Question

1. Neglecting the weight of the beam below, find the tension in the tie rope and the force components at the hinge in the terms of W. Repeat if the uniform beam weighs ½ W.

Answer 1

Answer:

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Answer 2

Given:

The angle between beam and rod = 50°

The angle between the rope and rod = 70°

To find:

a) Tension and force components at the hinge when the weight of beam =W

b) Tension and force components at the hinge when the weight of beam =W/2

Solution:

a) Torque due to the weight W about hinge =  - 0.8 WL sin 50°  

Torque due to tension in the string about hinge = TL sin 20°

The beam is in equilibrium so the net torque on it should be zero.

⇒ TL sin 20°  - 0.8 WL sin 50° = 0

 or T = 0.8W X 0.766 / 0.342

T = 1.792W

Let the horizontal and vertical force components be Fh and Fv respectively.

From the figure,

Fh - T sin 70 = 0

Substituting the value of T from above,

Fh = 1.684W

Similarly equating the vertical forces,

Fv - T cos 70 = 0

or Fv = 1.613W

b) Taking the weight of beam as W/2

Torque due to the weight = - 0.25WL sin 50

Equating the net torque to zero, we get:

T sin 20 = 1.05W sin 50

or T = 2.352W

There is no change in the horizontal component as the weight contributes to Fv only.

So, Fh = 1.684W

And Fv = T cos 70 + W +W/2

= 1.5 W + 2.352W X 0.342

= 2.304W