Engineering Mechanics

Chapter 16 : Virtual Work „„„„„ 355

Solution. Given : Weight (W) = 5 kN

Let P= Force which can hold the weight in equilibrium, and

y= Virtual upward displacement of the weight.
From the geometry of the system of pulleys, we find that when the virtual upward displacement
of the weight is y, the virtual downward displacement of the force is 2y.
∴ Virtual work done by the load
= + Wy = 5y ...(i)
...(Plus sign due to upward movement of the load)
and virtual work done by the effort

= – P × 2y = – 2 Py ...(ii)
... (Minus sign due to downward movement of the effort)
We know that from the principle of virtual work, that algebraic sum of the virtual works
done is zero. Therefore
5y – 2Py = 0

or
5
2.5 kN
2

y

P

y

== Ans.

Example 16.9. A weight of 1000 N resting over a smooth surface inclined at 30° with the
horizontal, is supported by an effort (P) resting on a smooth surface inclined at 45° with the horizontal
as shown in Fig. 16.19.

Fig. 16.19.
By using the principle of virtual work, calculate the value of effort (P).
Solution. Given: Weight (W) = 1000 N; Inclination of weight surface (α 1 ) = 30° and inclination
of effort surface (α 2 ) = 45°

Let x = Virtual vertical displacement of 1000 N weight, and
y = Virtual vertical displacement of the effort (P).
From the geometry of the system, we find that when the weight (1000 N) moves downwards,
the effort (P) will move upwards. We also find that the distance through which the 1000 N weight
will move downwards on the inclined surface will be equal to the distance through which the load P
will move upwards on the inclined surface.

∴ Distance through which the 1000 N weight moves on the inclined surface AC.

2

sin 30 0.5

xx

===x

° ...(i)

and distance through which the load P will move on the inclined surface BC

1.414

sin 45 0.707

yy

===y

° ...(ii)

Equating equations (i) and (ii),

2 x = 1.414 y or x = 0.707 y