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For part (a), is the normal force by the hinge on the bridge at an angle or is it horizontal?

For part (b), I know how to resolve forces horizontally and vertically, and to take torques about the hinge, but the information is still insufficient for me to figure out what the tension force is.

Any help would be much appreciated!

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If you take torques about the hinge, the direction of the hinge force is not important. However, I don't think it is horizontal as Ron suggests, neither generally nor in your specific problem. – Pygmalion May 20 '12 at 19:05
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@RonMaimon If it's horizontal I resign from my job. – Pygmalion May 20 '12 at 20:38
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As of the 2) you should have open a new question, because answer no longer matches the question. At point B you simply have the force of the rod BD to the rod AC. No friction. – Pygmalion May 20 '12 at 21:46
@Pygmalion Thank you! So unlike a hinge, the "link" here has no friction because it is akin to a model string, I suppose. Thanks again Pyg, I really do so so so appreciate your help. – Ryan May 20 '12 at 21:59
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Please do not issue edits which change the question into a completely new one. That is not fair to the people who have already written answers and risks confusing later readers. – dmckee♦ May 21 '12 at 19:45
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1 Answer

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(1)

As for the (a) the total force of the ground/hinge (e.g. thrust or normal force + friction) is generally neither vertical nor horizontal. EDIT: You can obtain the force of the ground/hinge by calculating the force of the rope first, and then add all three forces together to get zero.

As for the (b) you have three forces acting to beam, force of the ground, gravitational force and force of the rope. Since problem suggests "considering equilibrium", torques of these three forces must equal zero.

(2) Force at point B is simply the force of rod BD to rod AC (and vice versa). Effectively, you have three forces acting on rod AC. Note also that the force of the rod BD is along its direction (because it is limited by two joints at its ends and there is no force in between).

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"(e.g. normal force + friction)" At a hinge the engineers usually call this "thrust", and it can point in any direction. To put it in more formal physicist language treat it as a constraint force (the axle of the hinge is constrained to remain in the hinge...). – dmckee♦ May 20 '12 at 18:49
@Ryan The ground forces act through hinge. If you take torques about the hinge, the torque of the ground forces is zero. – Pygmalion May 20 '12 at 19:01
@dmckee thanks. I am not very familiar with English expressions. – Pygmalion May 20 '12 at 19:03
@Ryan so what's the problem? The torque of the gravitational force is $-\frac{L}{2}mg$, while the torque of the tension force is $L T \sin \phi$. Together they are zero. Just have to express $\alpha$ from geometry of the problem. – Pygmalion May 20 '12 at 19:21
@Pygmalion Oh yes I see. Haha I've been so muddled, I was caught up trying to use vector product to solve the torque, and was looking for T to help me find T. Doh. This is what happens when one is sleep-deprived during exams season. Thanks! – Ryan May 20 '12 at 20:54
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