
Chapter 6 – Friction
Page No 95:
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?
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?
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 (see the figure).
Question 3:
What is the average friction force when a person has a usual 1 km walk?
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?
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?
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.
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.
Page No 96:
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?
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?
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?
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?
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.
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 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:
(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
and
N = mg ï¼ Fcosθ
where m is the mass of the body
F is the contact force acting on the body
If we decrease the angle between this contact force and the vertical, then Fcosθ increases and the normal reaction force (N) as well as the frictional force (f) decrease.
Question 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:
(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 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 = μ Mg
(c) Mg ≤ F ≤ Mg
1+μ2(d) Mg ≥ F ≥ Mg
1μ2
Answer:
(c) Mg ≤ F ≤ Mg
1+μ2Let T be the force applied on an object of mass M.
If T = 0, F_{min} = Mg.
If T is acting in the horizontal direction, then the body is not moving.
∴
T=μ(mg) Fmax =(Mg)2+(T)2 =(Mg)2+(μMg)2Thus, we have:
Mg≤F≤Mg1+(μ)2
Question 4:
A scooter starting from rest moves with a constant acceleration for a time âˆ†t_{1}, then with a constant velocity for the next âˆ†t_{2} and finally with a constant deceleration for the next âˆ†t_{3} 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 âˆ†t_{1} and âˆ†t_{3} and 500 N for âˆ†t_{2}_{.}
Answer:
d) >500 N for time âˆ†t_{1} and âˆ†t_{3} and 500 N for âˆ†t_{2}_{.}
During the time interval âˆ†t_{2}, 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 âˆ†t_{1} and âˆ†t_{3}, 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 5:
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
Figure
(a) is upward
(b) is downward
(c) is zero
(d) the system cannot remain in equilibrium.
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 6:
Suppose all the surface in the previous problem are rough. The direction of friction of B due to A
(a) is upward
(b) is downward
(c) is zero
(d) depends on the masses of A and B.
Answer:
(a) is upward
The normal reaction force on the system (comprising of wall and contact surface of A and 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 f_{A} and f_{BA} (friction on B due to A) is in upward direction.
Question 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:
(c) is same for both cars
Given: both the cars have same initial speed.
Let the masses of the two cars be m_{1} and m_{2}_{.}
Frictional force on car with mass m_{1} = μm_{1} g
So, the deceleration due to frictional force =
μm1gm1=μgFrictional force on car with mass m_{2} = μm_{2} g
So, the deceleration due to frictional force =
μm2gm2=μgAs both the acceleration are same, from the second equation of motion
s=ut+12at2Thus, we can say that both the cars have same minimum stopping distance.
Question 8:
In order to stop a car in 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:
(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 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 of three are possible.
Answer:
(d) all of three are possible.
We know that
N = mg cos θËš
f_{max} = μN = μmg cos θ
where N = normal reaction force
f_{max} = frictional force
θ = angle of inclination
μ = coefficient of friction
When the block just begins to slide, it means
mg sin θ = f_{max}
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 − f_{max} = ma
mg sin 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.
Question 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:
(a) μ < μ’, M < M‘
Let T be the force applied by the boy on the block.
Free body diagram for the box:
The condition for preventing the slide is
f_{max} > T
μ’M’g > T (i)
Now see the free body diagram of a boy of mass M:
f_{max} = μmg
The condition for preventing the slide is
f_{max} > T
μmg > T
The condition for sliding the entire system (block and boy) is
f’ > f (block is not slide)
μ’M’g > μmg
μ’M‘ > μm
μ < μ’
m < M’
Page No 97:
Question 1:
Let F, F_{N} 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 > F_{N}
(b) F > f
(c) F_{N} > f
(d) F_{N} − f < F < F_{N} + f.
Answer:
(a) F > F_{N}
(b) F > f
(d) F_{N} − f < F < F_{N} + f
The system is in equilibrium condition when F = f.
Hence, the net horizontal force is zero.
f = μF_{N}
F > F_{N}
f = F_{N} and 0 ≤ μ ≤ 1
Therefore, we can say that F > f. So the net horizontal force is nonzero.
F > f, and so the net horizontal force is zero.
