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

## Question

## Thalia 64

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## Link to post

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