A traffic light hangs from a pole. The uniform aluminum pole is 7.50m long and project from a vertical pole.?

Question

and has a mass of 12.0kg. The vertical pole is 3.8m. The mass of the traffic light is 21.5 kg.
Determine a)the tension in the horizontal mass-less cable and b)the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole. I managed to find a picture of the same problem but with different values: http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460004546

If someone could please help me solve this by monday night that would be greatly appreciated as I am terrible at physics :(!! Thank you.

Answer 1

Well, it's Wednesday morning. I hope this answer still has some value to you.

I started by summing the torques at the hinge. There are three:
T(light)=5*9.8*7.5*cos(37)
T(cable)=-tension*6.31*sin(37)
T(pole)=8*97.5/2)*cos(37)

These all sum to zero.

I computed tension=140.5

The sum of the horizontal forces must be zero.
These are the tension in the cable and the horizontal reaction force at the hing. So the horizontal component is 140.5
The vertical forces are the weights of the pole and the light offset by the vertical reaction force at the hinge
the vertical component is (8+5)*9.8
=127.4

I checked the answer by computing the magnitude of the reaction force two ways. Using Pythagorean theorem
sqrt(140.5^2+127.4^2)
=189

I then summed the forces parallel to the pole:
5*9.8*sin(37)
8*9.8*sin(37)
140.5*cos(37)
=189

j

Answer 2

Aluminum Pole