

Question 10

You are standing with your bag in your hands, on the ice in the middle of a pond. The ice is so slippery that it can offer no friction. How can you come out of the ice?

Accepted Answer

By throwing the bag in one direction, we gain some velocity in the opposite direction as per the law of conservation of linear momentum. In this way we can come out of the ice easily.

Question 11

When two surfaces are polished, the friction coefficient between them decreases. But the friction coefficient increases and becomes very large if the surfaces are made highly smooth. Explain.

Accepted Answer

The coefficient of friction increases between two highly smooth surfaces because the atoms of both the materials come very closer to each and the number of bonds between them increase.

Question 12

In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased, the frictional force between the surface and the body will

Accepted Answer

(b) Decrease. According to the first law of limiting friction, f = μN, where f  is the frictional force.N is the normal reaction force, μ is the coefficient of static friction andN = mg － Fcosθwhere m is the mass of the body and F is the contact force acting on the body.

Question 13

While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure:

Accepted Answer

(b) Smaller friction.According to the first law of the limiting friction,f = μN , where f is the frictional force, μ is the coefficient of friction, N is the normal reaction force.When we take smaller steps on ice, the normal reaction force exerted by the ice is small. Therefore, the smaller steps ensure smaller friction.

Question 14

A body of mass M is kept on a rough horizontal surface (friction coefficient = μ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on A is F, where:

Accepted Answer

(c) Mg≤F≤Mga1+μ2​.Let T be the force applied on an object of mass M.If T = 0, Fmin​  = Mg.If T is acting in the horizontal direction, then the body is not moving.∴ T = μ(mg)Fmax​  = (Mg)2+T2)​ = Mg2+(μMg)2​ Thus, we have:Mg≤F≤Mga1+μ2​

Question 15

A scooter starting from rest moves with a constant acceleration for a time ∆t1​ , then with a constant velocity for the next ∆t2​  and finally with a constant deceleration for the next ∆t3​  to come to rest. A 500 N man sitting on the scooter behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is:

Accepted Answer

(d) >500 N for time ∆t1​ and ∆t3​ and 500 N for ∆t2​.During the time interval ∆t2​, the scooter is moving with a constant velocity, which implies that the force exerted by the seat on the man is 500 N (for balancing the weight of the man).During the time interval ∆t1​ and ∆t3​, the scooter is moving with constant acceleration and deceleration, which implies that a frictional force is also applied. Therefore, the net force exerted by the seat on the man should be >500 N.

Question 16

Consider the situation shown in figure. The wall smooth but the surface of A and B in contact are rough. The friction on B due to A in equilibrium.

Accepted Answer

(d) The system cannot remain in equilibrium.Since the wall is smooth and the surface of A and B in contact are rough, the net vertical force on the system is in the downward direction. Hence, the system cannot remain in equilibrium.

Question 17

Suppose all the surface in the previous problem are rough. The direction of friction of B due to A.

Accepted Answer

(a) Is upward.The normal reaction force on the system (comprising of wall and contact surface of Aand B) is provided by F. As can be seen from the figure, the weight of A and B is in the downward direction. Therefore, the frictional force fA and fBA (friction on B due to A) is in upward direction.

Question 18

Two cars of unequal masses use similar tyres. If they are moving at the same initial speed, the minimum stopping distance:

Accepted Answer

(c) Is same for both cars.Given: both the cars have same initial speed.Let the masses of the two cars be m1​ and m2​.Frictional force on car with mass m1​=μm1​ g So, the deceleration due to frictional force =m1​μm1​g​=μg Frictional force on car with mass m2​=μm2​ g So, the deceleration due to frictional force =m2​μm2​ g​=μg As both the acceleration are same, from the second equation of motion s=ut+21​at2 Thus, we can say that both the cars have same minimum stopping distance

Question 19

In order to stop a car in shortest distance on a horizontal road, one should:

Accepted Answer

(b) Apply the brakes hard enough to just prevent slipping.When we apply hard brakes just enough to prevent slipping on wheels, it provides optimum normal reaction force, which gives the maximum friction force between tyres of the car and the road.

Question 20

A block A kept on an inclined surface just begins to slide if the inclination is 30°. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40°.

Accepted Answer

(d) All of three are possible.We know thatN = mg cos θ˚fmax = μN = μmg cos θwhere N  =  normal reaction force          fmax​ = frictional force , θ =  angle of inclination , μ = coefficient of frictionWhen the block just begins to slide, it means mg sin θ = fmax​ mg sin θ = μmg cos θ          μ = tan θ and the coefficient of friction depends on the angle of inclination (θ) and does not depend on mass.Now consider the block sliding condition: mg sin θ − fmax = ma mgsin θ − μmg cos θ = ma∴ a = g(sin θ − μ cos θ)From the above equation it is clear that acceleration does not depend on the mass but depends on θ and μ.

Chapter 6 : Friction