Engineering Mechanics

(^354) „„„„„ A Textbook of Engineering Mechanics
Solution. Given: Length of the ladder (l) = 5 m and weight of the ladder (W) = 200 N
Let T= Tension in the rope PQ,
θ= Angle, which the ladder makes with the horizontal.
x= Virtual vertical displacement of the mid of the ladder Q (or in
other words, weight of the ladder), and
y= Virtual horizontal displacement of the rope PQ due to tension.
From the geometry of the figure, we find that
4
tan
3
θ=
We also find that when the mid point P of the ladder moves downwards (due to weight), it
causes top of the ladder B to move downwards and bottom of the ladder A to move towards left. It
causes tension (T) in the rope PQ. Moreover, when the virtual vertical displacement of the mid of the
ladder P (or weight of the ladder) is x, then the virtual horizontal displacement of the ladder,
3
0.75
tan 4 3 4
xxx
yx====
θ
∴ Virtual work done by the tension in the rope
= + Tx ...(i)
...(Plus sign due to tension)
and virtual work done by the 200 N weight of the ladder
= – 200 y ...(ii)
...(Minus sign due to downward movement of the weight)
We know that from the principle of virtual work, that algebraic sum of the total virtual works
done is zero. Therefore
Tx – 200 y=0
or
200 200 0.75
150 N
yx
T
xx
×
== = Ans.
16.9.APPLICATION OF PRINCIPLE OF VIRTUAL WORK ON LIFTING
MACHINES
We know that in the case of lifting machines, the effort moves downwards, whereas the load
moves upwards. In such cases, the virtual works done by the effort and that by the load are found out.
Now apply the principle of virtual work as usual.
Example 16.8. A weight (W) of 5 kN is raised by a system of pulleys as shown in Fig. 16.18
Fig. 16.18.
Using the method of virtual work, find the force P, which can hold the weight in equilibrium.