BrainDen.com - Brain Teasers
• 0

## Question

I'm going over my notes to get ready for the final and my notes for this problem are blurred and don't make sense to me. Anyone good at physics?

A traffic light is hanging from a pole. The uniform pole (AB) is 7.5 m long and has a mass of 12 kg. The mass of the traffic light is 21.5 kg. Determine the tension in the horizontal massless cable (CD) and the vertical and horizontal components of the force exerted by the pivot (A) on the pole.

Notes:

D is the point where the cable meets the diagonal pole.

B is the end of the diagonal pole.

AC = 3.8 m.

The angle formed by the cable and the diagonal pole is 37 degrees.

I can't upload the actual image so this is my best shot at it.

| C____________D __/ _B

| ---------------------- / __|

| ____________ /____ |

| ___________ /_____ |

| _________ /_______----

| _______ /________| __ |

| _____ /__________ ----

|_A__/

| __/

I have sum of torques = .5(7.5 m)(12 kg)(g)sin 127° + something that doesn't make sense.

I'm also unsure about how that works. I've been using torque = rFsin(theta) so I've been assuming the .5(7.5) is the center of gravity used for r. Is that right? I also don't understand why 127°. It would by the supplemental angle to the inside angle formed by AC and AD but why use that?

For the horizontal component, I have sum of forces = F of x + tension x cos 180° = 0

For the vertical component, I have sum of forces = F of y - weight of pole (AB) - weight of light = 0

Does that look right?

Edited by Thalia

## Recommended Posts

• 0

If you can tell me the answer then possibly i can reverse engineer.. I've done problem like these before. But i'm not very sure with the approach i'm taking.

##### Share on other sites
• 0

The translational equiilibrium looks right.

For the rotational equilibrium...

You have the torques of the weight of the traffic light and the weight of the rod. These balance the torque of the tension in the cable. 127 degrees is as good as 53, which is the angle made between r and the weight of the rod.

I think you'll be able to take out the tension in the cable with this.

And then you can get F of x.

##### Share on other sites
• 0

Thanks! I've got it figured out now. I think. . . Oh well. Chances are that homework sheet won't survive til next quarter to find out.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.