Question

A 26.4 kg beam is attached to a wall with a hinge and its far end is supported by a cable. The angle between the beam and the cable is 90o. If the beam is inclined at an angle of θ = 23.4° with respect to horizontal. What is the horizontal component of the force exerted by the hinge on the beam?

Answer 1

Answer:

The horizontal component of the force exerted by the hinge on the beam is 47.15 N.

Explanation:

Given data:

Weight of beam = 26.4 kg

Angle between the beam and the cable is 90°

Beam inclination with respect to horizontal with an angle, $\theta = 23.4\°$

We need to find the horizontal component of the force exerted by the hinge on the beam.

Solution:

Let 'L' be length of the beam, 'T' be tension in the cable , $F_{h}$ be horizontal component of force by the hinge, and $F_{v}$ be vertical component of force by the hinge.

Take counterclockwise torque as positive.

Let us find torques around the hinge.

Torque by tension is given as:

$\tau = T \times L$  

Torque by the force of gravity is given as:

$\tau_g= m g \frac{L}{2}\times cos \theta$

Torques by $F_{h}$ and $F_{v}$ are 0 as they act on the hinge itself.

Now, for equilibrium, net torque about the hinge is 0. So,

$\tau-\tau_g=0$

$T L - m g \frac{L}{2}\times \cos(\theta) = 0$

Dividing both sides by 'L', we get:

$T - m \frac{g}{2}\times \cos \theta = 0$

$T=m \frac{g}{2} \times cos \theta$ --------------------(1)

As per question, the cable makes 90° with the horizontal.

So, the net horizontal force is also zero. Therefore,

$F_{h} -T cos(90- \theta) = 0$

$F_h - T sin(\theta) = 0$

$F_h = T sin(\theta)$ --------------------------(2)

Plug the value of 'T' from equation (1) into equation (2). This gives,

$F_{h} = m \frac{g}{2} \times cos \theta \times sin \theta$

$F_{h} = 26.4 \times \frac{9.8}{2} \times cos(23.4) \times sin(23.4)$

$F_{h} = 47.15\ N$

Therefore, the horizontal component of the force exerted by the hinge on the beam is 47.15 N.