Problems
1. If cyclist moving with a speed of 4.9 m/s on a
level road can take a sharp circular turn of radius 4 m, then what is the
coefficient of friction between the cycle tyre and the road?
Solution:
To take a turn centripetal acceleration will be
provided by frictional force
∴
= μN
∴
= rμg ------ (1)
∴
μ = 
------ (2)
Here
radius of circular track,
r = 4 m
Velocity,
= 4.9 m/s
Gravitational
acceleration,
g = 9.8 m s−2
Substituting
above values in eqn (2) we get
μ = 0.6125
μ = 0.61
2. A motor cycle is going on an over bridge of
radius R. The driver maintains a constant
speed. As the motor cycle is ascending on the over bridge, what will be the
normal force on it?
Solution:
From
figure reaction of normal force on the motor cyclist
= 
∴ R = 
As motorcycle ascends on the over bridge, angle
of inclination of motorcycle with vertical θ
decreases and becomes zero at the highest point.
So cos θ
goes on increasing and becomes 1 at highest point.
Value of R
is continuously increasing while ascending.
3. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a
smooth horizontal table. If the particle moves in a circle with speed 
, then what will be the net force on the particle
(directed towards the centre)?
Solution:
When a particle connected to a string revolves in
a circular path around a centre, the centripetal force is provided by the
tension produced in the string.
Hence, in the given case, the net force on the
particle is the tension T, i.e.,
= T = 
Where
F is the net force acting on the
particle.
4. A stone of mass 0.25 kg tied to the end of a
string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min
in a horizontal plane. What is the maximum speed with which the stone can be
whirled around if the string can withstand a maximum tension of 200 N?
Solution:
Mass
of the stone, m = 0.25 kg
Radius
of the circle, r = 1.5 m
Maximum
tension in the string,
= 200 N
= 
∴ 
= 
= 
= 
=
34.64 m/s
Therefore,
the maximum speed of the stone is 34.64 m/s.
5. A weight W
hangs from a rope that is tied to two other ropes that are fastened to the ceiling
as shown in figure.
The upper ropes make angles θ and φ with
the horizontal. What are the values of T1 and
T2?
Solution:
For equilibrium,
Along the horizontal
direction
T1 cos
θ = T2
cos φ
T2 = 
------(1)
Along the vertical
direction
T1 sin
θ + T2 sin φ = W ------(2)
Substituting
the value of T2 in eqn.
(2)
T1 sin
θ + 
sin φ = W
T1
(sin θ cos φ + cos θ sin φ) = W
cos φ
T1 = 
Similarly, T2 = 
6. A mass of 5 kg is suspended in equilibrium, by
two light inextensible strings S1 and
S2, which make angle of
30° and 45° respectively with the horizontal. Then which of the following is
correct? (Take g = 10 m s−2)
(a) Tension in both the strings is same.
(b) Tension in S1 is
more than that in S2.
(c) Tension in S1
is less than that in S2.
(d) Sum of tension in both is equal to 50 N.
Solution:
From figure, we
concluded
T1 cos 30° = T2
cos 45°
T1 
= T2 
T2 = 
T2 > T1
∴ Tension in S1 is less than that in S2
7. A block of mass 200 kg is being pulled up by
men on an inclined plane at angle of 45° as shown in figure.
The coefficient of static friction is 0.5. Each man
can only apply a maximum force of 500 N. Find the minimum number of men
required to start moving the block up the plane.
Solution:
Mass of the block,
m = 200 kg
Coefficient of static
friction,
μs =
0.5
= 
Angle of incline plane,
θ = 45°
Maximum
force that each man be apply,
F
= 500 N
Let n
be the number of men required to start moving the block up the plane. Then,
nF =
mg sin θ + f
= mg sin θ + μsN
= mg sin θ + μs mg cos θ
= mg[sin θ + μs cos
θ]
= 200 × 10
= 200 × 10
= 
≃ 5
8. Which one of the following motions on a smooth
plane surface does not involve force?
(a) Accelerated motion in a straight line.
(b) Retarded motion in a straight line.
(c) Motion with constant momentum along a straight line.
(d) Motion along a straight line with varying velocity.
Solution:
Motion with constant momentum along a straight
line implies 
= constant.
So, according to Newton’s 2nd law
= 
= 
(constant)
= 0
∴ Motion with constant
momentum along a straight line is the correct statement.
9. The monkey B
is holding on to the tail of monkey A which is climbing up a rope. The masses
of the monkeys A and B are 5 kg and 2
kg respectively. If A can tolerate a tension of 30 N in its tail, what force
should it apply on the rope in order to carry the monkey B with it? (Take g = 10
m s−2).
