Mechanical Engineering Principles

FORCE, MASS AND ACCELERATION 147

Now try the following exercise

Exercise 63 Further problems on New-

ton’s laws of motion

(Takegas 9.81 m/s^2 , and express answers to

three significant figure accuracy)

1. A car initially at rest accelerates uni-
   formly to a speed of 55 km/h in 14 s.
   Determine the accelerating force required
   if the mass of the car is 800 kg.
   [873 N]


2. The brakes are applied on the car in ques-
   tion 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
   [1.87 kN, 6.55 s]


3. The tension in a rope lifting a crate ver-
   tically upwards is 2.8 kN. Determine its
   acceleration if the mass of the crate is
   270 kg. [0.560 m/s^2 ]


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 oppos-
   ing the motion of the ship, assuming the
   reduction in speed is uniform and the
   mass of the ship is 2000 t. [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 ten-
   sion in the cable supporting the cage dur-
   ing (a) the initial period of acceleration,
   (b) the period of constant speed travel,
   (c) the final retardation period.

[(a) 11.6 kN (b) 19.6 kN (c) 24.kN]

6. A miner having a mass of 80 kg is stand-
   ing 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) 464.8 N (b) 784.8 N (c) 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.

[(a) 1.09 m/s^2 (b) 43.6 N]

13.3 Centripetal acceleration

When an object moves in a circular path at con-

stant speed, its direction of motion is continually

changing and hence its velocity (which depends on

both magnitude and direction) is also continually

changing. Since acceleration is the (change in veloc-

ity)/(time taken) the object has an acceleration.

Let the object be moving with a constant angular

velocity ofωand a tangential velocity of magnitude

vand let the change of velocity for a small change of

angle ofθ(=ωt)beV(see Figure 13.3(a)). Then,

v 2 −v 1 =V.

v

2

V

• v 1 v 2

(b)

q

2

v 2

v 1

q = ωt

r

r

(a)

Figure 13.3

The vector diagram is shown in Figure 13.3(b) and

since the magnitudes ofv 1 andv 2 are the same, i.e.

v, the vector diagram is also an isosceles triangle.

Bisecting the angle betweenv 2 andv 1 gives:

sin

θ

2

=

V/ 2

v 2

=

V

2 v

i.e. V= 2 vsin

θ

2

( 13. 1 )