F_{N} > f
⇒F_{N} > μF_{N}
⇒μ < 1
Here, the given relation between F and f i.e
F > f and f = μF_{N}_{ } will not be satisfied So it cannot be said that the net horizontal force is zero or nonzero.
F_{N} − f < F < F_{N} + f
âˆµ f = μF_{N}
fμf<F<fμ+ff1μμ<F<f1+μμFor the above relation, we can say that F ≠ f and so the net horizontal force is nonzero.
Question 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 surface must be frictionless
(b) the force of friction between the bodies is zero
(c) the magnitude of normal force equal that of friction
(d) the bodies may be rough but they don’t slip on each other.
Answer:
(b) the force of friction between the bodies is zero
(d) the bodies may be rough but they don’t slip on each other
The contact force exerted by a body A on another body B is equal to the normal force between the bodies. Therefore, we can conclude that the force of friction between the bodies is zero or the bodies may be rough but they don’t slip on each other.
Question 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:
(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 the static friction.
All the above statements are correct. The static friction is sometimes less than the kinetic friction.
Question 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 Faxis.
(c) The graph is a straight line of slope 45° for small F and a straight line parallel to the Faxis for large F.
(d) There is a small kink on the graph.
Answer:
(c) The graph is a straight line of slope 45° for small F and a straight line parallel to the Faxis for large F.
(d) There is a small kink on the graph.
When force F is applied on the block, the force of friction f comes into play. As we increase the applied F, the static friction force adjusts itself to become (equal) to the applied force F and goes upto its maximum value equal to limiting friction force.After this ,it is treated as a constant force (i.e . now its value does not change until and unless the body starts moving). If the applied force F is greater than the limiting friction force, then the kinetic friction force comes into play at that time. The kinetic friction force is always less than the limiting friction force.
Question 5:
Consider a vehicle going on a horizontal road towards east. Neglect any force by the air. The frictional force 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:
(a) is towards east if the vehicle is accelerating
(b) is zero if the vehicle is moving with a uniform velocity
When the vehicle is accelerating, the force is applied (by the tyre on the road) in west direction .That causes a net resultant frictional force acting in east direction. Due to this force of friction only ,the car is moving in east direction.
When the vehicle is moving with a uniform velocity, the force of friction on the wheels of the vehicle by the road is zero.
Question 1:
A body slipping on a rough horizontal plane moves with a deceleration of 4.0 m/s^{2}. What is the coefficient of kinetic friction between the block and the plane?
Answer:
Let m be the mass of the body.
From the free body diagram,
R − mg = 0
(where R is the normal reaction force and g is the acceleration due to gravity)
⇒ R = mg (1)
Again ma − μ_{k}R = 0
(where μ_{k} is the coefficient of kinetic friction and a is deceleration)
or ma = μ_{k}R
From Equation (1),
ma = μ_{k}mg
⇒ a = μ_{k}g
⇒ 4 = μ_{k}g
⇒μk=4g=410=0.4Hence, the coefficient of the kinetic friction between the block and the plane is 0.4.
Question 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:
Friction force acting on the block will decelerate it.
Let the deceleration be ‘a‘.
Using free body diagram
R − mg = 0
(where R is the normal reaction force)
⇒ R = mg (1)
Again, ma − μ_{k}R = 0
(where μ_{k }is the coefficient of kinetic friction)
From Equation (1),
⇒ ma = μ_{k}mg
⇒ a = μ_{k}g = 0.1 × 10
= 1 m/s^{2}
Given:
initial velocity, u = 10 m/s
final velocity, v = 0 m/s (block comes to rest)
a = −1 m/s^{2} (deceleration)
Using equation of motion v^{2}
–u^{2} = 2as
(where s is the distance travelled before coming to rest)
s=v2u22aOn substituting the respective values, we get
=01022 1 =1002=50 mTherefore, the block will travel 50 m before coming to rest.
Question 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:
A block of mass m is kept on a horizontal table. If force is applied on the block, a friction force will be there: p → frictional force and F → applied force
So, friction force is equal to the applied force. One of the case is that the friction force is equals to zero when the applied force is equal to zero.
Question 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:
Free body diagram for the block is as follows:
From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ (1)
For the block, u = 0 m/s, s = 8 m and t = 2 s.
