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Can somebody please assist me in finding the direction of the normal force exerted on B? (How is normal force at B also normal to the rod? )

There is no hint provided from which I can conclude the direction of normal force. Can somebody please provide a hint so I can work it out?

P.S

Contact force = Normal force since friction is 0.

mathnoob123
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  • Length of AB can be found using sine rule in triangle ABC and then you can write torque about point A .(Consider AB as X and line perpendicular to AB as y axis for simplicity).Remember rod is also in translatory equilibrium.Write equation of motion along AB and line perpendicular to AB. – Abhash Jha May 16 '17 at 18:23
  • Yes, I used the same reasoning at first. But it's not applicable because the point at which the weight of the rod acts is unknown. This is because the midpoint of the rod is unknown(since the rod continues even after the circumference). – mathnoob123 May 16 '17 at 18:27
  • it is at a distance a from point A !! – Abhash Jha May 16 '17 at 18:30
  • Sir, this information is useless. That's pretty much is because all the other measurements are in "r". This problem can be solved by making an equation for a in terms of r but I don't think that's necessary because that's given for third part of the question. – mathnoob123 May 16 '17 at 18:31
  • @AbhashJha Can you please give a try at this question? – mathnoob123 May 16 '17 at 18:51
  • From answers looks like normal at B is perpendicular to rod and at A it is directed towards C – Abhash Jha May 16 '17 at 19:04
  • From answers? I assume you're talking about the mark scheme of this question. Can you please also explain how normal at B is perpendicular to rod? – mathnoob123 May 16 '17 at 19:05

2 Answers2

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It's a very good point, worth to analyze. At B, you can see the force exerted on the bowl by the rod is obviously normal to the rod as no friction force, acting tangentially, exists. Now, from the Newton's third law, the force exerted from the bowl over the rod is normal to the rod too.

mathnoob123
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Rafa Budría
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Assuming that

$$ 2r\cos(\theta) > a $$

we have three forces in equilibrium

$$ F_A = \{\cos(2\theta),\sin(2\theta)\}f_a\\ F_B = \{\cos(\theta+\pi/2),\sin(\theta+\pi/2)\}f_b\\ W = \{0,-1\}w $$

The condition

$$ F_A+F_B+W = 0 $$

gives

$$ f_a = w\tan(\theta)\\ f_b = w\left(\frac{\cos(2\theta)}{\cos(\theta)}\right) $$

NOTE

The force $F_A$ is not normal to the rod.

Cesareo
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