CHAPTER 14 FORCE, MASS AND ACCELERATION
EXERCISE 73, Page 165
1. A car initially at rest accelerates uniformly to a speed of 55 km/h in 14 s. Determine the
accelerating force required if the mass of the car is 800 kg.
Initial velocity, v1 = 0
Final velocity, v 2 = 55
km
m
h
1000
1
= 15.278 m/s
h
km 3600s
Time, t = 14 s
Since v 2 = v1 + at then acceleration, a =
v 2  v1 15.278  0

 1.09 m / s 2
t
12
Hence, accelerating force, F = ma = 800 kg 1.09 m / s 2 = 873 N
2. The brakes are applied on the car in question 1 when travelling at 55 km/h and it comes to rest
uniformly in a distance of 50 m. Calculate the braking force and the time for the car to come to
rest.
Initial velocity, v1 = 55 km/h = = 15.278 m/s (from above)
Final velocity, v 2 = 0
Distance travelled, s = 50 m
v 2 2  v12  2as
v 2 2  v12 0  15.2782

 2.33m / s 2
from which, acceleration, a =
2s
2  50
Hence, braking force, F = ma = 800 kg × 2.334 m / s 2
= 1867 N = 1.87 kN
Since v 2 = v1 + at then time to come to rest, t =
v 2  v1 15.278  0

= 6.55 s
a
2.334
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3. The tension in a rope lifting a crate vertically upwards is 2.8 kN. Determine its acceleration if the
mass of the crate is 270 kg.
T – mg = ma
i.e. 2800 – 270 × 9.81 = 270 × a
from which, acceleration, a =
2800  270  9.81
= 0.560 m / s 2
270
4. A ship is travelling at 18 km/h when it stops its engines. It drifts for a distance of 0.6 km and its
speed is then 14 km/h. Determine the value of the forces opposing the motion of the ship,
assuming the reduction in speed is uniform and the mass of the ship is 2000 t.
Initial velocity, v1 = 18 km/h = 18
Final velocity, v 2 =
km
m
h
18
1000
1

= 5 m/s
h
km 3600s 3.6
14
= 3.889 m/s
3.6
Distance travelled, s = 0.6 km = 600 m
v 2 2  v12  2as
from which, acceleration, a =
v 2 2  v12 3.8892  52

 - 8.23 103 m / s 2
2s
2  600
Hence, force opposing motion, F = ma = 2000 1000 kg  8.23103 m / s 2
= 16459 N = 16.5 kN
5. A cage having a mass of 2 t is being lowered down a mineshaft. It moves from rest with an
acceleration of 4 m/s 2 , until it is travelling at 15 m/s. It then travels at constant speed for 700 m
and finally comes to rest in 6 s. Calculate the tension in the cable supporting the cage during
(a) the initial period of acceleration, (b) the period of constant speed travel, (c) the final
retardation period.
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(a) Initial tension in cable, T1 = mg – ma = m(g – a) = 2000(9.81 – 4)
= 11620 N = 11.6 kN
(b) Tension in cable during constant speed, T2 = mg – m a 2 = mg – 0 = 2000 × 9.81
= 19620 N = 19.6 kN
(c) Tension in retardation period, T3 = mg – m a 3
where a 3 =
v  u 0  15

  2.5 m / s 2
t
6
Hence, tension, T3 = mg – m a 3 = 19620 – 2000(- 2.5)
= 24620 N = 24.6 kN
6. A miner having a mass of 80 kg is standing in the cage of problem 5. Determine the reaction
force between the man and the floor of the cage during (a) the initial period of acceleration,
(b) the period of constant speed travel, and (c) the final retardation period.
(a) Reaction during initial acceleration = mg – m a1 = m(g – a1 ) = 80(9.81 – 4)
= 464.8 N
(b) Reaction during constant speed = mg – m a 2 = mg – 0 = 80 × 9.81
= 784.8 N
(c) Reaction in retardation period = mg – m a 3 = 80(9.81 - - 2.5) = 80 (9.81 + 2.5)
= 984.8 N
7. During an experiment, masses of 4 kg and 5 kg are attached to a thread and the thread is passed
over a pulley so that both masses hang vertically downwards and are at the same height. When
the system is released, find (a) the acceleration of the system, and (b) the tension in the thread,
assuming no losses in the system.
Let m1 = 4 kg and m 2 = 5 kg
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Consider the motion of mass ‘ m 2 ’
m2 g – T = m2 a
i.e.
from which,
5g – T = 5a
a=
5g  T
T
g
5
5
(1)
Consider the motion of mass ‘ m1 ’
T - m1 g = m1 a
from which,
a=
T  m1g T  4g T

 g
m1
4
4
Equating equations (1) and (2) gives: g 
i.e.
i.e.
T
T
= g
5
4
2g =
9T
= 2 × 9.81
20
T T 5T  4T 9T
 