According to the equation of motion
s=ut+12at2
s=0+12a228=2aa = 4 m/s^{2}
Again,
μ_{k}R + ma − mg sin θ = 0
(where μ_{k }is the coefficient of kinetic friction)
From Equation (1):
μ_{k}mg cos θ + ma − mg sin θ = 0
⇒ m (μ_{k}g cos θ + a − g sin θ) = 0
⇒ μ_{k} × 10 × cos 30° = g sin 30° − a
⇒μk×1032=10×124⇒53μk=1⇒μk=153=0.11Therefore, the coefficient of kinetic friction between the block and the surface is 0.11.
Question 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:
Free body diagram of the block for this case is as follows:
From the adove diagram:
F − ma − μ_{k}R + mg sin 30° = 0
4 − 4a − μ_{k}R + 4g sin 30° = 0 (1)
R − 4g cos 30° = 0 (2)
⇒ R = 4g cos 30° = 0
Substituting the values of R in Equation (1) we get
4 − 4a − 0.11 × 4g cos 30° + 4g sin 30° = 0
⇒44a0.11×4×10×32+4×10×12=04 − 4a − 3.81 + 20 = 0
4 − 4a − 3.18 + 20 = 0
a ≈ 5 m/s^{2}
For the block, u = 0, t = 2 s and a = 5 m/s^{2}.
According to the equation of motion,
â€‹ â€‹
s=ut+12at2
=0+125×22=10 mTherefore, the block will move 10 m.
Question 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) To make the block move up the incline, the applied force should be equal and opposite to the net force acting down the incline.
Applied force = μR + 2g sin 30° (1)
(where μ is the coefficient of static friction)
R = mg cos 30°
Substituting the respective values in Equation (1), we get
=0.2×9.83+2×9.8×123.39 + 9.8
≈13 N
With this minimum force, the body moves up the incline with a constant velocity as the net force on it is zero.
(b) Net force acting down the incline is given by
F = 2g sin 30° − μR
=2×9.8×123.99=6.41 NBecause F = 6.41 N, the body will move down the incline with acceleration, hence the force required is zero.
Question 7:
Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
Answer:
Using the free body diagram,
g = 10 m/s^{2}, m = 2 kg, θ = 30 and μ = 0.2
R − mg cos θ − F sin θ = 0
⇒ R = mg cos θ + F sin θ (1)
and
mg sin θ + μR − F cos θ = 0
⇒ mg sin θ + μ(mg cos θ + F sin θ) − F cos θ = 0
⇒ mg sin θ + μmg cos θ + μF sin θ − F cos θ = 0
⇒F=mg sin θ+μ mg cos θμ sin θcos θ (θ=30°) =2×10×12+0.2×2×10×320.2×1232 =13.4640.76=17.7 N=17.5 NTherefore, while pushing the block to move up on the incline, the required force is 17.5 N.
Question 8:
In a childrenpark an inclined plane is constructed with an angle of incline 45° in the middle part (figure 6−E1). Find the acceleration of boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 19 m/s^{2}.
Figure
Answer:
Let m be the mass of the boy.
From the above diagram:
R − mg cos 45° = 0
R=mg cos 45°=mg2 1Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°
=m×10×120.6×m×10×12=m5232=m×2×2The acceleration of the boy
=ForceMass
=m22m=22 m/s2
Question 9:
A body starts slipping down an incline and moves half metre in half second. How long will it take to move the next half metre?
Answer:
Let a be the acceleration of the body sliding down.
From the above diagram:
R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
ma + mg sin θ − μR = 0
⇒
a=mg(sinθμ cosθ)m=g(sinθμ cosθ)For the first half metre, u = 0, s = 0.5 m and t = 0.5 s.
According to the equation of motion,
v = u + at
= 0 + (0.5)4 = 2 m/s
s=ut+12at2
0.5=0+12a0.52⇒a=4 m/s2For the next half metre, u = 2 m/s, a = 4 m/s^{2} and s = 0.5.
⇒0.5=2t+124t2⇒ 2t^{2} + 2t − 0.5 = 0
⇒ 4t^{2} + 4t − 1 = 0
⇒t=4±16+162×4 =1.6568=0.2027Therefore, the time taken to cover the next half metre is 0.21 s.
Question 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.
Answer:
Let
f be the applied force,
R be the normal reaction force and
F be the frictional force.
