Mechanical Engineering Principles

146 MECHANICAL ENGINEERING PRINCIPLES

The weight of an object is the force acting vertically
downwards due to the force of gravity acting on the
object. Thus:

weight=force acting vertically downwards

=mass×gravitational field

= 1. 6 × 9. 81 = 15 .696 N

Problem 5. A bucket of cement of mass

40 kg is tied to the end of a rope connected

to a hoist. Calculate the tension in the rope

when the bucket is suspended but stationary.

Take the gravitational field,g,as9.81N/kg

(or 9.81 m/s^2 ).

Thetensionin the rope is the same as the force
acting in the rope. The force acting vertically down-
wards due to the weight of the bucket must be equal
to the force acting upwards in the rope, i.e. the
tension
Weight of bucket of cement,

F=mg= 40 × 9. 81 = 392 .4N

Thus,the tension in the rope = 392.4 N

Problem 6. The bucket of cement in

Problem 5 is now hoisted vertically upwards

with a uniform acceleration of 0.4 m/s^2.

Calculate the tension in the rope during the

period of acceleration.

With reference to Figure 13.1, the forces acting on
the bucket are:

(i) a tension (or force) ofTacting in the rope

(ii) a force ofmgacting vertically downwards, i.e.
the weight of the bucket and cement

Acceleration

Force due

to acceleration

F= ma

Weight,

mg

T

Figure 13.1

The resultant forceF=T−mg;

hence, ma=T−mg

i.e. 40 × 0. 4 =T− 40 × 9. 81

from which, tension,T= 408 .4N

By comparing this result with that of Problem 5,

it can be seen that there is an increase in the

tension in the rope when an object is accelerating

upwards.

Problem 7. The bucket of cement in

Problem 5 is now lowered vertically

downwards with a uniform acceleration of

1.4 m/s^2. Calculate the tension in the rope

during the period of acceleration.

With reference to Figure 13.2, the forces acting on

the bucket are:

(i) a tension (or force) of T acting vertically

upwards

(ii) a force ofmgacting vertically downwards, i.e.

the weight of the bucket and cement

Acceleration

Weight,

mg

T F= ma

Figure 13.2

The resultant force, F=mg−T

Hence, ma=mg−T

from which,tension,T=m(g−a)

= 40 ( 9. 81 − 1. 4 )

= 336 .4N

By comparing this result with that of Problem 5, it

can be seen that there is a decrease in the tension

in the rope when an object is accelerating down-

wards.