## Friction Short Answer Type Questions

**Question 1. A block of mass M rests on an inclined plane. If the coefficient of friction between the block and the plane is μ, then the block will slide down the plane under its own weight when the angle of inclination is**

- θ >tan
^{-1 }(μ) - θ >tan
^{-1 }(1/μ) - θ <tan
^{-1 }(μ) - θ <tan
^{-1 }(1/μ)

**Answer:** The option 1 is correct.

**Question 2. A block of mass 5 kg rests on a table. 8 N horizontal force is applied to push the block. Find out the force of friction between the block and the table. Given, μ _{s }= 0.3 and μ_{k} = 0.2, between the block and the table and g = 10m · s^{-2}**

**Answer:**

Normal reaction = 5×10 = 50N

So, limiting friction = μ_{s} x normal reaction = 0.3 x 50 = 15

Hence, on the application of 8 N horizontal force, the block remains at rest

So, frictional force = applied force = 8 N

**Question 3. A block is placed on a horizontal surface. The block is pushed by applying a force F which acts at an angle of θ with the vertical. How much force will be required to move the block if the frictional coefficient μ? Discuss the fact if tanθ < μ.**

**Answer:**

Normal reaction force, N = (mg+ Fcosθ)

∴ Minimum required force (horizontal) to move the block, p_{min} = μ(mg + Fcosθ)

Now, \(\frac{\text { required force }\left(P_{\min }\right)}{\text { applied horizontal force }(P)}=\frac{\mu(m g+F \cos \theta)}{F \sin \theta}\)

If \(\tan \theta<\mu\) then \(\frac{P_{\min }}{P}>\frac{m g \tan \theta+\tan \theta \cdot F \cos \theta}{F \sin \theta}>\frac{m g}{F \cos \theta}+1>1\) [because cosθ >0]

i.e., P<P_{min}

∴ The block remains static.

**Question 4.**

- Establish the relation between the angle of friction and the angle of repose.
- Explain with reason, whether the coefficient of friction between two surfaces can be zero.

**Answer:**

1. \(\vec{F}\) = limiting friction, \(\vec{N}\) = normal reaction, \(\vec{R}\) = resultant of \(\vec{F}\) and \(\vec{N}\).

The angle λ, between \(\vec{R}\) and \(\vec{N}\) is called the angle of friction.

∴ tanλ = \(\frac{F}{N}\) = μ = coefficient of friction

When a body just starts to move downward due to its own weight along an inclined plane, the angle of inclination of the plane at that moment is called the angle of repose.

Now, coefficient of friction, \(\mu=\frac{F}{N}=\frac{W \sin \theta}{W \cos \theta}=\tan \theta\)

∴ \(\mu=\tan \lambda=\tan \theta \quad \text { or, } \theta=\lambda\)

i.e., the angle of friction and the angle of repose are equal.

2. Coefficient of friction, μ = \(\frac{F}{N}\); for p to be zero F has to be zero. In practice, no matter how smooth two surfaces may be, there will always be some friction between the two surfaces whenever one is made to move over the other. Hence, μ can never be zero.

**Question 5. An object of weight W rests on an inclined plane. The coefficient of friction between the object and the plane is μ. For what value of the angle of inclination θ, will the object move downward with uniform speed under its own weight?**

**Answer:**

The forces acting on the body along with their components are shown.

When the body is about to slide down along the incline then, \(W \sin \theta=\mu R=\mu W \cos \theta\)

or, \(\tan \theta=\mu \quad \text { or, } \theta=\tan ^{-1} \mu\)

**Question 6. To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at 45° and the block is released from rest. The block takes a time t in moving a distance d. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time t/2 in moving down the same distance d. The coefficient of friction is**

- 3/4
- 5/4
- 1/2
- 1/2

**Answer:**

For a smooth surface, the block takes a time t/2 in moving a distance d with an acceleration gsinθ.

∴ d = \(\frac{1}{2} g \sin \theta \cdot \frac{t^2}{4}=\frac{1}{8} g \sin \theta \cdot t^2\)…(1)

For rough surfaces, the block now takes a time t in moving down the same distance with an acceleration (gsinθ-f) or (gsinθ – μgcosθ).