4 5
20
20
and tension in thread, T =
From (1), acceleration, a = g 
(2)
20
 2  9.81 = 43.6 N
9
T
43.6
 9.81 
= 1.09 m / s 2
5
5
196
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EXERCISE 74, Page 166
1. Calculate the centripetal force acting on a vehicle of mass 1 tonne when travelling round a bend
of radius 125 m at 40 km/h. If this force should not exceed 750 N, determine the reduction in
speed of the vehicle to meet this requirement.
Centripetal acceleration =
where v = 40
v2
r
km
m
1h
1000

= 11.11 m/s
h
km 3600s
Hence, centripetal acceleration, a =
and r = 125 m
11.112
= 0.988 m/s 2
125
Centripetal force = ma = 1000 kg × 0.988 m/s 2 = 988 N
If centripetal force ≤ 750 N ≤ ma 2
then
a2 =
v2
750 N
 0.75m / s 2  2
1000 kg
r
v 2 2  0.75 125 and v2  0.75 125 = 9.682 m/s
i.e.
9.682 m/s = 9.682
m 1km 3600s


= 34.86 km/h
s 1000 m
1h
Hence the speed reduces form 40 km/h to 34.86 km/h
2. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and 150 m. If
the speed of the boat is constant at 34 km/h, determine the change in acceleration when leaving
one arc and entering the other.
34 km/h = 34
km 1000 m
1h


= 9.444 m/s
h
1km 3600s
Acceleration, a1 
v 2 9.4442

 0.891m / s 2
r1
100
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Acceleration, a 2 
v 2 9.4442

 0.595 m / s 2
r2
150
Change of acceleration = a1  a 2 = 0.891 – 0.595 = 0.296 m / s 2
i.e. change in acceleration = 0.3 m / s 2
3. An object is suspended by a thread 400 mm long and both object and thread move in a horizontal
circle with a constant angular velocity of 3.0 rad/s. If the tension in the thread is 36 N, determine
the mass of the object.
mv 2
Centripetal force (i.e. tension in thread) =
= 36 N
r
The angular velocity,  = 3.0 rad/s and radius, r = 400 mm = 0.4 m.
Fr
mv 2
Since linear velocity v = r, v = 3.0  0.4 = 1.2 m/s, and since F =
, then m = 2
v
r
i.e. mass of object, m =
36  0.4
= 10 kg
1.2 2
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EXERCISE 75, Page 168
1. Calculate the mass moment of inertia of a thin rod, of length 0.5 m and mass 0.2 kg, about its
centroid.
Mass moment of inertia of a thin rod about its centroid,
I=
mL2 0.2  0.52

= 4.167 10 3 kg m 2 or 0.004167 kg m 2
12
12
2. Calculate the mass moment of inertia of the thin rod of Problem 1, about an end.
Mass moment of inertia of a thin rod about an end,
mL2 0.2  0.52

= 0.01667 kg m 2
3
3
I=
3. Calculate the mass moment of inertia of a solid disc of uniform thickness about its centroid. The
diameter of the disc is 0.3 m and its thickness is 0.08 m. The density of its material of
construction is 7860 kg/m 3 .
Mass moment of inertia of a solid disc of uniform thickness about its centroid,
I = R 2 t 
R2
0.152
 7860 kg / m3   0.152  0.08 
= 0.50 kg m 2
2
2
4. If a hole of diameter 0.2 m is drilled through the centre of the disc of Problem 3, what will be its
mass moment of inertia about its centroid?
Mass moment of inertia of a solid disc of uniform thickness about its centroid,
I =   R 2  R
2
2
1

R
t
2
1
 R 22 
2
2
 7860 kg / m3   (0.152  0.12 )  0.08 
(0.152  0.12 )
2
= 0.401 kg m 2
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EXERCISE 76, Page 168
Answers found from within the text of the chapter, pages 162 to 168.
EXERCISE 77, Page 169
1. (c) 2. (b) 3. (a) 4. (d) 5. (a) 6. (b) 7. (b) 8. (a) 9. (a) 10. (d) 11. (d) 12. (c) 13. (b)
200
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