The coefficient of static friction is given by
u=tan λ=FR(where λ is the angle of friction)
When F = μR, F is the limiting friction (maximum friction). When applied force increases and the body still remains still static then the force of friction increases up to its maximum value equal to limiting friction (μR).
F<μR∴tan λ=FR≤μRR⇒ tan λ ≤ μ
⇒ λ ≤ tan^{−1} μ
Question 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.
Figure
Answer:
From the above diagrams:
T + ma − mg = 0
T + 0.5a − 0.5 g = 0 (1)
μR + ma + T_{1} − T = 0
μR + 1a + T_{1} − T = 0 (2)
μR + 1a − T_{1} = 0
μR + a = T_{1} (3)
From Equations (2) and (3) we have
μR + a = T − T_{1}
⇒ T − T_{1} = T_{1}
⇒ T = 2T_{1}
So, Equation (2) becomes
μR + a + T_{1} − 2T_{1} = 0
⇒ μR + a − T_{1} = 0
⇒ T_{1} = μR + a
= 0.2g + a (4)
and Equation (1) becomes
2T_{1} + 0.5a − 0.5g = 0
⇒T1=0.5g0.5a2 =0.25g0.25a (5)From Equations (4) and (5)
0.2g + a = 0.25g − 0.25a
⇒a=0.051.25×10 =0.4×10 m/s2 g=10 m/s2Therefore,
(a) the acceleration of each 1 kg block is 0.4 m/s^{2},
(b) the tension in the string connecting the 1 kg blocks is
T_{1} = 0.2g + a + 0.4 = 2.4 N
â€‹ and
(c) the tension in the string attached to the 0.5 kg block is
T = 0.5g − 0.5a
= 0.5 × 10 − 0.5 × 0.4
= 4.8 N.
Page No 98:
Question 12:
If the tension in the string in figure (6−E3) is 16 N and the acceleration of each block is 0.5 m/s^{2}, find the friction coefficients at the two contact with the blocks.
Figure
Answer:
From the free body diagram:
μ_{1}R + m_{1}a − F = 0
μ_{1}R + 1 − 16 = 0 (R = mg cos θ)
⇒ μ_{1}(2g) + (−15) = 0
μ1=1520=0.75Again,
μ_{2}R_{1} + ma = F − mg sin θ = 0
μ_{2}R_{1} + 4 × 0.5 = 16 − 4g sin 30° = 0
R_{1} = mg cos θ (θ = 30°)
⇒μ2203+2+1620=0⇒μ2=2203=117.32 =0.057=0.06Therefore, the friction coefficients at the two contacts with blocks are
μ_{1} = 0.75 and μ_{2} = 0.06.
Question 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.
Figure
Answer:
Consider that a 15 kg object is moving downward with an acceleration a.
From the above diagram,
T + m_{1}a − m_{1}g = 0
T + 15a − 15g = 0
⇒ T = 15g − 15a (1)
Now,
T_{1} − m_{2}g − m_{2}a = 0
T_{1} − 5g − 5a = 0
⇒ T_{1} = 5g + 5a (2)
Again,
T − (T_{1} + 5a + m_{2}R) = 0
⇒ T − (5g + 5a + 5a +m_{2}R) = 0 (3)
(where R = μg)
From Equations (1) and (2),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90 [g = 10 m/s^{2}]
⇒ a = 3.6 m/s^{2}
From Equation (3),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.
From Equation (2),
T_{1} = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
= 68 N in the right string.
Question 14:
The friction coefficient between a road and the type of a vehicle is 4/3. Find the maximum incline the road may have so that once had 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:
Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0
s = 5 m,
μ=43, g = 10 m/s^{2}
Let the maximum angle of incline be
θ.
Using the equation of motion
a=v2u22s=01022×5 =10 m/s2From the free body diagram
R − mg cos θ = 0
⇒ R = mg cos θ (1)
Again,
ma + mg sin θ − μ R = 0
⇒ ma + mg sin θ − μmg cos θ = 0
⇒ a + g sin θ − μg cos θ = 0
⇒10+10 sin θ43×10 cos θ=0⇒ 30 + 30 sin θ − 40 cos θ = 0
⇒ 3 + 3 sin θ − 4 cos θ = 0
⇒ 4 cos θ − 3 sin θ = 3
⇒41sin2 θ=3+3 sin θOn squaring, we get
16 (1 − sin^{2} θ) = 9 + 9 sin^{2} θ + 18 sin θ
25 sin^{2} θ + 18 sin θ − 7 = 0
⇒sin θ=18+182425 72×25 =18+3250=1450=0.28 Taking positve sign only⇒θ=sin1 0.28=16°Therefore, the maximum incline of the road, θ = 16°.