∴ d = \(\frac{1}{2}(g \sin \theta-\mu g \cos \theta) \cdot t^2\)…(2)

Comparing equations (1) and (2), \(\frac{1}{4} \sin \theta=\sin \theta-\mu \cos \theta\)

or, \(\mu=\tan \theta-\frac{1}{4} \tan \theta=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 45^{\circ}=\frac{3}{4}\)

The option 1 is correct.

**Question 7. Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal. The maximum weight of block A for which the system will be stationary is**

- \(\frac{W \tan \theta}{\mu}\)
- \(\mu W \tan \theta\)
- \(\mu W \sqrt{1+\tan ^2 \theta}\)
- \(\mu W \sin \theta\)

**Answer:**

The normal reaction of the table on block B = W

So the effective frictional force acting on the left = μW

Again when the maximum weight of A is W_{0}, it acts downwards.

Therefore, when in equilibrium, the cord between the knot and the wall will be along the resultant of μW and W_{0}.

In that case, \(\tan \theta=\frac{W_0}{\mu W} \quad \text { or, } W_0=\mu W \tan \theta\)

The option 2 is correct.

**Question 8. A block of mass m _{2} is placed on a horizontal table and another block of mass m_{1} is placed on top of it. An increasing horizontal force F = αt is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is μ_{1} and that between the lower block and the table is μ_{2}, then what is the maximum possible value of fi μ_{1}/μ_{2}?**

- \(\frac{m_2}{m_1}\)
- \(1+\frac{m_2}{m_1}\)
- \(\frac{m_1}{m_2}\)
- \(1+\frac{m_1}{m_2}\)

**Answer:**

∴ \(\mu_2\left(m_1+m_2\right) g\) ≥ \(\mu_1 m_1 g\)

or, \(\frac{m_1+m_2}{m_1}\) ≥ \(\frac{\mu_1}{\mu_2}\)

or, \(\frac{\mu_1}{\mu_2}\) ≤ \(1+\frac{m_2}{m_1} or, \left(\frac{\mu_1}{\mu_2}\right)_{\text {max }}=1+\frac{m_2}{m_1}\)

The option 2 is correct.

**Question 9. Given two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is **

- 100 N
- 80 N
- 120 N
- 150 N

**Answer:**

On the plane of contact of A and B, the upward frictional force acting on A = μF = W_{A} = 20 N.

So the downward frictional force on B on that plane = 20 N

This 20 N force and the weight of B is balanced by μ_{2}F (the upward frictional force applied by the wall on B)

∴ μ_{2}F = 20 + W_{A} = 20 + 100 = 120 N

The option 3 is correct.

**Question 10. Two masses m _{1 }= 5 kg and m_{2} = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown. The coefficient of friction of the horizontal surface is 0.15. The minimum weight m that should be put on top of m_{2} to stop the motion is**

- 43.3 kg
- 10.3 kg
- 18.3 kg
- 27.3 kg

**Answer:**

Let the mass m_{2} come to rest if a mass m is placed on it. For the mass m_{2} to come to rest, an acceleration (due to friction) should act in the opposite direction of its motion i.e., when f≤ μN

Also f = T = 5 g [when m and m_{2} are at rest]

∴ m_{1}g ≤ μ(m +m_{2})g or, 5 ≤ 0.15(m +10)

or, 23.33 <≤ m

The option 4 is correct.

**Question 11. A system consists of three masses m _{1}, m_{2}, and m_{3} connected by a string passing over a pulley P. The mass m_{1} hangs freely and m_{2} and m_{3} are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass ml is (Assume m_{1} = m_{2} = m_{3} = m)**

- \(\frac{g(1-g \mu)}{9}\)
- \(\frac{1 g \mu}{3}\)
- \(\frac{g(1-2 \mu)}{3}\)
- \(\frac{g(1-2 \mu)}{2}\)

**Answer:**

Acceleration = \(\frac{\text { net force in the direction of motion }}{\text { total mass of system }}\)

= \(\frac{m_1 g-\mu\left(m_2+m_3\right)} g= {m_1+m_2+m_3}=\frac{g(1-2 \mu)}{3}\)

The option 3 is correct.

