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Question: Draw the Reaction Force at point P.

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My Attempt and Reasoning: Since the Reaction or the Normal Force is normal to the surface the solution according to me should be-

enter image description here

However, the correct answer is this- enter image description here

I don't understand why is this true? The question mentioned that the Force F acts in such a manner so that the system in equilibrium. In that case, the Part D. seems to be correct as it is opposite to the direction of the vectorial sum of the Force F and W. In general should one know whether a system is in equilibrium to find out the direction of the Reaction Force or is there a direct rule that allows us to find the direction?

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    $\begingroup$ If the system is in equilibrium then all forces must sum to zero because if there were a net force then there must be an associated net acceleration. So $\mathbf{F} + \mathbf{W} + \mathbf{R} = 0$. $\endgroup$ – John Rennie Apr 1 '16 at 8:27
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    $\begingroup$ Remember friction that holds it all up - without it, the whole thing would fall. $\endgroup$ – Steeven Apr 1 '16 at 8:50
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For three non co-linear forces the lines of action of the three force must meet at a point otherwise there would be a net torque and hence the system would not be in equilibrium.

Vector addition of the three forces produces a triangle which means that the resultant (sum) of the three force is zero.

enter image description here

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If the system is in mechanical equlibrium, then it implies that: $$\sum \vec{F}_{i} = 0$$

  • So part D is correct, since $\vec{F}+\vec{R}+\vec{W} = 0$, as you already figured out.

  • The part C in which there is no opposing force for $W$ would cause the rod to accelerate downward.

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The answer D is correct. The Reaction force is vector addition of Normal force and the friction force.
I think you are considering C as the answer because its the normal force. but the normal force will act from point of contact, so that option is wrong
I hope that I answered your question, If you find any mistake or question, feel free to comment

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Let me provide an engineer’s- point of view-answer.

Fixed support and reactions:

The beam PQ has to be a stable structure, so first of all we must ensure that it is provided with the proper support. In your example, we call this kind of support as ‘fixed support’. You can find some details about the types of supports in the above link:

https://en.wikipedia.org/wiki/Support_%28structure%29

So ‘fixed support’ makes our beam stable by ‘exerting’ the proper reactions- forces $Ry$ and $Rx$ but also a torque-moment Mp.

Now as already mentioned in other answers, in order of static equilibrium, must: $ΣFx=o <=> Rx=F$ and $ΣFy=0 <=> Ry=W$ and so: $R(→) = Rx(→)+ Ry(→)$

and also:

$ΣMp=o <=> Mp=-(F*h-W*d)$.

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    $\begingroup$ thank you for the detailed answer. I just wanted to know whether the Reaction Force is always chosen to be opposite to the vectorial sum of all other forces (as in this case). $\endgroup$ – model_checker Apr 1 '16 at 19:06

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