Solution:
Let us make the free body diagram for both
monkeys separately. As the monkey A,
applies a force F downward on the
rope, the rope applies a force F
upward on the monkey A.
For A, F − 5g – T = 5a
------ (1)
For B, T
− 2g = 2a
------ (2)
Dividing the two eqn.
= 
2(F − 5g – T ) = 5(T − 2g)
2F − 10g – 2T
= 5T −
10g
2F
= 7T or
T = 
Now from eqn
(2), for a = 0
= 2g = 20 N; = 30 N is given
Hence, 20 N ≤
30 N
20 N ≤ 
≤ 30 N
(20 N) ≤ 
F ≤ (30 N)
70 N ≤ F ≤ 105 N
10. A gramophone record is revolving with an
angular velocity (ω). A coin is
placed at a distance r from the centre of the record. The
coefficient of static friction is μ.
The coin will revolve with the record if..?
Solution:
The coin will revolve with the record, if the
centripetal force ≤ the force of friction
mrω2 or
r ≤ 
11. Block A
of mass m and block B of mass 2 m are
placed on a fixed triangular wedge by means of a massless, inextensible string
and a frictionless pulley as shown in the figure. The wedge is inclined at 45°
to the horizontal on both the sides. If the coefficient of friction between the
block A and the wedge is 
and that
between the block B and the wedge is 
. Both A and B are released from rest, what will
be the acceleration of A?
Solution:
The free body diagram of the blocks is shown in
the figure
The equation of
motion of block B is
2mg sin 45° − μ2N2 – T = 2ma
or 2mg sin 45° − 
2mg cos 45° − T = 2ma
or 2a
= 
g − 
or a = 
g −
But (mB −
mA)g
sin θ = 
will be less than
(μA mA + μB mB)g cos θ = 
Therefore the masses will not move and hence
acceleration of A or B will be Zero.
12. A body of 100 kg is placed on a truck. The
coefficient of static friction between the body and the truck is 0.2. The truck
suddenly decreases its speed from 90 km/h to 36 km/h in 5s. Which of the
following is correct?
(a) The block does not move.
(b) The block slips forward and hit the driver's
cabin.
(c) Nothing can be said about the block.
(d) Block shifts backward.
Solution:
Force of static
friction = μmg
= 0.2 × 100 × 10
= 200 N
Deceleration = 
= 
= 3 m s−2
Force on the mass = 100 kg × 3 m s−2
= 300 N
This is greater than the force due to static friction.
Therefore due to pseudo force, the block of mass m will go forward and hit the driver's
cabin.
13. A boy stands on a weighing machine inside a
lift. When the lift is going down with acceleration 
, the machine shows a reading 30 kg. When the
lift goes upwards with acceleration , the reading would be?
Solution:
While going down, machine reading is 30 kg.
mg – N = m
mg
– 30g = 
3 = 30g
m
= 
= 40 kg
While
going up, the reaction would be
N' – mg
= m
N' = 5
= 
(40 kg)g
N' = 5 kgf
14. The rear side of a truck is open and a box of
mass 20 kg is placed on the truck 4 m away from the open end. If μ is 0.15 and g = 10 m s−2 and
the truck starts from rest with an acceleration of 2 m s−2 on
a straight road, then the box will fall off the truck when it is at a distance
of 
metre from
the starting point. What is the value of ?
Solution:
a = 2 m s−2, N
= mg
ma – μN = m
(ma – μmg) = m
= a – μg
= 2 m s−2 – (0.15)(10 m s−2)
= 0.5 m s−2
For the box’s motion
w.r.t. truck,
= 
· t + 
· t2
4 = 0 + 
(0.5) t2
t2 = 
t = 4 s
After 4 s, the box will fall off the truck.
During 4 s, the truck travels,
= at2
= (2)(4)2
= 16 m.
15. An insect crawls up a hemispherical surface
very slowly as shown in the figure.
The coefficient of friction between the insect
and the surface is . If the line joining the centre of the
hemispherical surface to the insect makes an angle α, with the vertical,
what is the maximum possible value of α?
Solution:
The insect I is under the action
of two forces.
N denotes
the normal reaction while mg denotes
the weight of the insect.
The weight mg can be resolved
into two perpendicular components. For equilibrium
N = mg
cos α, f = mg sin α
where f
denotes force of friction.
But f = μN ∴ μ(mg cos α) = mg sin α
or μ
= tan α
= 
cot α = 3
16. A weightless thread can bear tension up to
3.7 kg weight. A stone of mass 500 g is tied to it and revolves in a circular
path of radius 4 m in vertical plane. If g = 10
m s−2, then what will be the maximum angular velocity of the
stone?
