Table of Contents
QUESTIONS FOR SHORT ANSWER
1 A 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?
Answer: No
Explanation: It is not always necessary for the friction coefficient to be less than 1. While many common surfaces have a coefficient of friction (μ) between 0 and 1, values greater than 1 are also possible. In some cases, the friction coefficient can exceed 1, meaning the frictional force is greater than the normal reaction force. For example, in certain situations the coefficient of friction is given as 4/3, which is greater than 1, and even a value of 1.0 is observed in some cases. Theoretically, the coefficient of friction is defined as μ = f/N, so if the limiting friction force (f) is greater than the normal force (N), then μ becomes greater than 1. In cases of high surface adhesion or extremely smooth contact, the friction coefficient can become very large due to strong molecular attraction between surfaces.
2 A Why is it easier to push a heavy block from behind than to press it on the top and push?
Answer: Because friction is lower when pushing from behind
Explanation: The ease of moving a block depends on the frictional force which is directly proportional to the normal force (f = μN). When the block is pushed from behind, the normal force is equal to its weight (Mg). However, when the block is pressed on the top and pushed, an additional downward force is applied, which increases the normal force (N = Mg + applied force). This increase in normal force leads to an increase in limiting friction, making it harder to move the block. Therefore, pushing from behind requires less force because the frictional resistance is smaller.
3. A What is the average friction force when a person has a usual 1 km walk?
Answer: Zero
Explanation: When a person walks 1 km starting from rest and ending at rest, the total change in momentum is zero. During walking, static friction between the shoes and ground provides the necessary force for motion, including both acceleration and deceleration. According to the impulse-momentum theorem, the average net force depends on the change in momentum over time. Since the initial momentum and final momentum are both zero, the total change in momentum is zero. Hence, the time-averaged net external force, which includes friction effects in the horizontal direction, is also zero.
4. Why is it difficult to walk on solid ice?
Solution:
Walking depends on the frictional force between a person’s feet and the ground. To move forward, a person must push the ground backward; the ground then exerts an equal and opposite frictional force forward on the person, providing the necessary external force for acceleration.
Solid ice is extremely smooth and has a very low coefficient of friction, meaning it offers very little resistance to slipping. Because the maximum static friction available (fs = μN) is so small on ice, the horizontal force exerted by a normal walking step often exceeds this limit, causing the foot to slip. This makes it difficult to generate enough forward force to walk effectively or maintain balance.
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?
Solution:
No to both questions.
Accelerating: For a vehicle to accelerate, there must be a net external force acting on it. On a horizontal road, this force is provided by the friction between the tires and the road. While putting more petrol in the engine increases the torque and can make the wheels spin faster, without friction, there is no external force to push the car forward. The wheels would simply spin in place.
Stopping: Similarly, to stop a moving car, an external force must act opposite to the motion to cause deceleration. Applying brakes will stop the wheels from rotating, but on a frictionless surface, there is no force between the tires and the road to slow the car’s motion. The car would continue to move at constant velocity.
6. Spring fitted doors close by themselves when released. You want to keep the door open… 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 sandwich 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.
Solution:
The ability of an object to stop the door depends on the maximum static friction it can generate. Friction is given by f = μN, where μ is the coefficient of friction and N is the normal force.
The heavy stone: The normal force acting on the stone is equal to its weight (mg). Although the stone is heavier, the friction produced is still not enough to oppose the strong closing force of the spring. As a result, the door pushes the stone and it slides along the floor.
The small wooden wedge: When the wood is placed in the narrow gap, the closing force of the spring pushes it tightly into position. This creates a large normal reaction force due to the wedging effect. Because friction depends on normal force, the frictional force becomes very large (f = μN), enough to resist the spring force and keep the door open.
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?
Solution: The coefficient of friction should be small, so that minimal horizontal force is transferred to the coin and it remains nearly at rest due to inertia.
(b) Should the coin be light or heavy?
Solution: The coin should be heavy, because a heavier object has greater inertia and resists changes in its state of motion more effectively.
(c) Why does the experiment fail if the card is gently pushed?
Solution: When the card is pushed gently, the force acts for a longer time, allowing friction to accelerate the coin along with the card. Instead of staying at rest, the coin moves with the card and does not fall straight into the glass.
Based on the principles of mechanics and friction discussed in the sources, here are the solutions to Questions 8 and 9:
8. Can a tug of war be ever won on a frictionless surface?
Solution:
No, a tug of war cannot be won on a frictionless surface.
Reasoning:
In a tug of war, winning depends on the ability of a team to exert a horizontal external force on the ground. When a team pushes backward against the ground, the ground provides an equal and opposite static frictional force that allows them to stay anchored and pull the rope effectively.