**Question 12. A block A of mass m _{1} rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m_{2} is suspended. The coefficient of kinetic friction between the block and the table is μ_{k}. When the block A is sliding on the table, the tension in the string is**

- \(\frac{\left(m_2+\mu_k m_1\right) g}{\left(m_1+m_2\right)}\)
- \(\frac{\left(m_2-\mu_k m_1\right) g}{\left(m_1+m_2\right)}\)
- \(\frac{m_1 m_2\left(1+\mu_k\right) g}{\left(m_1+m_2\right)}\)
- \(\frac{m_1 m_2\left(1-\mu_k\right) g}{\left(m_1+m_2\right)}\)

**Answer**:

The frictional force acting on block A = μ_{k}m_{1}g

Hence, the resultant force on the blocks = m_{2}g-μ_{k}m_{1}g

Their acceleration, a = \(\frac{m_2 g-\mu_k m_1 g}{m_1+m_2}\)

If the tension in the string is T, then for block B, \(m_2 g-T=m_2 a\)

∴ T = \(m_2(g-a)=m_2 g\left(1-\frac{m_2-\mu_k m_1}{m_1+m_2}\right)\)

= \(m_2 g \frac{m_1+\mu_k m_1}{m_1+m_2}\)

= \(\frac{m_1 m_2\left(1+\mu_k\right) g}{m_1+m_2}\)

The option 3 is correct.

**Question 13. Which one of the following statements is incorrect?**

- Frictional force opposes the relative motion
- The limiting value of static friction is directly proportional to the normal reaction
- Rolling friction is smaller than sliding friction
- The coefficient of sliding friction has dimensions of length

**Answer: **4. The coefficient of friction is a dimensionless quantity.

The option 4 is correct.

**Question 14. The inclination θ of a rough plane is increased gradually. The ply on the plane just comes into motion when inclination θ becomes 30°. Find the coefficient of friction. If the inclination is further increased to 45° then find the acceleration of the body along the plane, (g = 10 m · s ^{-2})**

**Answer:**

Here, angle of repose =30°

∴ Coefficient of friction, μ = tan30° = 1/√3

If the angle of inclination is 45°, the acceleration of the body,

a = \(g\left(\sin 45^{\circ}-\mu^{\prime} \cos 45^{\circ}\right)=10 \times \frac{1}{\sqrt{2}}\left(1-\mu^{\prime}\right)\)

= \(\frac{10}{\sqrt{2}}\left(1-\frac{1}{\sqrt{3}}\right) \approx 2.99 \mathrm{~m} \cdot \mathrm{s}^{-2} \quad\left[because \mu^{\prime} \approx \mu\right]\)

**Question 15. Give the magnitude and direction of the net force acting on a car moving with a constant velocity of 30 km · h ^{-1} on a rough road.**

**Answer:**

The velocity does not change, so, there is no acceleration.

The relation F = ma shows that, as a = 0, the net force F = 0.

Here, the force applied by the engine is balanced exactly by the friction of the road.

**Question 16. Two bodies A, B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is 0.15. A force of 200N is applied horizontally to A.**

- What is the reaction of the partition?
- The action-reaction forces between A, and B?
- What happens when the wall is removed?

**Does the answer 2. Change when bodies are in motion? [Ignore the difference between μ _{s }and μ_{k}]**

**Answer:**

We take the rounded-off value, g = 10 m · s^{-2}

Normal reaction on A, N_{A} = its weight (mg)

= 5 x 10 = 50 N

Normal reaction on B, N_{B} = 10 x 10 = 100 N

So the limiting friction on A, f_{A} = μN_{A} = 0.15 x 50 = 7.5 N

and limiting friction on B, f_{B} = μN_{B} = 0.15 x 100 = 15 N

(As the applied force, 200 N, is higher, the force of friction attains its limiting value)