Solution:
Mass
of stone = 500 g
Radius = 4 m
g = 10 m s−2
For vertical motion, tension in the string is
given by:
= − mg
where m is the mass of the block attached to the string
Maximum tension is T =
3.7 × 9.8 N
=
= 
= 329.8
= 18.14 m/s
we know = ωr
ω is the angular velocity
ω = 
= 4.53 rad/sec
17. Two fixed frictionless inclined planes making
an angle 30° and 60° with the vertical are shown in the figure. Two blocks A
and B are placed on the two planes. What is the relative vertical acceleration
of A with respect to B?
Solution:
The acceleration of the body down the smooth
inclined plane is a = g sin θ
It is along the inclined plane.
where θ is the angle of inclination
∴ The vertical
component of acceleration a is
= (g sin θ)sin θ
= g sin2 θ
For block A,
= g sin2
60°
For
block B,
= g sin2
30°
The relative vertical acceleration of A with respect to B is
= −
= g sin2 60° − g sin2 30°
= g
= 
= 4.9 m s−2
in vertical direction
18. A person sitting in an open car moving at
constant velocity throws a ball vertically up into air. Where will the ball
fall?
Solution:
Since the velocity is constant, the horizontal
component of velocity for the car and the ball are the same.
They cover equal horizontal distances in the same
time interval.
Therefore the ball will land in the hands of the
person in the car.
19. A man is standing stationarily
with respect to a horizontal conveyor belt that is accelerating with 1 m s–2, as
shown in figure. What is the net force on the man? If the coefficient of static
friction between the man’s shoes and the belt is 0.2, up to what acceleration
of the belt can the man continue to be stationary relative to the belt? (Mass
of the man = 65 kg.)
Solution:
Mass
of the man, m = 65
kg
Acceleration
of the belt, a
= 1 m s–2
Coefficient
of static friction, μ = 0.2
The net force F,
acting on the man is given by Newton’s second law of motion as:
Fnet = ma
= 65 × 1
= 65 N
The man will continue to be stationary with
respect to the conveyor belt until the net force on the man is less than or
equal to the frictional force fs, exerted by the belt,
i.e.,
Fnet = fs
ma' = μmg
a' = 0.2 × 10
= 2 m s–2
Therefore, the maximum acceleration of the belt
up to which the man can stand stationarily is 2 m s−2.
20. A stream of water flowing horizontally with a
speed of 15 m s–1 gushes out of a tube of cross-sectional area
10–2 m2, and hits a vertical wall nearby. What is
the force exerted on the wall by the impact of water, assuming it does not
rebound?
Solution:
Speed
of the water stream,
= 15 m/s
Cross-sectional
area of the tube,
A = 10–2 m2
Volume of water coming out from the pipe per
second,
V = A
= 15 × 10–2 m3/s
Density
of water,
ρ = 103 kg/m3
Mass of water flowing out through the pipe per
second,
= ρ × V
= 150 kg/s
The water strikes the wall and does not rebound.
Therefore, the force exerted by the water on the wall is given by Newton’s
second law of motion as:
F = Rate of change of momentum
=
=
= 150 × 15
= 2250 N
21. An aircraft executes a horizontal loop at a
speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Solution:
Speed
of the aircraft, = 720 km/h
= 720 × 
= 200 m/s
Acceleration
due to gravity, g
= 10 m/s2
Angle
of banking, θ = 15°
For radius r, of the loop, we have the relation:
tan θ =
r = 
= 
= 
= 14925.37 m
= 14.92 km
22. A 70 kg man stands in contact against the
inner wall of a hollow cylindrical drum of radius 3 m rotating about its
vertical axis with 200 rev/min. The coefficient of friction between the wall
and his clothing is 0.15. What is the minimum rotational speed of the cylinder
to enable the man to remain stuck to the wall (without falling) when the floor
is suddenly removed?
Solution:
The cylinder being vertical, the normal reaction of the wall
on the man acts horizontally and provides the necessary centripetal force which
is given below,
R = mr
------ (1)
The
frictional force F, acting upwards
balances his weight, i.e.,
F
= mg ------
(2)
The man
will remain stuck to the wall and continue to rotate with the cylinder without
slipping,
when μ
≥ 
F
≤ μR
Substituting
equations (1) and (2),
mg ≤ μ mr
≥
ω ≥ 
------(3)
∴ The minimum angular speed of rotation of the cylinder drum
is given by
ωmin ≥ ------ (4)
Here, μ = 0.15, r = 3 m, g = 9.8 m s−2
Substituting in equation (4),
ωmin = 
rad s−1
=
4.67 ≃ 5 rad s−1
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