On a frictionless surface, there is no horizontal external force from the ground. Hence, neither team can generate the necessary traction to remain fixed or pull the rope effectively. As a result, no team can win the tug of war.
9. Why do tyres have a better grip of the road while going on a level road than while going on an incline?
Solution:
The grip of tyres depends on the maximum static friction, given by fs = μN, where N is the normal force between the tyre and the road.
On a level road: The normal force is equal to the weight of the vehicle, N = Mg.
On an incline: The normal force reduces to N = Mg cosθ.
Since cosθ < 1, the normal force on an incline is less than that on a level road. Since friction is directly proportional to the normal force, the available grip is reduced on an incline. Therefore, tyres have better grip on a level road than on an inclined surface.
Based on the principles of mechanics and the properties of friction described in the sources, here are the solutions to Questions 10 and 11:
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?
Solution:
Since the ice is frictionless, you cannot walk or run to the shore because there is no external horizontal force (static friction) to provide acceleration.
To move, you must use Newton’s Third Law of Motion and the principle of conservation of momentum.
Method: Throw your bag horizontally away from the shore.
When you throw the bag, you exert a force on it. The bag exerts an equal and opposite reaction force on you. Since there is no friction, this reaction force will cause you to move in the opposite direction (towards the shore).
As there is no friction to stop your motion, you will continue sliding until you reach the edge of the pond.
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.
Solution:
This behavior is due to the change in the nature of interaction between surfaces as their smoothness increases.
Initially (polishing stage):
Surfaces have microscopic irregularities (peaks and valleys). Friction mainly arises due to mechanical interlocking of these irregularities. Polishing reduces these irregularities, decreasing interlocking and therefore reducing the coefficient of friction.
At very high smoothness:
When surfaces become extremely smooth, they come into very close molecular contact over a large area.
Molecular adhesion:
At this stage, strong attractive forces between molecules of the two surfaces become significant. These adhesive forces act like bonding between surfaces, making relative motion difficult. As a result, the friction coefficient increases and can become very large.
OBJECTIVE I
1. 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:
(a) increase
(b) decrease
(c) remain the same
(d) may increase or decrease.
Answer:
The correct option is (b) decrease.
Solution:
The contact force exerted by a surface on a body is the resultant of two perpendicular components:
• The normal force (N), acting vertically upward.
• The frictional force (f), acting horizontally along the surface.
The magnitude of the resultant contact force is:
Fc = √(N² + f²)
Let θ be the angle between the contact force and the vertical (normal direction).
Then, using trigonometry:
• f = Fc sinθ
• N = Fc cosθ
Since the magnitude of the contact force (Fc) is constant, a decrease in θ leads to a decrease in sinθ.
Therefore, the frictional force f = Fc sinθ decreases when θ decreases.
Based on the provided source material, here are the questions, answers, and solutions for Questions 2 and 3 of the OBJECTIVE I section:
2. While walking on ice, one should take small steps to avoid slipping. This is because smaller steps ensure:
(a) larger friction
(b) smaller friction
(c) larger normal force
(d) smaller normal force
Answer:
The correct option is (b) smaller friction.
Solution:
While walking, the foot applies a force on the ground which has a horizontal component. This horizontal component must be balanced by the static frictional force to prevent slipping.
When a person takes smaller steps, the leg remains more vertical. This reduces the horizontal component of the force exerted on the ground. As a result, the required friction to prevent slipping is reduced.
If the required friction is less than the maximum static friction available on ice, slipping does not occur. Hence, smaller steps ensure smaller friction is needed and reduce the chances of slipping.
3. 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:
(a) F = Mg
(b) F = μMg
(c) Mg ≤ F ≤ Mg√(1 + μ²)
(d) Mg ≥ F ≥ Mg√(1 − μ²)
Answer:
The correct option is (c) Mg ≤ F ≤ Mg√(1 + μ²).
Solution:
The force exerted by the surface is the resultant of two perpendicular components:
• Normal force: N = Mg
• Frictional force: f
The resultant contact force is:
F = √(N² + f²)
Since the body does not move, friction is static and varies from zero up to its limiting value:
f ≤ μN = μMg
Minimum value: When f = 0,
F = Mg
Maximum value: When f = μMg,
F = √(Mg)² + (μMg)² = Mg√(1 + μ²)
Therefore, the force by the surface lies in the range:
Mg ≤ F ≤ Mg√(1 + μ²)
4. A scooter starting from rest moves with a constant acceleration for a time Δt₁, then with a constant velocity for the next Δt₂ and finally with a constant deceleration for the next Δt₃ 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:
(a) 500 N throughout the journey
(b) less than 500 N throughout the journey
(c) more than 500 N throughout the journey
(d) > 500 N for time Δt₁ and Δt₃ and 500 N for Δt₂
Answer:
The correct option is (d).
