Two rigid steel beams XY and YZ are fixed at their lower ends and are hinged at Y. Each beam is inclined at 50° to the horizontal, as shown.

Question 23

Two rigid steel beams XY and YZ are fixed at their lower ends and are hinged at Y. Each beam is inclined at 50° to the horizontal, as shown. A weight of 4.0 × 10⁴ N hangs from Y. The structure is in equilibrium.

What is the force exerted by each beam on the hinge at Y?

A 2.6 × 10⁴ N              B 3.1 × 10⁴ N              C 5.2 × 10⁴ N              D 6.2 × 10⁴ N

Reference: Past Exam Paper – November 2017 Paper 12 Q14

Solution:

Answer: A.

Since the structure is in equilibrium, the resultant force on it should be zero.

The weight of the hinge is downwards at Y. So, each of the beam will act a force on the hinge such that when the resultant is considered, it is found to be zero.

Let the force exerted by each beam be F.

Since the beams are 50° to the horizontal, the angle at Y in the triangle is (180 – 50 – 50 =) 80°. This is the angle between the two forces. Drawing a vertical shows that the angle on each side is 40°.

Now, consider the forces vertically.

Vertical component of each force exerted by beam = F cos40

For equilibrium,

Sum of upward (component of) forces = Downward force

F cos40 + F cos40 = 4.0 × 10⁴

2F cos40 = 4.0 × 10⁴

F = (4.0 × 10⁴) / 2 cos40

 
Force F = 2.6×10⁴ N