# Torque help

A sign with weight 800 N is hung from point B at the end of a uniform hoizontal rod of weight 300 N as diagrammed below. The rod is attached to the vertical wall by a hinge at point A. The supporting cable is attached to the rod at B and to the wall at C. The cable makes a 42 degree angle with the rod.

A.Find the magnitude of the total force R exerted on the hinge at point A. note that the total force has both horizontal and vertical components.

B.The Cable has a tensile strength of 3216N. Find the minimum possible angle that the cable can make with the rod if it is to support the rod and sign without breaking.

Torque=Fsin(theta)r

I know exactly how to do A. I describe the torque equation, equal to zero and solve for P, and then use that to find Rx and Ry, and then R itself. But to create a valid equation I need to know the length of the bar, don't I? It's really confounded me.

B has stumped me completely.

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Doc Al
Mentor
But to create a valid equation I need to know the length of the bar, don't I?
Do you? Call the length L and see what happens.

Then...

Torque= (Tsin42)(L) - (1100)(L)=0

I need to know either the tension or the length.

Doc Al
Mentor
Then...

Torque= (Tsin42)(L) - (1100)(L)=0
That equation is not quite right. Where does the weight of the rod act?
I need to know either the tension or the length.
No you don't. Simplify that equation.

The weight act normal to the bottom of the rod, producing a negative torque, which is why I put it in my equation. Is that not correct?

Doc Al
Mentor
The weight act normal to the bottom of the rod, producing a negative torque, which is why I put it in my equation. Is that not correct?
What's the torque due to the weight of the rod? Where along the rod does the weight act?

In the center in at the very right. I suppose if I change the axis of roation to the right side, the equation would be Torque=Ry(L)-300(L/2), but that gets me no where.

Doc Al
Mentor
Yes, the weight of the rod acts at the center of the rod (the rod's center of mass). But you don't have to change your axis of rotation.

There are two forces creating negative torque: the weight of the sign (800 N) and the weight of the rod (300 N). Since those forces act at different points, figure out the torque they produce separately (then add them up to get the total, of course).