Can someone help with a physics problem about tension and horizontal & vertical forces of a sign hanging from a beam?  A shop sign weighing 250 N is supported on a uniform 150 N beam.  Find...

Can someone help with a physics problem about tension and horizontal & vertical forces of a sign hanging from a beam?

 

A shop sign weighing 250 N is supported on a uniform 150 N beam.  Find the tension in the guy wire and the horizontal & vertical forces exerted by the hinge on the beam.   The length from the vertical beam to the wire that the sign is hanging from is 1.70 m.  The length from the vertical beam to the location where the wire is attached to the horizontal beam is 1.35 m  This wire is attached at a 35 degree angle to the horizontal beam and is attached to the vertical beam as well.

Asked on by peache

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ndnordic | High School Teacher | (Level 2) Associate Educator

Posted on

This is a problem in equilibrium and statics. You must look at both the forces, and the torques, acting to solve this problem. All the forces must total zero, and all the torques must also total zero.

Looking first at the forces, you have:

The force of the hinge in the y-direction  F(hy)

The force of the hinge in the x-direction  F(hx)

The force from the wire tension in the y-direction  F(ty)

The force from the wire tension in the x-direction  F(tx)

The weight of the beam = 150 N

The weight of the sign = 250 N

Summing these forces in the y-direction you get:

F(hy) + F(ty) - 150 - 250 = 0

Then F(hy) +F(ty) = 400 N

Summing in the x-direction:

F(hx)-F(tx) = 0 or F(hx) = F(tx)

Now consider the torques, with the point of rotation where the wire is attached to the beam:

clockwise torque of .35 m * 250 N

clockwise torque of .35 m * (150*.35/1.35) (this is the portion of the beam itself to the right of the point of attachment)

counterclockwise torque of 1.35 * F(hy)

summing:  1.35 F(hy) - .35*250 - .35 (.35*150/1.35) = 0

Solving for F(hy) = 74.9 N

Go back to the first equation and solve for F(ty)

F(ty) = 400 - F(hy) = 325.1N

Since the sin(35) = F(ty)/F(t), then F(t) = 566.8 N

since tan35 = F(hy)/F(hx) then F(hx) = 106.97N

Summarizing:

F(ty) = 325.1 N

F(hy) = 74.9 N

F(t) = 566.8 N

F(hx = 106.97 N

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