Question 15:
The friction coefficient between an athelete’s shoes and the ground is 0.90. Suppose a superman wears these shoes and races for 50 m. There is no upper limit on his capacity of running at high speeds. (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:
To reach the 50 m distance in minimum time, the superman has to move with maximum possible acceleration.
Suppose the maximum acceleration required is ‘a‘.
∴ ma − μR = 0 ⇒ ma = μ mg
⇒ a = μg = 0.9 × 10 = 9 m/s^{2}
(a) As per the question, the initial velocity,
u = 0, t = ?
a = 9 m/s^{2}, s = 50 m
From the equation of motion,
s=ut+12at2
50=0+129t2⇒t=103 s(b) After covering 50 m, the velocity of the athelete is
v = u + at
=0+9×103 m/s=30 m/sThe superman has to stop in minimum time. So, the deceleration, a = − 9 m/s^{2} (max)
R = mg
ma = μR (maximum frictional force)
ma = μmg
⇒ a = μg
= 9 m/s^{2} (deceleration)
u_{1} = 30 m/s, v = 0
⇒t=v1u1a =030a =30a=103 s
Question 16:
A care is going at a speed of 21.6 km/hr when it encounters at 12.8 m long slope of angle 30° (figure 6−E5). The friction coefficient between the road and the tyre is
1/23. 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/s^{2}.
Figure
Answer:
When the driver applies hard brakes, it signifies that maximum force of friction is developed between the tyres of the car and the road.
So, maximum frictional force = μR
From the free body diagram,
R − mg cos θ = 0
⇒ R = mg cos θ (1)
and
μR + ma − mg sin θ = 0 (2)
⇒ μ mg cos θ + ma − mg sin θ = 0
where θ = 30Ëš
⇒μg cos θ+a10×12=0⇒a=5123×1032 =510×34 =20410×34 =104 =2.5 m/s2s = 12.8 m
u = 6 m/s
∴ Velocity at the end of incline
ν=u2+2as =62+22.5 12.8 =36+64 =10 m/s=36 km/hTherefore, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h.
Question 17:
A car starts from rest on a half kilometre 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.
Answer:
Let a be the maximum acceleration of the car for crossing the bridge.
From the above diagram,
ma = μR
(For more accelerations the tyres will slip)
ma = μmg
a = μg = 1 × 10 = 10 m/s^{2}
To cross the bridge in minimum possible time, the car must be at its maximum acceleration.
u = 0, s = 500 m, a = 10 m/s^{2}
From the equation of motion,
s=ut+12at2Substituting respective values
500=0+1210t2⇒t=100=10 sTherefore, if the car’s acceleration is less than 10 m/s^{2}, it will take more than 10 s to cross the bridge. So, one cannot drive through the bridge in less than 10 s.
Question 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 μ_{1}, and that between the block of mass 4.0 kg and incline is μ_{2}. Calculate the acceleration of the 2.0 kg block if (a) μ_{1} = 0.20 and μ_{2} = 0.30, (b) μ_{1} = 0.30 and μ_{2} = 0.20. Take g = 10 m/s^{2}.
Figure
Answer:
(a) From the free body diagram
R = 4g cos 30°
⇒R=4×10×32 =203 N 1μ_{2}R + m_{1}a − p − m_{1}g sin θ = 0
μ_{2}R + 4a − p − 4g sin 30° = 0
⇒ 0.3
× (40) cos 30° + 4a − p − 40 sin 30° = 20 (2)
R_{1} = 2g cos 30°
=103 (3)
p + 2a − μ_{1}R_{1} − 2g sin 30° = 0 (4)
From Equation (2),
63+4ap20=0From Equation (4),
p+2a+2310=1063+6a+30+23=0⇒6a=3083 =3013.85=16.15⇒a=16.156 =2.69=2.7 m/s2(b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that a = 2.4 m/s^{2}.
