The assignment is to eventually calculate the forces on the BD beam of this picture:


My issue is that on the solution sheet, the professor drew a free body diagram (FBD) for just the beams, with the force of the block hanging from the string, mg, acting downwards exactly at the point $C$. I'm guessing he is implicitly ignoring the radius of the pulley?

Can anyone explain if this is a property of the system, or if it's just a sloppily worded problem? I mean, generally the force mg should generate some sort of torque about $C$, shouldn't it?

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    $\begingroup$ Always assume massless, frictionless and dimensionless unless stated otherwise. $\endgroup$ Dec 31 '13 at 10:40
  • $\begingroup$ By C, I assume you mean G? $\endgroup$ Dec 31 '13 at 10:40
  • $\begingroup$ No, he just writes C:s that look like G:s :) $\endgroup$ Dec 31 '13 at 10:41

The force does not depend on the radius of the pulley, but your professor is wrong to draw the force as straight down.

The standard assumption is a massless string and frictionless pulley. In that case, the force on the pulley from the string is solely a normal force and is at all points related to the tension and radius of curvature of the string. The tension is constant (because the rope is massless) and the radius of curvature is constant over the $1/4$ of a circle over which the rope contacts the pulley. Therefore, the force from the string on the pulley is at a 45-degree angle down and to the right and goes through the center of the pulley. Its magnitude is $\sqrt{2} m g$.

Another way to see the same thing is to account for the forces on the string. There is a force $mg$ down from the mass. Then there is a force $mg$ to the right from the wall. That means there must be a force $\sqrt{2}m g $ up and to the left from the pulley so the net force on the string is zero.

If you change the radius of the pulley, you slide the point where the force is applied along a line that is parallel to the force. Thus all torques are unaffected, and the radius of the pulley doesn't matter.

However, if we do not assume that the top part of the string is perfectly horizontal, then the direction of the force depends on the angle of the top part of the string, which could then depend on the radius of the pulley, and this would make the force on $BD$ dependent on the radius of the pulley.


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