Solve Torque Confusion: Find D to Snap Cable in Fig. 12-35

[Puchinita5]

Homework Statement

In Fig. 12-35, a uniform beam of weight 500 N and length 3.0 m is suspended horizontally. On the left it is hinged to a wall; on the right it is supported by a cable bolted to the wall at distance D above the beam. The least tension that will snap the cable is 1200 N. (a) What value of D corresponds to that tension? (b) To prevent the cable from snapping, should D be increased or decreased from that value?

Homework Equations

http://edugen.wiley.com/edugen/courses/crs1650/art/images/halliday8019c12/image_t/tfg035.gif

The Attempt at a Solution

ok i don't actually need anyone to solve this question. I am having trouble with a minor concept.

To begin, I wanted to set the Torques equal to zero. When calculating the Torque for the tension however, i am confused about the angle between r and F. From all my studies, I feel like the angle between r x F should be the obtuse angle outside the triangle. As in, if I were to keep extending the lever out towards the right, the angle between this and the tension angle. However, When I set my torque equation, i get a very small number for the angle, which suggests that it has given me the interior angle. But conceptually i don't see how this could be. I don't know if that makes sense. But if anyone has any clarification for me I would really appreciate it.

I had:

(1200)(3)sin(theta)-(500)(1.5)=0
which gives me theta=12.02...

 

[RoyalCat]

\tan{\theta}=\frac{D}{l} where l is the length of the beam.

Knowing the angle required to produce the critical tension, you can now easily use the above trigonometric relation to find what the critical angle theta is, and from that, find your critical height, D.

Also, I don't quite understand the source of your confusion, but one relation comes to mind:
\sin{\theta}\equiv \sin{(\pi-\theta)}

 

[Puchinita5]

i more confused just as the concept of torque rather than this problem specifically. Thought this is the problem I'm on so i thought I would use it to clarify. Since Torque=rxF=rFsin(theta), I'm confused as to WHICH angle theta would be...I would think it should be large angle from the horizontal and going COUNTERclockwise up to the tension force/chord. However when i work out this problem, it works only if I use the interior angle between the horizontal going up clockwise to the tension force. I don't understand why this is.

 

[RoyalCat]

i more confused just as the concept of torque rather than this problem specifically. Thought this is the problem I'm on so i thought I would use it to clarify. Since Torque=rxF=rFsin(theta), I'm confused as to WHICH angle theta would be...I would think it should be large angle from the horizontal and going COUNTERclockwise up to the tension force/chord. However when i work out this problem, it works only if I use the interior angle between the horizontal going up clockwise to the tension force. I don't understand why this is.

Ooh, you got yourself confused there, you did.
The \sin{\theta} in the definition of torque refers to the angle between the lever arm and the force. Taking into consideration the identity I posted above, that means you could also use the angle that compliments it to 180º, which could prove useful in cases where its sine is more easily calculated.

What it MEANS, though, is that you're taking the component of the force perpendicular to the lever arm and multiplying it by the length of the lever arm. You then look at where the moment is pointing to see whether it's clockwise or counterclockwise.

Another way of looking at it is that you're taking the force, and multiplying it by the component of the lever arm perpendicular to the force (This is an important realization if you want to calculate moments quickly without getting dragged down by the trigonometry)