Question 19:
Two masses M_{1} and M_{2} 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:
From the free body diagram
R_{1} = M_{1}g cos θ (1)
R_{2} = M_{2}g cos θ (2)
T + M_{1}g sin θ − M_{1}a − μR_{1} = 0 (3)
T − M_{2}g + M_{2}a + μR_{2} = 0 (4)
From Equation (3),
T + M_{1}g sin θ − M_{1} a − μM_{1}g cos θ = 0 (5)
From Equation (4),
T − M_{2} g sin θ + M_{2} a + μM_{2} g cos θ = 0 (6)
From Equations (5) and (6),
g sin θ(M_{1} + M_{2}) − a(M_{1} + M_{2}) − μg cos θ(M_{1} + M_{2})
⇒ a(M_{1} + M_{2}) = g sin θ(M_{1} + M_{2}) = μg cos θ(M_{1} + M_{2})
⇒ a = g(sin θ − μ cos θ)a − g(sin θ − μ cos θ)
∴ The acceleration of the block (system) = g(sin θ − μcos θ)
The force exerted by the rod on one of the blocks is tension, T.
T = −M_{1}g sin θ + M_{1}a + μM_{1}g cos θ
T = −M_{1}g sin θ + M_{1}(g sin θ − μg cos θ) + μM_{1}g cos θ = 0
Question 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:
Let P be the force applied to slide the block at an angle θ.
From the free body diagram,
R + P sin θ − mg = 0
⇒ R = −P sin θ + mg (1)
μR = P cos θ (2)
From Equation (1),
μ(mg − P sin θ)−P cos θ = 0
⇒ μmg = μP sin θ + P cos θ
⇒P=μ mgμ sin θ+cos θThe applied force P should be minimum, when μ sin θ + cos θ is maximum.
Again, μ sin θ + cos θ is maximum when its derivative is zero:
ddθμ sin θ+cos θ=0⇒ μ cos θ − sin θ = 0
θ = tan^{−1} μ
So,
P=μ mgμ sin θ+cos θDividing numerator and denominator by cos θ, we get
=μ mg/cos θμ sin θcos θ+cos θcos θ P=μ mg sec θ1+μ tan θ =μ mg sec θ1+tan2 θ=μ mgsec θ =μ mg1+tan2 θ=μ mg1+μ2(using the property
1+tan2θ=sec2θ)
Therefore, the minimum force required is
μ mg1+μ2 at an angle θ = tan^{−1} μ.
Question 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.
Figure
Answer:
Let T be the maximum force exerted by the man on the rope.
From the free body diagram,
R + T = Mg
⇒ R = Mg − T (1)
Again,
R_{1} − R − mg = 0
⇒ R_{1} = R + mg (2)
and
T − μR_{1} = 0
From Equation (2),
T − μ(R + mg) = 0
⇒ T − μR − μ mg = 0
⇒ T − μ(Mg − T) − μmg = 0
T − μMg + μt − μmg = 0
⇒ T (1 + μ) = μMg + μmg
⇒T=μ M+mg1+μTherefore, the maximum force exerted by the man is
μM+mg1+μ.
Question 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 block 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/s^{2}.
Answer:
Consider the free body diagram.
(a) For the mass of 2 kg, we have:
R_{1} − 2g = 0
⇒ R_{1} = 2 × 10 = 20
2a + 0.2 R_{1} − 12 = 0
⇒ 2a + 0.2 (20) = 12
⇒ 2a = 12 − 4
⇒ a = 4 m/s^{2}
Now,
4a − μR_{1} = 0
⇒ 4a = μR_{1} = 0.2 (20) = 4
⇒ a_{1} = 1 m/s^{2}
The 2 kg block has acceleration 4 m/s^{2} and the 4 kg block has acceleration 1 m/s^{2}.
(ii) We have:
R_{1} = 2g = 20
Ma = μR_{1} = 0
a = 0
And,
Ma + μmg − F = 0
4a + 0.2 × 2 × 10 − 12 = 0
⇒ 4a + 4 = 12
⇒ 4a = 8
⇒ a = 2 m/s^{2}
Question 23:
Find the accelerations a_{1}, a_{2}, a_{3} 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/s^{2}.