1. We take the combination of A and B as a single body. It is at rest due to the presence of the wall. If R is the reaction of the wall, the force equation in the horizontal direction is, 200-f_{A}-f_{B}-R=0

or, R = 200- 7.5- 15 = 177.5 N

2. Let F_{AB} = action force of A on B; F_{BA} = reaction force of B on A; F_{AB} and F_{BA} are equal and opposite. From the equilibrium of A, we get 200-f_{A}-F_{BA} = 0

or, F_{BA} = 200-7.5 = 192.5 N

So, F_{AB} = -192.5 N

3. Now, if the wall is removed, the reaction force R is absent; due to the applied force, the combination of A and B would move forward with an acceleration a (say). Then the force equation in the horizontal direction would be, 200 -f_{A}-f_{B} = (m_{A}+m_{B})a

or, a = \(\frac{200-7.5-15}{5+10}=\frac{177.5}{15}=11.83 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Now, the force equation for the body A is, 200-f_{A}-F’_{BA} = m_{A}a

[F’_{BA} = new value of force by B on A]

or, F’_{BA} = 200-7.5-5 x 11.83 – 133.3 N

Similarly, F’_{AB} = -133.3 N

So due to the accelerated motion, the action-reaction pair changes to a much reduced value.

**Question 17. Define:**

- Stopping distance of a vehicle,
- Reaction time during a free fall.

**Answer:**

- Let at a certain moment, the brakes are applied on a vehicle running on a road. If the vehicle then travels a distance x until it comes to rest, then x is called the stopping distance of the vehicle.
- Many natural events demand immediate action from the observers. An observer observes an event, registers it in his brain, and then takes the required action. The time spent between the first observation and the action taken is called his/her reaction time.
- For example, suppose an observer sees an object falling freely from the ceiling of a room towards a glass plate kept on a table. He/she would try either to intercept the falling object or to remove the plate from the table. In this context, if the reaction time is greater than the time of fall, then the glass would break.

**Question 18. Give two methods of reducing friction. Show that kinetic friction is less than static friction.**

**Answer:**

- Application of lubricating oils, and
- Use of ball bearings and roller bearings, between the surfaces in contact, are two of many methods that can reduce friction.

- When a force, applied to move a body over a surface, just exceeds the limiting friction, the body starts to move. Now, even if the applied force is not increased any more, the body is observed to accelerate.
- This means that the resultant force is positive, i.e., the kinetic friction during the motion is less than the applied force. This indicates that the kinetic friction falls below the limiting friction (the limiting value of static friction) as soon as the body starts to move.

**Question 19. What is the acceleration of the block and trolley system shown given below, if the coefficient of friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m · s ^{-2}). Neglect the mass of the string.**

**Answer:**

The 3 kg block and the 20 kg trolley both have the same magnitude of acceleration.

Hence, for 3 kg block, 30 – T = 3a…..(1)

where T is tension in the string and a is acceleration.

For the trolley, T – f = 20a

Now for friction, f = μR = 0.04 x 20 x 10 = 8N

where, f = force of friction

μ = coefficient of friction between trolley and the surface

and R = normal force on the trolley

Hence, T – 8 = 20a…..(2)

Solving equations (1) and (2), we get, a = \(\frac{22}{23}\) m s = 0.96 m s and T = 27.2 N

**Question 20. A block of mass 3 kg slides down an incline of angle 30° with acceleration g/4. Complete the free-body diagram and find the coefficient of kinetic friction.**

**Answer:**

The forces on the block of mass m are as shown.

Taking components along the inclined plane, \(m g \sin 30^{\circ}-f=\frac{m g}{4}\)[because a=g]

or, f = \(\frac{m g}{4}\)

There is no acceleration perpendicular to inclin plane, so R = \(m g \cos 30^{\circ}=\frac{\sqrt{3}}{2} m g\)

Then coefficient of friction, \(\mu=\frac{f}{R}=\frac{m g}{4\left(m g \frac{\sqrt{3}}{2}\right)}=\frac{1}{2 \sqrt{3}}\)