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Question

Can somebody help me in determine the direction of frictional forces at $A,B,C$?

This is my reasoning. If friction was not present due to the weight, the rod AB will fall down and hence the point A should move to the right. Hence $F_a$ is to the left. By resolution of horizontal force, $F_c$ is towards right. By taking moments at O for the sphere, $F_b$ should be downwards. These are wrong.

mathnoob123
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  • What are your thoughts? (Also, this question, although mathematical in some regards, is better suited for Physics Stack) – infinitylord May 31 '17 at 02:16
  • Please see the edit. – mathnoob123 May 31 '17 at 02:19
  • Your reasoning and results for $F_a$ and $F_c$ seem correct to me. However, I would suspect that $F_b$ points upwards, as the rod wants to fall downwards under gravity, and friction (due to contact with the sphere), keeps it up. Again, though, this question is better asked on Physics Stack. – infinitylord May 31 '17 at 02:25
  • Yes, the main problem that arises in the question is due to $F_b$. How can you justify $F_c$ and $F_b$ by taking moments for the sphere at $O$? Wouldn't they both give a resultant moment? – mathnoob123 May 31 '17 at 02:26
  • Make sure you're talking about the correct force of friction. The force of friction from the rod on the sphere is downward. The force of friction from the sphere on the rod is upwards. – eyeballfrog May 31 '17 at 02:33
  • @eyeballfrog Your statement makes sense to me mathematically maybe because I am biased(to solve this question) but can you clear this through physics?(I am aware that this is the wrong place to ask but if it's possible please answer) I mean what will make the sphere to rotate anti clockwise for which a downward force is needed? – mathnoob123 May 31 '17 at 02:36
  • The frictional force at C from the ground points to the right. This generates a counterclockwise torque on the sphere. The downward frictional force from the rod produces a clockwise torque that counteracts this. – eyeballfrog May 31 '17 at 02:47
  • Okay thank you very much. If possible please look at this question https://math.stackexchange.com/q/2303564/335742 – mathnoob123 May 31 '17 at 02:49

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Frictional forces are always in the plane of contact. The frictional forces at A and C are horizontal while at B it is tangent to the sphere. Forces perpendicular to the plane of contact come from the rigidity of the elements in contact. As the ground under the sphere is considered to be rigid, whatever downward force, here from the weight of the sphere, is applied at C is resisted. That is independent of friction.

The rod wants to slip right at A. Because of the friction at A, the rod pivots about that point. There must be frictional force at A to the left to counteract the projected force of gravity. The rod wants to fall down, so makes a downward force at B. That is a frictional force which is counteracted by an upward force at B from the sphere. The sphere experiences a clockwise torque because of the force from the rod. That torque needs to be resisted at C, so the force on the sphere is to the right at C. The net horizontal force on the sphere plus rod needs to be zero, so the frictional force on the rod at A and sphere at C must be equal and opposite.

Ross Millikan
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