Figure
Answer:
Given:
μ_{1} = 0.2
μ_{2} = 0.3
μ_{3} = 0.4
Using the free body diagram, we have:
(a) When the 10 N force is applied to the 2 kg block, it experiences maximum frictional force.
Here,
μ_{1}R_{1} = μ_{1} × m_{1}g
μ_{1}R_{1} = μ_{1} × 2g = (0.2) × 20
= 4 N (From the 3 kg block)
Net force experienced by the 2 kg block = 10 − 4 = 6 N
∴
a1=62=3 m/s2But for the 3 kg block (Fig. 3), the frictional force from the 2 kg block, i.e, 4 N, becomes the driving force and the maximum frictional force between the 3 kg and 7 kg blocks.
Thus, we have:
μ_{2} = R_{2} = μ_{2}m_{2}g = (0.3) × 5 kg
= 15 N
Therefore, the 3 kg block cannot move relative to the 7 kg block.
The 3 kg block and the 7 kg block have the same acceleration (a_{2} = a_{3}), which is due to the 4 N force because there is no friction from the floor.
∴a2=a3=410=0.4 m/s2(b) When the 10 N force is applied to the 3 kg block, it experiences maximum frictional force of (15 + 4) N, i.e., 19 N, from the 2 kg block and the 7 kg block.
So, it cannot move with respect to them.
As the floor is frictionless, all the three bodies move together.
∴a1=a2=a3=1012=56 m/s2(c) Similarly, it can be proved that when the 10 N force is applied to the 7 kg block, all three blocks move together with the same acceleration.
∴ a1=a2=a3=56 m/s2
Question 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.
Figure
Answer:
From the free body diagrams of the two blocks, we have
R_{1} = mg …(i)
F = μR_{1}+T …(ii)
T − μR_{1} = 0 ..(iii)
From equations (i) and (ii), we have
F − μmg = T …(ii)
From equations (i) and (iii), we have
T = μmg
Putting T = μmg in equation (ii), we have
F = μmg + μmg = 2μmg
(b) From the free body diagram of upper block, we have
2F − T − μmg = ma ….(i)
From the free body diagram of lower block, we have
T = Ma + μmg
Putting the value of T in (i), we get
2F − Ma − μmg − μmg = ma
Putting F = 2μmg, we get
2(2μmg) − 2μmg = a(M + m)
⇒ 4μmg − 2μmg = a(M + m)
⇒a=2 μmgM+m in opposite directions.
Page No 99:
Question 25:
Suppose the entire system of the previous questions is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).
Answer:
From the free body diagram, we have:
R_{1} + ma − mg = 0
⇒ R_{1} = m (g − a)
= mg − ma …(i)
Now,
F − T − μR_{1} = 0 and
T − μR_{1} = 0
⇒ F − [μ (mg − ma)] − μ(mg − ma) = 0
⇒ F − μ mg − μma − μmg + μma = 0
⇒ F = 2 μmg − 2 μma
= 2 μm (g − a)
(b) Let the acceleration of the blocks be a_{1}.
R_{1} = mg − ma ….(i)
And,
2F −T − μR_{1} = ma_{1} …(ii)
Now,
T = μR_{1} + Ma_{1}
= μmg − μma + Ma_{1}
Substituting the value of F and T in equation (ii), we get:
2[2μm(g − a)] − (μmg − μma + Ma_{1}) − μmg + μma = ma_{1}
⇒ 4μmg − 4μma − 2μmg + 2μma= ma_{1} + Ma_{1}
⇒a1=2μm gaM+mThus, both the blocks move with same acceleration a_{1} but in opposite directions.
Question 26:
Consider the situation shown in figure (6−E9). Suppose a small electric field E exists in the space in the vertically 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?
[Hint: The force on a charge Q bye the electric field E is F = QE in the direction of E.]
Answer:
From the free body diagram:
R_{1} + QE − mg = 0
where
R_{1} is the normal reaction force
Q is the charge
E is the small electric field
⇒ R_{1} = mg − QE (1)
F − T − μR_{1} = 0
F − T = μR_{1}
where F is the maximum horizontal force required
From Equation (1),
F − T − μ(mg − QE) = 0
F − T = μ(mg − QE)
⇒ F − T − μmg + μQE = 0 (2)
T − μR_{1} = 0
⇒ T = μR_{1} = μ (mg − QE) (3)
From Equation (2),
F − μmg + μ QE − μmg + μQE = 0
⇒ F − 2 μmg + 2μQE = 0
⇒ F = 2μmg − 2μQE
⇒ F = 2μ (mg − QE)
Therefore, the maximum horizontal force that can be applied is 2μ (mg − QE).