Solution:
The force exerted by the seat on the man is the resultant of:
• Normal force (N) = 500 N (balances the weight of the man)
• Frictional force (f) = provides horizontal acceleration
During Δt₁ (acceleration) and Δt₃ (deceleration):
There is non-zero acceleration, so a horizontal frictional force is required. Hence, the resultant force is:
F = √(N² + f²)
Since f ≠ 0, the total force is greater than 500 N.
During Δt₂ (constant velocity):
Acceleration is zero, so no horizontal friction is needed. Thus:
F = N = 500 N
5. Consider the situation shown in figure (6-Q1). The wall is smooth but the surfaces of A and B in contact are rough. The friction on B due to A in equilibrium:
(a) is upward
(b) is downward
(c) is zero
(d) the system cannot remain in equilibrium
Answer:
The correct option is (d) the system cannot remain in equilibrium.
Solution:
For equilibrium, all forces in the system must balance.
The wall is smooth, so it provides no frictional force. Therefore, it cannot supply any vertical support to balance the weight of the system.
Since there is no external vertical force available to balance the weight of blocks A and B, the system will not be able to remain in equilibrium and will slide downward.
6. Suppose all the surfaces in the previous problem are rough. The direction of friction on B due to A:
(a) is upward
(b) is downward
(c) is zero
(d) depends on the masses of A and B
Answer:
The correct option is (a) is upward.
Solution:
Block B is in contact with block A, and a horizontal force presses the system against a smooth wall (as in the previous setup).
The weight of block B acts vertically downward due to gravity. To maintain equilibrium or prevent downward motion, an opposing force must act on B in the upward direction.
Since the surfaces between A and B are rough, friction can act between them. The frictional force exerted by A on B must oppose the tendency of B to move downward under gravity.
Therefore, the friction on B due to A acts upward.
7. Two cars of unequal masses use similar tyres. If they are moving at the same initial speed, the minimum stopping distance:
(a) is smaller for the heavier car
(b) is smaller for the lighter car
(c) is same for both cars
(d) depends on the volume of the car
Answer:
The correct option is (c) is same for both cars.
Solution:
The stopping distance is given by:
s = u² / (2a)
The retarding force is provided by friction:
f = μN = μmg
Using Newton’s second law:
a = f / m = μmg / m = μg
The mass cancels out, so the deceleration is independent of mass.
Thus, stopping distance becomes:
s = u² / (2μg)
Since both cars have the same initial speed and same tyres (same μ), their stopping distance is identical regardless of mass.
8. In order to stop a car in the shortest distance on a horizontal road, one should:
(a) apply the brakes very hard so that the wheels stop rotating
(b) apply the brakes hard enough to just prevent slipping
(c) pump the brakes (press and release)
(d) shut the engine off and not apply brakes
Answer:
The correct option is (b) apply the brakes hard enough to just prevent slipping.
Solution:
To stop a car in the shortest possible distance, the maximum decelerating force must be used. This force is provided by friction between the tyres and the road.
There are two types of friction involved:
• Static friction (maximum just before slipping)
• Kinetic friction (when wheels skid)
Static friction is greater than kinetic friction. Therefore:
If the brakes are applied so hard that the wheels lock and start skidding, kinetic friction acts and the stopping force is reduced.
If the brakes are applied just enough to prevent slipping, the tyres remain in rolling condition and the system experiences maximum static (limiting) friction, giving maximum deceleration.
Hence, the shortest stopping distance is achieved when wheels are just on the verge of slipping.
9. 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°.
(a) mass of A > mass of B
(b) mass of A < mass of B
(c) mass of A = mass of B
(d) all the three are possible
Answer:
The correct option is (d) all the three are possible.
Solution:
A block begins to slide on an incline when the component of its weight along the slope equals the limiting static friction.
Along the incline:
Mg sinθ = μs N
Normal force:
N = Mg cosθ
Substituting:
Mg sinθ = μs Mg cosθ
The mass cancels out:
μs = tanθ
This shows that the angle of sliding depends only on the coefficient of friction, not on mass.
Therefore, the change from 30° to 40° indicates a change in frictional properties, not mass. Hence, all mass relations between A and B are possible.
Based on the provided source material, here is the question, answer, and solution for Question 10 of the OBJECTIVE I section:
10. A boy of mass M is applying a horizontal force to slide a box of mass M’ on a rough horizontal surface. The coefficient of friction between the shoes of the boy and the floor is μ and that between the box and the floor is μ’. In which of the following cases it is certainly not possible to slide the box?
(a) μ < μ’, M < M’
(b) μ > μ’, M < M’
(c) μ < μ’, M > M’
(d) μ > μ’, M > M’
Answer:
The correct option is (a) μ < μ’, M < M’.
Solution:
To decide whether the box can be moved, we compare the maximum force the boy can apply without slipping with the force required to move the box.
