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?

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,

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 , be horizontal component of force by the hinge, and 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:

Torque by the force of gravity is given as:

Torques by and are 0 as they act on the hinge itself.

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

Dividing both sides by 'L', we get:

 --------------------(1)

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

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

--------------------------(2)

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

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

question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation:

true tan speed = ang vel x radius