Question 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?
Figure
Answer:
When a block slips on a rough horizontal table, the maximum frictional force acting on it can be found from the free body diagram (see below).
R = mg
(i) (ii)
From the free body diagram,
F − μR = 0
⇒ F = μR = μmg
But the table is at rest, so the frictional force at the legs of the table is also μR. Let this be f, so from the free body diagram
f − μR = 0
⇒ f = μR = μmg
Therefore, the total frictional force on the table by the floor is μmg.
Question 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 μ_{1} and that between the bigger block and the ground is μ_{2}.
Answer:
Let us the acceleration of the block of mass M be a and let it be towards right. Therefore, the block of mass m must go down with acceleration 2a. As both the blocks are in contact, it (block of mass m) will also have acceleration a towards right. Hence, it will experience two inertial forces as shown in the free body diagram given below.
(Free body diagram 1)
From the free body diagram 1, we have:
(Free body diagram2)
R_{1} − ma = 0
⇒ R_{1} = ma ….(i)
Again,
2ma + T − Mg + μ_{1}R_{1} = 0
⇒ T = Mg − (2 + μ_{1}) ma ….(ii)
Using the free body diagram 2, we have:
T + μ_{1}R_{1} + Mg − R_{2} = 0
Substituting the value of R_{1} from (i), we get:
R_{2} = T + μ_{1} ma + mg
Substituting the value of T from (ii), we get:
R_{2} = (Mg − 2ma − μ_{1}ma) = μ_{1} ma + Mg + ma
∴ R_{2} = Mg + Ma − 2ma ….(iii)
Again using the free body diagrams −2,
T + T − R − Ma − μ_{2}R_{2} = 0
⇒ 2T − Ma − ma − μ2(Mg + mg − 2ma) = 0
Substituting the values of R_{1} and R_{2} from (i) and (iii), we get:
2T = (M + m)a + μ_{2} (Mg + mg − 2ma) ….(iv)
From equations (ii) and (iv), we have:
2T = 2mg − 2(2 + μ_{1}) ma
= (M + m) a + μ_{1}(Mg + mg − 2ma)
⇒ 2mg − μ_{2} (M + m)g = a[M + m − 2μ_{2}m + 4m + 2μ_{1}m]
Therefore, the acceleration of the block of mass M in the given situation is given by
a=2m+m2M+mgM+m5+2μ1μ2
Question 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:
Net force on the block
=202+1520.5×40 = 25 − 20 = 5 N
∴tan θ=2015=43⇒θ=tan143=53°Therefore, the block will move at 53° angle with the 15 N force.
Question 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 (figure 6−E11). 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 wall with equal force. (b) Find the normal force exerted by either wall on the person. Take g = 10 m/s^{2}.
Figure
Answer:
Thus, we have:
μR + μR = mg
⇒ 2μR = 40 × 10
⇒R=40×102×0.8=250 NNormal force = 250 N
Question 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 block is μ/2. Find the time elapsed before the smaller blocks separates from the bigger block.
Figure
Answer:
Let a_{1} and a_{2} be the accelerations of masses m and M, respectively.
Also, a_{1} > a_{2} so that mass m moves on mass M.
Let after time t, mass m is separated from mass M.
Using the equation of motion
During this time, mass m covers
vt+12a1t2and
sm=vt+12a2t2.
For mass m to separate from mass M, we have:
vt+12a1t2=vt+12a2t2+1 ….ii
From the free body diagram, we have:
ma1+μ2R=0⇒ma1=μ2mg=μ2m×10⇒a1=5μAgain,
Ma2+μM+mgμ2mg=0⇒ 2Ma_{2} + 2μ (M + m)g − μmg = 0
⇒ 2Ma_{2} = μmg − 2μmg − 2μmg
⇒a2=μmg2μMg2MSubstituting the values of a_{1} and a_{2} in equation (i), we get:
t=4MlM+mμg
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity