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Concepts of Physics
Exercises
Friction
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?
Solution:
Deceleration of the body: a = 4 m/s2
Frictional force on the object: f = ma = 4m newton
Frictional force on the object: f = μN = μmg
4m = μmg ⇒ μ = 4
𝑔=
4
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?
Solution:
Initial speed of the block: u = 10 m/s
Coefficient of friction: μ = 0.1
Final speed of the block: v = 0 m/s
Deceleration of the block: a = μg = 0.1 x 10 = 1 m/s2
Distance travelled by the block before coming to rest: s = 𝑣2−𝑢2
2𝑎=
0−100
2 𝑥 (−1)= 50 m
Friction
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.
Solution:
Mass of the block: m
Coefficient of static friction: μ
External force acting on the block: F = 0
As there is no external force on the block, the block does not have any tendency to move.
As there is no tendency of motion (impending relative motion), there is no frictional (opposing) force.
Friction
4. A block slides down an inclined surface of inclination 30𝑜 with the horizontal. Starting from rest it covers 8 m in the
first two seconds. Find the coefficient of kinetic friction between the two.
Solution:
Angle of inclination of the plane: θ = 300
Initial speed of the block: u = 0 m/s
Distance covered by the block: s = 8 m
Time taken by the block: t = 2 sec
Acceleration of the block down the rough inclined plane: a = g (sin θ – μ cos θ)
a = 10 sin 30 − 𝜇 cos 30 = 10 1
2− 𝜇
3
2
Distance covered by the block: s = 1
2at2 ⇒ 8 = 0.5 x 10
1
2− 𝜇
3
2x 4
8
20=
1
2− 𝜇
3
2⇒ 𝜇
3
2= 0.5 – 0.4 ⇒ μ =
0.2
1.732= 0.11
Friction
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.
Solution:
External force on the block: F1 = 4 N (down the incline)
Sum of gravitational and frictional forces on the block: F2 = mg (sin θ – μ cos θ)
F2 = 4 x 10 (sin 30 −0.2
3cos 30) = 40
1
2−
0.2
3
3
2= 40 (0.5 – 0.1) = 40 x 0.4 = 16 N
Resultant force on the block: F = F1 + F2 = 4 + 16 = 20 N
Acceleration of the block: a = 𝐹
𝑚=
20
4= 5 m/s2
Distance travelled by the block: s = 1
2at2 = 0.5 x 5 x 4 = 10 m
Friction
6. A body of mass 2 kg is lying on a rough inclined plane of inclination 30𝑜. 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.
Solution:
Mass of the body: m = 2 kg; Angle of inclination: θ = 300
Coefficient of friction: μ = 0.2
Force required to move the body up the inclined plane: F = mg (sin θ + μ cos θ)
F = 2 x 9.8 (sin 30 + 0.2 cos 30) = 19.6 (0.5 + 0.2 x 0.86) =13 N
Gravitational force on the block: mg sin θ = 2 x 9.8 x sin 30 = 9.8 N (down the inclined plane)
Frictional force on the block: f = μN = μmg cos θ = 0.2 x 2 x 9.8 x cos 30 = 3.4 N (up the inclined plane)
As gravitational force is greater than the frictional force, the block slides down the incline without the help of any
external force.
Friction
7. Repeat part (a) of problem 6 if the push is applied horizontally and not parallel to the incline.
Solution:
Component of applied force along the inclined plane: F cos θ
Component of applied force perpendicular to inclined plane: F sin θ
Forces perpendicular to the plane: N = mg cos θ + F sin θ
Frictional force on the block: f = μN = μ (mg cos θ + F sin θ)
For moving the block up the plane: F cos θ = mg sin θ + f = mg sin θ + μ (mg cos θ + F sin θ)
F cos θ = mg sin θ + μmg cos θ + μF sin θ
F cos 𝜃 − 𝜇 sin 𝜃 = mg sin θ + μmg cos θ
F = mg (sin θ + μ cos θ)
cos 𝜃−𝜇 sin 𝜃=
2 𝑥 9.8 (0.5+0.2 𝑥 0.86)
0.86 −0.2𝑥0.5=
19.6 𝑥 0.672
0.76= 17.4 N
Friction
8. In a children-park an inclined plane is constructed with an angle of incline 45𝑜 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.
Solution:
Acceleration due to gravity: g = 10 m/s2
Angle of inclination: θ = 450
Coefficient of friction: μ = 0.6
Acceleration of an object down the inclined plane: a = g (sin θ – μ cos θ)
a = 10 (sin 45 – 0.6 cos 45) = 10 1
2−
3
5
1
2=
10
21 −
3
5=
10
2x
2
5= 2 𝟐 m/s2
Friction
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?
Solution:
Acceleration of body down the inclined plane: a = g sin θ
Displacement of the body: s = 1
2at2 ⇒ 0.5 = 0.5 g sin θ (0.5)2 ⇒ g sin θ =
1
0.25= 4
Velocity of the body: v = at = g sin θ t = 4 x 0.5 = 2 m/s
Displacement of the body: s = vt + 1
2at2 = vt +
1
2g sin θ t2
0.5 = 2t + 0.5 x 4t2 ⇒ 2t2 + 2t – 0.5 = 0 ⇒ t = −1
2+
1
2= −0.5 + 0.71 = 0.21 sec
Friction
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−1μ.
Solution:
For equilibrium of the body: R cos λ = N (vertical) ---- (1)
For equilibrium of the body: R sin λ = f (horizontal) ---- (2)
(2)/(1): tan λ = 𝑓
𝑁= 𝜇 ⇒ λ = tan-1 𝜇
Since the body is stationary (static friction): λ ≤ tan-1 𝝁
Friction
11. Consider the situation shown in figure. 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.
Solution:
For the hanging block: mg – T1 = ma ⇒ 0.5 x 10 – T1 = 0.5a ⇒ 5 – T1 = 0.5a ---- (1)
For the 1 kg blocks: T1 – f = ma ⇒ T1 − 2μmg = 2ma ⇒ T1 − 2 x 0.2 x 1 x 10 = 2 x 1 x a ⇒ T1 – 4 = 2a ---- (2)
(1) + (2): 5 – 4 = 2.5a ⇒ a = 1
2.5= 0.4 m/s2
for the first 1 kg block: T2 – f = ma ⇒ T2 - μmg = ma ⇒ T2 – 0.2 x 1 x 10 = 1 x 0.4 ⇒ T2 = 2.4 N
From (1): 5 – T1 = 0.5 x 0.4 ⇒ T1 = 5 – 0.2 = 4.8 N
Friction
12. If the tension in the string in figure is 16 N and the acceleration of each block is 0.5 m/s2, find the friction
coefficient at the two contacts with the blocks.
Solution:
For the 4 kg block: mg sin θ – T – f = ma ⇒ mg sin θ – T – μ2mg cos θ = ma
4 x 10 sin 30 – 16 – μ2 x 4 x 10 x cos 30 = 4 x 0.5
20 – 16 - 20 3 μ2 = 2 ⇒ 20 3 μ2 = 2 ⇒ μ2 = 1
10 3= 0.06
For the 2 kg block: T – f = ma ⇒ T – μ1mg = ma
16 – 2 x 10 μ1 = 2 x 0.5 ⇒ 16 – 20 μ1 = 1 ⇒ μ1 = 15
20= 0.75
Friction
13. The friction coefficient between the table and the block shown in figure is 0.2. Find the tensions in the two strings.
Solution:
For the 15 kg block: mg – T1 = ma ⇒ 15 x 10 – T1 = 15a ⇒ 150 – T1 = 15a ---- (1)
For the 5 kg block (hanging): T2 – mg = ma ⇒ T2 – 5 x 10 = 5a ⇒ T2 – 50 = 5a ---- (2)
For the 5 kg block (on the table): T1 – T2 – f = ma ⇒ T1 – T2 – μmg = ma
T1 – T2 – 0.2 x 5 x 9.8 = 5a ⇒ T1 – T2 – 10 = 5a ---- (3)
(1) + (2) + (3): a = 3.6 m/s2
From (1): 150 – T1 = 15 x 3.6 ⇒ T1 = 96 N
From (2): T2 – 50 = 5 x 3.6 ⇒ T2 = 68 N
Friction
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.
Solution:
Coefficient of friction between the road and the tyre: μ = 4
3
Speed of the vehicle down the road: v = 36 kmph = 36 x 5
18= 10 m/s
Distance travelled by the vehicle: s = 5 m
Acceleration of the vehicle down the road: a = −g (sin θ – μ cos θ) = −10 (sin θ −4
3cos θ)
Distance travelled: s = 𝑣2−𝑢2
2𝑎=
100 −0
2 𝑥 10 sin 𝜃−4
3cos 𝜃
= 5
sin 𝜃 −4
3cos 𝜃 = 1 ⇒ sin θ = 1 +
4
3cos 𝜃 ⇒ 1 − 𝑐𝑜𝑠2 𝜃 = 1 +
4
3cos 𝜃
1 − 𝑐𝑜𝑠2 𝜃 = 1 + 16
9cos2 θ +
8
3cos 𝜃 ⇒ −
25
9cos2 θ =
8
3cos 𝜃 ⇒ cos 𝜃 = −
24
25= 160
Friction
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. 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?
Solution:
Coefficient of friction between the shoe and the ground: μ = 0.9
Distance travelled: s = 50 m
Maximum acceleration provided by the shoe: a = μg = 0.9 x 10 = 9 m/s2
(a) Time taken to cover the given distance: s = 1
2at2 = 0.5 x 9t2 = 4.5 t2
50 = 4.5t2 ⇒ t2 = 100
9⇒ t =
𝟏𝟎
𝟑sec
(b) Time taken to stop will be same as the acceleration and deceleration are same.
Friction
16. A car is going at a speed of 21.6 km/hr when it encounters a 12.8 m long slope of angle 30𝑜. 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 of the car: u = 21.6 kmph
Length of the inclined plane: s = 12.8 m
Angle of inclination: θ = 300
Coefficient of friction between the road and the tyres: μ = 1
2 3
Acceleration of the car down the inclined plane: a = g (sin θ – μ cos θ) = 10 (sin 30 −1
2 3cos 30)
a = 10 1
2−
1
4=
10
4= 2.5 m/s2
Final speed of the car: v2 = u2 + 2as = 36 + 2 x 2.5 x 12.8 = 100 ⇒ v = 10 m/s = 10 x 18
5= 36 kmph
Friction
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:
Length of the bridge: s = 0.5 km = 500 m
Coefficient of friction: μ = 1
Acceleration of the car: a = μg = 1 x 10 = 10 m/s2
Time taken to cross the bridge: s = 1
2at2
500 = 0.5 x 10t2 ⇒ t2 = 100 ⇒ t = 10 sec
Friction
18. Figure shows two blocks in contact sliding down an inclined surface of inclination 30o. 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
Solution:
Acceleration of block down the inclined plane: a = g (sin θ – μ cos θ)
(a) Acceleration of the 4 kg block: 10 (sin 30 – 0.3 cos 30) = 10 1
2− 0.3
3
2= 5 (1 – 0.3 3) = 2.4 m/s2
Acceleration of the 2 kg block: 10 (sin 30 – 0.2 cos 30) = 10 1
2− 0.2
3
2= 5 (1 – 0.2 3) = 3.3 m/s2
Since the acceleration of the 4 kg block is less, the blocks move together.
a = 𝐹𝑛𝑒𝑡
𝑀=
4+2 x 10 x sin 30−0.3 𝑥 4 𝑥 10 cos 30−0.2 𝑥 2 𝑥 10 𝑥 cos 30
4+2=
30−103
2(1.2+0.4)
6= 2.7 m/s2
Friction
Solution:
Acceleration of block down the inclined plane: a = g (sin θ – μ cos θ)
(b) Acceleration of the 4 kg block: 10 (sin 30 – 0.2 cos 30) = 10 1
2− 0.2
3
2= 5 (1 – 0.2 3) = 3.3 m/s2
Acceleration of the 2 kg block: 10 (sin 30 – 0.3 cos 30) = 10 1
2− 0.3
3
2= 5 (1 – 0.3 3) = 2.4 m/s2
Since the acceleration of the 4 kg block (3.3 m/s2) is greater than that of 2 kg block (2.4 m/s2), the 2 kg block moves
with an acceleration of 2.4 m/s2
Friction
19. Two masses M1 and M2 are connected by a light rod and the system is slipping down a rough incline of angle θwith 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.
Solution:
For the block of mass M1:
Perpendicular to the plane: N – mg cos θ = m (0) ⇒ N = mg cos θ
Along the inclined plane: mg sin θ – f = ma
mg sin θ – μN = ma ⇒ mg sin θ – μmg cos θ = ma ⇒ ma = mg (sin θ – μ cos θ)
Acceleration of a block down the rough inclined plane: a = g (sin θ – μ cos θ)
Since both the blocks have same acceleration, the force exerted by the rod on the blocks is zero
Friction
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?
Solution:
Let a force of ‘F’ is applied on the block at an angle θ to the horizontal
For equilibrium in vertical direction: N + F sin θ = Mg
The block will move in horizontal direction: F cos θ ≥ f
Limiting condition: F cos θ = f = μN = μ (Mg – F sin θ) ⇒ F (cos θ + μ sin θ) = μMg ⇒ F = 𝜇𝑀𝑔
cos 𝜃 +𝜇 sin 𝜃
For the force to be minimum, denominator is maximum: 𝑑
𝑑𝜃(cos 𝜃 + 𝜇 sin 𝜃) = 0
−sin θ + μ cos θ = 0 ⇒ μ cos θ = sin θ ⇒ μ = tan θ ⇒ θ = tan-1 (μ)
Minimum force: Fmin = 𝜇𝑀𝑔
1
1+𝜇2+ 𝜇
𝜇
1+𝜇2
= 𝜇𝑀𝑔
1+𝜇2
1+𝜇2
= 𝝁𝑴𝒈
𝟏+𝝁𝟐
Friction
21. The friction coefficient between the board and the floor shown in figure is 𝜇. Find the maximum force that the man
can exert on the rope so that the board does not slip on the floor.
Solution:
Let the applied force be ‘F’
From the FBD of the man: Mg – N1 – F = M(0) ⇒ Mg = N1 + F ---- (1)
From the FBD of the board:
Vertical direction: N2 – mg – N1 = m (0) ⇒ N2 = mg + N1 = mg + Mg – F ---- (2)
Horizontal direction: F – f = m (0) ⇒ F = f = μN2
F = μ(mg + Mg – F) ⇒ F (1 + μ) = μg (m + M) ⇒ F = 𝝁𝒈(𝒎 + 𝑴)
𝟏 + 𝝁
Friction
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.
Solution:
Coefficient of friction between the blocks: μ = 0.2
Limiting friction between the blocks: f = μN = μmg = 0.2 x 2 x 10 = 4 N
(a) 12 N force is applied on the upper block
Net force on the upper block: Fnet = 12 – 4 = 8 N
Acceleration of the upper block: a1 = 𝐹𝑛𝑒𝑡
𝑚1=
8
2= 4 m/s2
Acceleration of the lower block: a2 = 𝐹𝑛𝑒𝑡
𝑚2=
4
4= 1 m/s2
(b) 12 N force is applied on the lower block
Common acceleration of the blocks: a = 𝐹
𝑀=
12
6= 2 m/s2
Pseudo force on 2 kg block: Fp = ma = 2 x 2 = 4 N
Since the pseudo force is equal to limiting friction, there is no relative slipping between the blocks
Friction
23. Find the accelerations a1, a2, a3 of the three blocks shown in figure 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.
Solution:
Limiting friction between 2 kg and 3 kg blocks: f12 = μ1m1g = 0.2 x 2 x 10 = 4 N
Limiting friction between 3 kg and 7 kg blocks: f23 = μ2 (m1 + m2)g = 0.3 (2 + 3) 10 = 15 N
(a) 10 N force is applied on 2 kg block:
Since the applied force is greater than limiting friction, the block will move
Net force on the 2 kg block: Fnet = 10 – 4 = 6 N
Acceleration of the 2 kg block: a1 = 𝐹𝑛𝑒𝑡
𝑚1=
6
2= 3 m/s2
Net force on the 3 kg block: Fnet = 4 N
Since the floor is frictionless and the force on 3 kg block (4 N) is less than frictional force between 3 kg and 7 kg
blocks (15 N), both these blocks move together.
Acceleration of the 3 kg and 7 kg blocks: a2 = a3 = 𝐹𝑛𝑒𝑡
𝑚2=
4
10= 0.4 m/s2
Friction
Solution:
(b) 10 N force applied on 3 kg block
Since the applied force (10 N) is less than the limiting friction between 3 kg and 7 kg blocks (15 N), there is no relative
slipping between the blocks
Common acceleration of the blocks: a1 = a2 = a3 = 𝐹
𝑀=
10
2+3+7=
10
12=
𝟓
𝟔m/s2
(c) 10 N force applied on 7 kg block
Since the applied force (10 N) is less than the limiting friction between 3 kg and 7 kg blocks (15 N), there is no relative
slipping between the blocks
Common acceleration of the blocks: a1 = a2 = a3 = 𝐹
𝑀=
10
2+3+7=
10
12=
𝟓
𝟔m/s2
Friction
24. The friction coefficient between the two blocks shown in figure 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 acceleration of the two masses.
Solution:
Friction coefficient between the blocks: μ
(a) For the equilibrium of the smaller block:
Vertical direction: N1 – mg = m(0) ⇒ N1 = mg
Horizontal direction: F – T – f = m(0) ⇒ F = T + μN1 = T + μmg ---- (1)
For the equilibrium of the bigger block:
Vertical direction: N2 – (m + M)g = M(0) ⇒ N2 = (m + M)g
Horizontal direction: T – f = M(0) ⇒ T = f = μmg ---- (2)
From (1) & (2): F = μmg + μmg = 2 μmg
Friction
Solution:
Given: applied force is doubled
(b) Since the string is inextensible, both the blocks will have same acceleration in opposite directions
For the upper block: Fnet = ma
2F – T – f = ma ⇒ 2F – T – μmg = ma ---- (3)
For the lower block: Fnet = Ma
T – f = Ma ⇒ T = f + Ma = μmg + Ma ---- (4)
(3) + (4): 2F – μmg = μmg + Ma + ma
2F = 2 μmg + (M + m) a
2 (F – μmg) = (M + m) a ⇒ a = 2 (𝐹 − 𝜇𝑚𝑔)
(𝑀+𝑚)=
2 (2𝜇𝑚𝑔− 𝜇𝑚𝑔)
(𝑀+𝑚)=
𝟐𝝁𝒎𝒈
(𝑴+𝒎)
Friction
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)
Solution:
Given: the entire system is kept in an elevator which is moving down with an acceleration ‘a’
Since the elevator is accelerating, it is a non-inertial frame of reference
So, a pseudo force acts on the system in upward direction
Effective acceleration of the system: a1 = g – a
(a) Maximum horizontal force: F = 2 μm(g – a)
(b) Acceleration of the masses: 𝟐𝝁𝒎(𝒈−𝒂)
(𝑴+𝒎)
Friction
26. Consider the situation shown in figure. 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?
Solution:
Friction coefficient between the blocks: μ
(a) For the equilibrium of the smaller block:
Vertical direction: N1 +EQ – mg = m(0) ⇒ N1 = mg – EQ
Horizontal direction: F – T – f = m(0) ⇒ F = T + μN1 = T + μ (mg – EQ) ---- (1)
For the equilibrium of the bigger block:
Vertical direction: N2 – (m + M)g = M(0) ⇒ N2 = (m + M)g
Horizontal direction: T – f = M(0) ⇒ T = f = μ (mg – EQ) ---- (2)
From (1) & (2): F = μ (mg – EQ) + μ (mg – EQ) = 2μ (mg – EQ)
Friction
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?
Solution:
Limiting friction between the block and the table: f = μN = μmg
Since the block is moving, the applied force is greater than (equal to) limiting friction.
Direction of frictional force on the block: left
Direction of frictional force on the table top: right
Direction of frictional force on the legs of the table: left (since the table is stationary)
Magnitude of frictional force on the legs of the table: μmg (since the table is stationary)
Friction
28. Find the acceleration of the block of mass M in the situation of figure. The coefficient of friction between the two
blocks is μ1 and that between the bigger block and the ground is μ2.
Solution:
Let the tension in the string be T
a1: acceleration of smaller block; a2: acceleration of bigger block
Constraint equation:
σ ത𝑇. ത𝑎 = 0 ⇒ Ta2 cos 00 + Ta2 cos 00 + Ta1 cos 1800 = 0 ⇒ 2Ta2 = Ta1 ⇒ a1 = 2a2 ---- (1)
From the free body diagram of smaller block:
Horizontal direction: normal force = pseudo force = ma2 = R1
Frictional force on the block: f = μ1ma2
Vertical direction: mg – T – f = ma1 ⇒ mg – T − μ1ma2 = ma1
mg – T − μ1ma2 = m (2a2) ⇒ mg – T = ma2 (μ1 + 2) ⇒ T = mg − ma2 (μ1 + 2) ---- (2)
Friction
Solution:
From the free body diagram of bigger block:
Vertical direction: R – Mg – T – f = M(0) ⇒ R = Mg + T + μ1ma2
R = Mg + T + μ1ma2 = Mg + mg − ma2 (μ1 + 2) + μ1ma2 ⇒ R = Mg + mg − 2ma2 ---- (3)
Horizontal direction: 2T – f1 – R1 = Ma2 ⇒ 2T – μ2R – ma2 = Ma2
2(mg − ma2 (μ1 + 2)) − μ2 (Mg + mg − 2ma2) = a2 (m + M)
2mg – 2ma2μ1 – 4ma2 − μ2Mg − μ2mg + 2ma2μ2 = a2 (m + M)
a2 (2mμ2 – 4m – 2mμ1) + 2mg − μ2Mg − μ2mg = a2 (m + M)
[2m − μ2(M + m)]g = a2 [(m + M) – (2mμ2 – 4m – 2mμ1)]
[2m − μ2(M + m)]g = a2 [M + m(5 + 2(μ1 – μ2)] ⇒ a2 = 𝟐𝒎−𝝁𝟐 𝑴+𝒎 𝒈
𝑴+𝒎[𝟓+𝟐 𝝁𝟏−𝝁𝟐 ]
Friction
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.
Solution:
Pushing force on the block: 40 N
Normal force from the wall: R = 40 N
Limiting friction on the block: f = μR = 0.5 x 40 = 20 N
Let the 15 N force be acting on the block into the plane (horizontal & parallel to the wall)
Gravitational force on the block: Fg = mg = 2 x 10 = 20 N
Magnitude of net force on the block: F = 𝐹𝑔2 + 15 2 = 202 + 152 = 25 N
Direction of net force on the block: tan θ = 20
15=
4
3⇒ θ = tan-1 4
3= 530
So, the block moves in a direction making an angle 530 with 15 N force.
Friction
30. A person (40 kg) is managing to be at rest between two vertical walls by pressing on 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 frictions 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. Take g = 10 m/s2.
Solution:
(a) The person pushes the two walls with equal force.
Otherwise, there will be an unbalanced force in horizontal direction that gives rise to torque.
Since the person is in equilibrium between the walls, net force in horizontal direction is zero.
(b) For equilibrium of the person: frictional force = gravitational force
𝜇NB + μNA = mg ⇒ 2μN = mg ⇒ N = 𝑚𝑔
2𝜇=
40 𝑥 10
2 𝑥 0.8=
400
1.6= 250 newton
Friction
31. Figure 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 υ. 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.
Solution:
Let the smaller block fall off the bigger block in time ‘t’
Let a1 be the acceleration of smaller block and a2 be that of bigger block
Acceleration of the smaller block: a1 = −𝜇
2g
Acceleration of the bigger block: a2 = 𝐹𝑛𝑒𝑡
𝑀=
𝜇
2𝑚𝑔−𝜇 𝑀+𝑚 𝑔
𝑀=
−𝜇
2𝑚𝑔−𝜇𝑀𝑔
𝑀= −
𝜇𝑔
2𝑀𝑚 + 2𝑀
Distance covered by the smaller block: s = vt + 1
2a1t
2
Distance covered by the bigger block: S = vt + 1
2a2t
2
The smaller block will fall off the larger block: s = S + 𝑙
vt + 1
2a1t
2 = vt + 1
2a2t
2 + 𝑙 ⇒ −𝜇
4gt2 = −
𝜇𝑔
4𝑀𝑚 + 2𝑀 t2 + 𝑙
𝑔𝑡2𝜇
4
1
𝑀𝑚 + 2𝑀 − 1 = 𝑙 ⇒
𝑔𝑡2𝜇
4𝑀(m + 2M – M) = 𝑙 ⇒ t =
𝟒𝑴𝒍
(𝑴+𝒎)𝝁𝒈
Friction