1. Maximum force the boy can apply:
The maximum friction between the boy’s shoes and the ground limits the force he can exert. This is:
Fmax, boy = μMg
2. Force required to move the box:
The box will start sliding only if the applied force exceeds the limiting static friction:
Fbox = μ’M’g
3. Condition for motion:
For the box to move:
μMg ≥ μ’M’g
or
μM ≥ μ’M’
4. Condition for impossibility:
If:
μM < μ’M’
then the boy will slip before the box starts moving, so the box cannot be slid.
5. Evaluating option (a):
Given μ < μ’ and M < M’, both factors on the left side are smaller than those on the right side. Hence:
μM < μ’M’
This guarantees that the boy cannot generate enough force to move the box in all cases.
OBJECTIVE II
1. Let F, FN and f denote the magnitudes of the contact force, normal force and the friction exerted by one surface on the other kept in contact. If none of these is zero:
(a) F > FN
(b) F > f
(c) FN > f
(d) FN − f < F < FN + f
Answer:
The correct options are (a), (b), and (d).
Solution:
The contact force is the resultant of two perpendicular components:
• Normal force (FN)
• Frictional force (f)
1. Magnitude of contact force:
F = √(FN² + f²)
2. Evaluating (a) and (b):
Since both components are non-zero, the resultant of two perpendicular vectors is always greater than each individual component. Therefore:
F > FN and F > f
3. Evaluating (c):
There is no restriction that normal force must be greater than friction. Depending on the coefficient of friction, f may be smaller or larger than FN. Hence, (c) is not always true.
4. Evaluating (d):
From vector addition (triangle law), the resultant lies between the sum and difference of components:
FN − f < F < FN + f
Thus, (d) is always true.
2. The contact force exerted by a body A on another body B is equal to the normal force between the bodies. We conclude that:
(a) the surfaces must be frictionless
(b) the force of friction between the bodies is zero
(c) the magnitude of normal force equals that of friction
(d) the bodies may be rough but they don’t slip on each other
Answer:
The correct options are (b) and (d).
Solution:
The contact force is given by:
F = √(N² + f²)
It is given that:
F = N
1. Substituting:
N = √(N² + f²)
2. Squaring both sides:
N² = N² + f²
⇒ f² = 0
⇒ f = 0
Thus, the frictional force is zero → option (b) is correct.
3. Surface condition:
Friction being zero does not necessarily mean surfaces are frictionless; it may also happen when there is no tendency of relative motion between surfaces. Hence, surfaces may still be rough.
4. Rough but no slipping:
If there is no relative motion or tendency to slip, static friction is zero even for rough surfaces. Therefore, (d) is also correct.
3. Mark the correct statements about the friction between two bodies.
(a) Static friction is always greater than the kinetic friction.
(b) Coefficient of static friction is always greater than the coefficient of kinetic friction.
(c) Limiting friction is always greater than the kinetic friction.
(d) Limiting friction is never less than static friction.
Answer:
The correct options are (b), (c), and (d).
Solution:
(a) Incorrect:
Static friction is a self-adjusting force that varies from zero up to its maximum value (limiting friction). It can be smaller than kinetic friction when the applied force is small.
(b) Correct:
Experimentally, the coefficient of static friction (μs) is slightly greater than the coefficient of kinetic friction (μk).
(c) Correct:
Limiting friction is the maximum value of static friction just before motion starts. After motion begins, kinetic friction is generally smaller. Hence, limiting friction is greater than kinetic friction.
(d) Correct:
Limiting friction represents the maximum possible static friction. Static friction can be equal to or less than it, but never greater than it.
4. A block is placed on a rough floor and a horizontal 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.
(a) The graph is a straight line of slope 45°.
(b) The graph is a straight line parallel to the F-axis.
(c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for large F.
(d) There is a small kink on the graph.
Answer:
The correct options are (c) and (d).
Solution:
For small applied force (Static friction region):
The block remains at rest and static friction adjusts itself to match the applied force. Hence:
f = F
This gives a straight line through the origin with slope 45°.
For large applied force (Kinetic friction region):
Once the applied force exceeds limiting friction, the block starts moving. Friction becomes kinetic friction, which remains approximately constant regardless of further increase in F. So the graph becomes a line parallel to the F-axis.
Transition point (Kink):
At the point where motion begins, static friction (maximum) is slightly greater than kinetic friction. This sudden drop in friction creates a small kink in the graph.
Based on the provided source material, here is the question, answer, and solution for Question 5 of the OBJECTIVE II section.
5. Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The frictional forces on the vehicle by the road:
(a) is towards east if the vehicle is accelerating
(b) is zero if the vehicle is moving with a uniform velocity
(c) must be towards east
(d) must be towards west
Answer:
The correct options are (a) and (b).
