Question

A sign is supported by a uniform horizontal boom of length 3.30 m and weight 78.0 N. A cable, inclined at an angle of 23° with the boom, is attached at a distance of 2.44 m from the hinge at the wall. The weight of the sign is 136.0 N. What is the tension in the cable and what are the horizontal and vertical forces Fx and Fy exerted on the boom by the hinge? Comment on the magnitude of Fy.

Answer 1

T=605.8N

Fx=-557.64N

Fy=22.7N

The magnitude of the Fy is in the upward direction and can bear the sign from falling

Step-by-step explanation:

From the principle of moment which states that anticlockwise moment must be equal to clockwise moment and upward forces must be equal to downward forces.

Moment is the product of force and the perpendicular distance in line of the action of the force. Force is that which tends to change a body's state of rest or uniform motion in a straight line.

T=?

angle is 23

weight of the boom will be acting at the center of gravity which is half od the length(1.65m)

weight of the boom is 78N

weight of the sign =136N

We therefore take moments about the fixed end of the boom , we call it B

2.44*Tsin 23°=78*1.65+136*3.3

2.44*Tsin23=577.665

Tsin23=236.74

T=236.74/sin23

T=605.8N

let's find the sum of both horizontal component and vertical component of the forces acting on the boom.

we designate downward forces as negative and upward forces as positive. Also, forces going left as negatives, while forces to the right are designated as positive.

E(Fx,Fy)=(0,-78)+605.8(-cos 23,sin23)+(0,-136)

-557.64N,22.7N

Fx=-557.64N

Fy=22.7N

The magnitude of the Fy is in the upward direction and can bear the sign from falling