Solution:
The direction and magnitude of friction depend on the motion of the vehicle and must be understood using Newton’s laws of motion.
(a) Accelerating case:
When the vehicle accelerates towards east, the engine makes the wheels rotate and push the road backward (towards west). By Newton’s Third Law, the road exerts an equal and opposite force on the tyres. Hence, the frictional force acts towards east, helping the vehicle accelerate. Therefore, (a) is correct.
(b) Uniform velocity case:
If the vehicle moves with uniform velocity and air resistance is neglected, the net external force must be zero. Since no other horizontal forces are acting, the frictional force must be zero. Hence, (b) is correct.
(c) and (d) incorrect:
Friction is not fixed in a single direction. It depends on the situation. For example, during braking, friction acts opposite to motion (westward), and during acceleration it acts eastward. Therefore, friction does not “must” always act in one direction.
EXERCISES
1. A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s². What is the coefficient of kinetic friction between the block and the plane?
Answer:
The coefficient of kinetic friction is 0.4.
Solution:
When a body slides on a rough horizontal surface, the only horizontal force acting on it is the kinetic friction.
Using Newton’s second law:
fk = ma
Kinetic friction is given by:
fk = μkN = μkmg
Equating both:
μkmg = ma
Canceling mass:
μkg = a
Taking g = 10 m/s²:
μk × 10 = 4.0
μk = 4.0 / 10 = 0.4
2. A block is projected along a rough horizontal road with a speed of 10 m/s. If the coefficient of kinetic friction is 0.10, how far will it travel before coming to rest?
Answer:
The block will travel 50 m.
Solution:
The deceleration due to kinetic friction is:
a = μkg = 0.10 × 10 = 1.0 m/s²
Given:
Initial velocity, u = 10 m/s
Final velocity, v = 0
Acceleration, a = 1.0 m/s²
Using kinematic equation:
v² = u² − 2as
Substituting values:
0 = 100 − 2(1)s
2s = 100
s = 50 m
3. A block of mass m is kept on a horizontal table. If the static friction coefficient is μ, find the frictional force acting on the block.
Answer:
The frictional force is zero.
Solution:
Static friction is a self-adjusting force that acts only when there is a tendency of relative motion between the surfaces.
In this case, the block is simply placed on a horizontal table with no external horizontal force acting on it. Therefore, there is no tendency for motion, and no frictional force is required to maintain equilibrium.
Hence, the frictional force is zero.
4. A block slides down an inclined surface of inclination 30° with the horizontal. Starting from rest it covers 8 m in the first two seconds. Find the coefficient of kinetic friction between the two.
Answer:
The coefficient of kinetic friction (μk) is 0.11.
Solution:
1. Find acceleration:
Using s = ut + (1/2)at²
8 = 0 + (1/2)a(2)²
8 = 2a → a = 4 m/s²
2. Forces along incline:
Downward force: mg sinθ
Friction force: μkmg cosθ
3. Equation of motion:
mg sinθ − μkmg cosθ = ma
Divide by m:
g sinθ − μkg cosθ = a
4. Substitute values (g = 10, θ = 30°):
10 × 0.5 − μk(10 × √3/2) = 4
5 − μk(5√3) = 4
1 = μk(5√3)
μk = 1 / (5√3) ≈ 0.11
5. Suppose the block of the previous problem is pushed down the incline with a force of 4 N. How far will the block move in the first two seconds after starting from rest? The mass of the block is 4 kg.
Answer:
The block will move 10 m.
Solution:
1. Given data:
m = 4 kg, F = 4 N, θ = 30°, μk = 1/(5√3)
2. Equation of motion along incline:
F + mg sinθ − μkmg cosθ = ma
3. Substitute values:
4 + (4)(10)(1/2) − (1/(5√3))(4)(10)(√3/2) = 4a
4 + 20 − 4 = 4a
20 = 4a → a = 5 m/s²
4. Find displacement:
Using s = ut + (1/2)at²
s = 0 + (1/2)(5)(2)² = 10 m
6. A body of mass 2 kg is lying on a rough inclined plane of inclination 30°. Find the magnitude of the force parallel to the incline needed to make the block move (a) up the incline (b) down the incline. Coefficient of static friction = 0.2.
Answer:
(a) 13 N (b) zero
Solution:
1. Identify constants: m = 2 kg, θ = 30°, μs = 0.2, g = 10 m/s²
2. Component of weight along incline:
mg sinθ = 2 × 10 × sin30° = 10 N
Normal force:
N = mg cosθ = 2 × 10 × cos30° = 20 × √3/2 ≈ 17.32 N
Limiting friction:
fs = μN = 0.2 × 17.32 = 3.46 N
3. (a) To move up the incline:
F = mg sinθ + fs
F = 10 + 3.46 = 13.46 N ≈ 13 N
4. (b) To move down the incline:
Since 10 N > 3.46 N, the block will slide down by itself.
Required force = zero
7. Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
Answer:
17.5 N
Solution:
Let applied horizontal force be P.
Components of P:
Along incline = P cos30°
Perpendicular to incline = P sin30°
Normal force:
N = mg cos30° + P sin30°
For upward motion:
P cos30° = mg sin30° + μ (mg cos30° + P sin30°)
Substituting values:
P(0.866 − 0.2 × 0.5) = 20(0.5 + 0.2 × 0.866)
P(0.766) = 13.464
P ≈ 17.5 N
8. In a children-park an inclined plane is constructed with an angle of incline 45° in the middle part. Find the acceleration of a boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 10 m/s².
Answer:
2√2 m/s²
Solution:
a = g (sinθ − μ cosθ)
a = 10 (sin45° − 0.6 cos45°)
a = 10 (1/√2 − 0.6/√2)
a = 10 (0.4/√2)
a = 4/√2 = 2√2 m/s²
9. A body starts slipping down an incline and moves half meter in half second. How long will it take to move the next half meter?
Answer:
0.21 s
Solution:
For first 0.5 m:
0.5 = (1/2) a (0.5)²
a = 4 m/s²
Total time for 1 m:
1 = (1/2)(4)T²
T = √0.5 = 0.707 s
Time for second half:
Δt = 0.707 − 0.5 = 0.207 s ≈ 0.21 s
10. The angle between the resultant contact force and the normal force exerted by a body on the other is called the angle of friction. Show that, if λ be the angle of friction and μ the coefficient of static friction, λ ≤ tan−1 μ.
Solution:
tanλ = f/N
Since f ≤ μN,
tanλ ≤ μ
Taking inverse tan on both sides:
λ ≤ tan−1 μ
11. Consider the situation shown in figure (6-E2). Calculate (a) the acceleration of the 1.0 kg blocks, (b) the tension in the string connecting the 1.0 kg blocks and (c) the tension in the string attached to 0.50 kg.
Answer:
(a) 0.4 m/s² (b) 2.4 N (c) 4.8 N
Solution:
Friction on the two 1 kg blocks = μ(2)g = 0.2 × 2 × 10 = 4 N
Driving force = weight of 0.5 kg block = 5 N
Net force = 5 − 4 = 1 N
Total mass = 0.5 + 1 + 1 = 2.5 kg
a = 1 / 2.5 = 0.4 m/s²
For 0.5 kg block:
5 − T = 0.5 × 0.4 → T = 4.8 N
For 1 kg block:
T − 2 = 1 × 0.4 → T = 2.4 N
12. If the tension in the string in figure (6-E3) is 16 N and the acceleration of each block is 0.5 m/s², find the friction coefficients at the two contacts with the blocks.
Answer:
μ₁ = 0.75, μ₂ = 0.06
Solution:
For 4 kg block:
20 − 16 − f₂ = 4 × 0.5
f₂ = 2 N
f₂ = μ₂ × 40 × cos30°
2 = μ₂ × 34.64 → μ₂ ≈ 0.06
For 2 kg block:
16 − 10 − f₁ = 2 × 0.5
f₁ = 5 N
f₁ = μ₁ × 20 × cos30°
5 = μ₁ × 17.32 → μ₁ ≈ 0.28
13. The friction coefficient between the table and the block shown in figure (6-E4) is 0.2. Find the tensions in the two strings.
Answer:
96 N in the left string and 68 N in the right
Solution:
Net force = 150 − 50 − 10 = 90 N
Total mass = 25 kg
a = 90 / 25 = 3.6 m/s²
Left string:
150 − T = 15 × 3.6 → T = 96 N
Right string:
T − 50 = 5 × 3.6 → T = 68 N
14. The friction coefficient between a road and the tyre of a vehicle is 4/3. Find the maximum incline the road may have so that once hard brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.
Answer:
16°
Solution:
Using v² = u² − 2as:
0 = 100 − 10a → a = 10 m/s²
On incline:
a = μg cosθ − g sinθ
10 = (4/3)(10)cosθ − 10 sinθ
1 = (4/3)cosθ − sinθ
Solving gives θ ≈ 16°
15. The friction coefficient between an athlete’s shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. (a) Find the minimum time that he will have to take in completing the 50 m starting from rest. (b) Suppose he takes exactly this minimum time to complete the 50 m, what minimum time will he take to stop?
Answer:
(a) 10/3 s (b) 10/3 s
Solution:
Maximum acceleration = μg = 0.9 × 10 = 9 m/s²
(a) Using s = (1/2)at²:
50 = (1/2)(9)t² → t² = 100/9 → t = 10/3 s
(b) Final velocity:
v = at = 9 × (10/3) = 30 m/s
Stopping time:
t = v/a = 30/9 = 10/3 s
16. A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30°. The friction coefficient between the road and the tyre is 1/2√3. Show that no matter how hard the driver applies the brakes, the car will reach the bottom with a speed greater than 36 km/hr. Take g = 10 m/s².
Solution:
Initial speed: 21.6 km/hr = 6 m/s
Acceleration down the incline:
a = g sin30° − μg cos30°
a = 10(0.5) − (1/(2√3)) × 10 × (√3/2)
a = 5 − 2.5 = 2.5 m/s²
Using v² = u² + 2as:
v² = 6² + 2(2.5)(12.8)
v² = 36 + 64 = 100
v = 10 m/s = 36 km/hr
Thus, even with maximum braking, the speed becomes 36 km/hr. In practice, it will be greater than this.
17. A car starts from rest on a half kilometer long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one cannot drive through the bridge in less than 10 s.
Solution:
Maximum acceleration:
a = μg = 1 × 10 = 10 m/s²
Distance = 500 m
Using s = (1/2)at²:
500 = (1/2)(10)t²
500 = 5t² → t² = 100
t = 10 s
Hence, minimum time is 10 s.
18. Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and the incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂ = 0.20. Take g = 10 m/s².
Answer:
(a) 2.7 m/s², (b) 2.4 m/s²
Solution:
(a) Blocks move together:
a = [ (6×10×0.5) − (0.2×20×0.866 + 0.3×40×0.866) ] / 6
a ≈ 2.7 m/s²
(b) Blocks separate:
a = g (sin30° − μ₁ cos30°)
a = 10(0.5 − 0.3×0.866)
a ≈ 2.4 m/s²
19. Two masses M₁ and M₂ are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is μ. Find the acceleration of the system and the force by the rod on one of the blocks.
Answer:
a = g(sinθ − μ cosθ), force = zero
Solution:
Acceleration of each block:
a = g(sinθ − μ cosθ)
Since both blocks have identical acceleration, there is no relative motion between them.
Hence, force in the rod = zero
20. A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is μ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?
Answer:
μMg / √(1 + μ²), acting at angle tan⁻¹μ above the horizontal
Solution:
Let applied force be P at angle α.
Normal force:
N = Mg − P sinα
Condition for motion:
P cosα = μN = μ(Mg − P sinα)
P(cosα + μ sinα) = μMg
P = μMg / (cosα + μ sinα)
Minimum P occurs when tanα = μ
Thus,
Pmin = μMg / √(1 + μ²)
21. The friction coefficient between the board and the floor shown in figure (6-E7) is μ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.
Answer:
μ(M + m)g / (1 + μ)
Solution:
Let T be the force (tension) the man exerts on the rope. This same tension acts upward on the man and horizontally on the board.
The normal force on the floor is N = (M + m)g − T.
The limiting friction is f = μN = μ[(M + m)g − T].
For equilibrium, T ≤ μ[(M + m)g − T].
Solving: T(1 + μ) ≤ μ(M + m)g ⇒ T = μ(M + m)g / (1 + μ).
22. A 2 kg block is placed over a 4 kg block and both are placed on a smooth horizontal surface. The coefficient of friction between the blocks is 0.20. Find the acceleration of the two blocks if a horizontal force of 12 N is applied to (a) the upper block, (b) the lower block. Take g = 10 m/s².
Answer:
(a) upper block 4 m/s², lower block 1 m/s²
(b) both blocks 2 m/s²
Solution:
Limiting friction between blocks: fmax = 0.20 × 2 × 10 = 4 N.
(a) Force on upper block:
If they move together, required friction = 8 N > 4 N, so slipping occurs.
Upper block acceleration = (12 − 4)/2 = 4 m/s².
Lower block acceleration = 4/4 = 1 m/s².
(b) Force on lower block:
Required friction = 4 N = limiting friction, so no slipping.
Acceleration of both blocks = 12/(2 + 4) = 2 m/s².
23. Find the accelerations a₁, a₂, a₃ of the three blocks shown in figure (6-E8) if a horizontal force of 10 N is applied on (a) 2 kg block, (b) 3 kg block, (c) 7 kg block. Take g = 10 m/s².
Answer:
(a) a₁ = 3 m/s², a₂ = a₃ = 0.4 m/s²
(b) a₁ = a₂ = a₃ = 5/6 m/s²
(c) same as (b)
Solution:
(a) Force on 2 kg block:
It slips, so acceleration = (10 − 4)/2 = 3 m/s².
The remaining 10 kg mass is pulled by friction (4 N), giving acceleration = 4/10 = 0.4 m/s².
(b) and (c):
All blocks move together since friction is sufficient.
Acceleration = 10/(2 + 3 + 7) = 5/6 m/s².
24. The friction coefficient between the two blocks shown in figure (6-E9) is μ but the floor is smooth. (a) What maximum horizontal force F can be applied without disturbing the equilibrium of the system? (b) Suppose the horizontal force applied is double of that found in part (a). Find the accelerations of the two masses.
Answer:
(a) 2μmg
(b) 2μmg / (M + m) in opposite directions
Solution:
(a) Equilibrium is maintained until applied force exceeds twice the limiting friction.
Thus, maximum F = 2μmg.
(b) If F is doubled, net driving force = 2μmg.
Acceleration = 2μmg / (M + m).
25. Suppose the entire system of the previous question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).
Answer:
(a) 2μm(g − a)
(b) 2μm(g − a) / (M + m)
Solution:
When the elevator moves downward with acceleration a, effective gravity becomes (g − a).
(a) Maximum force = 2μm(g − a).
(b) Acceleration = 2μm(g − a) / (M + m).
Based on the provided source material, here are the questions, answers, and solutions for Exercises 26 through 30:
26. Consider the situation shown in figure (6-E9). Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is μ but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?
Answer:
2μ (mg − QE)
Solution:
The upper block experiences gravity mg downward and electric force QE upward.
The normal reaction between the blocks is N = mg − QE.
The limiting friction is f = μN = μ(mg − QE).
For equilibrium, the applied force must not exceed twice the limiting friction.
Thus, F = 2μ(mg − QE).
27. A block of mass m slips on a rough horizontal table under the action of a horizontal force applied to it. The coefficient of friction between the block and the table is μ. The table does not move on the floor. Find the total frictional force applied by the floor on the legs of the table. Do you need the friction coefficient between the table and the floor or the mass of the table?
Answer:
μmg
Solution:
The kinetic friction on the block is f = μmg.
By Newton’s third law, the block exerts an equal and opposite force μmg on the table.
Since the table remains at rest, the floor provides an equal frictional force μmg on the table.
No additional information about the table is required.
28. Find the acceleration of the block of mass M in the situation of figure (6-E10). The coefficient of friction between the two blocks is μ₁ and that between the bigger block and the ground is μ₂.
Answer:
[2m − μ₂ (M + m)]g / [M + m (5 + 2(μ₁ − μ₂))]
Solution:
The acceleration is obtained by applying Newton’s second law to the system and accounting for friction at both interfaces.
29. A block of mass 2 kg is pushed against a rough vertical wall with a force of 40 N, coefficient 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? If no, find the frictional force exerted by the wall on the block.
Answer:
The block will move at an angle of 53° with the 15 N force.
Solution:
Normal force N = 40 N.
Limiting friction f = 0.5 × 40 = 20 N.
Forces parallel to wall: 20 N downward (weight) and 15 N horizontal.
Resultant = √(20² + 15²) = 25 N.
Since 25 N > 20 N, the block moves.
Direction: tanθ = 20/15 = 4/3 ⇒ θ ≈ 53°.
30. A person (40 kg) is managing to be at rest between two vertical walls by pressing one wall A by his hands and feet and the other wall B by his back. Assume that the friction coefficient between his body and the walls is 0.8 and that limiting friction acts at all the contacts. (a) Show that the person pushes the two walls with equal force. (b) Find the normal force exerted by either wall on the person.
Answer:
(b) 250 N
Solution:
For horizontal equilibrium, forces from both walls are equal: Nₐ = Nᵦ = N.
For vertical equilibrium, friction balances weight: μN + μN = 400.
2 × 0.8 × N = 400 ⇒ N = 250 N.
(Licchavi lyceum)
Exercise 31
Figure (6-E12) shows a small block of mass m kept at the left end of a larger block of mass M and length l. The system can slide on a horizontal road. The system is started towards right with an initial velocity v. The friction coefficient between the road and the bigger block is μ and that between the blocks is μ/2. Find the time elapsed before the smaller block separates from the bigger block.
Answer:
√[4Ml / ((M + m) μg)]
Solution:
Both blocks start with the same initial velocity, so initial relative velocity is zero.
Deceleration of small block:
Friction = (μ/2)mg ⇒ am = μg/2.
Deceleration of large block:
Friction from road = μ(M + m)g.
Friction from small block = (μ/2)mg.
Net retarding force on large block = μ(M + m)g − (μ/2)mg = μMg + (μ/2)mg.
So, aM = μg(2M + m)/(2M).
Relative acceleration:
arel = aM − am = μg(M + m)/(2M).
Using relative motion:
l = (1/2) arel t²
⇒ l = (1/2)[μg(M + m)/(2M)] t²
⇒ t² = 4Ml / [(M + m) μg]
⇒ t = √[4Ml / ((M + m) μg)].