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The problem is as follows:

The figure from below shows a system at equilibrium. The bar is homogeneous and uniform and has a weight of $14\,N$ and the block which is labeled $Q$ has a weight equal to $28\,N$. Find the reaction force in Newtons experienced by the joint attached to the wall.

Sketch of the problem

The alternatives are given as follows:

$\begin{array}{ll} 1.&14\sqrt{2}\,N\\ 2.&7\sqrt{2}\,N\\ 3.&14\,N\\ 4.&21\,N\\ 5.&28\,N\\ \end{array}$

I'm not sure exactly how to assign the equilibrium condition in this problem and where the forces are acting in the bar. Can someone help me with the right approach and and free body diagram for this thing?.

What I've attempted to do was to use the equilibrium condition as:

$\sum ^n_{i=1}\tau_{i}=0$

By following this

$T\cos53^{\circ}\cdot 6a + (T-mg)\cdot 2a - (14 + 28) = 0$

Then the reaction in the joint would be given by:

$R=T\sin 53^{\circ}$

or will be given by a vertical reaction?

$R= mg-T-T\cos53^{\circ}$

Which of these reactions are referring in this problem?

Am I understanding this correctly?. How exactly should be understood the reaction. Can someone help me here?. I would like someone can explain me the meaning of the reaction experienced by the joint this part is very important hence a free body diagram will help me a lot here.

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    $\begingroup$ This question would be more appropriate for the Physics Stack Exchange. $\endgroup$
    – Sam
    Mar 23, 2020 at 9:58

2 Answers 2

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Calling

$$ \cases{ \vec H=h(1,0)\\ \vec V=v(0,1)\\ \vec Q=q(0,-1)\\ \vec W=w(0,-1)\\ \vec R=r(0,1)\\ \vec T=t(-\sin(\theta),\cos(\theta))\\ OQ=(2a,0)\\ OT=(6a,0)\\ OW=(3a,0) } $$

we have the equilibrium conditions

$$ \cases{ \vec H + \vec V + \vec Q + \vec R + \vec T + \vec W = 0\\ (\vec R+\vec Q)\times OQ+\vec W\times OW + \vec T\times OT =0\\ t+r-q=0 } $$

thus we obtain

$$ \left\{ \begin{array}{rcl} h-t \sin (\theta )&=&0 \\ r+v-w-q+t \cos (\theta )&=&0 \\ q-r-t&=&0 \\ 2 a q-2 a r+3 a w-6 a t \cos (\theta )&=&0 \\ \end{array} \right. $$

and solving

$$ \cases{ h= \frac{3 w \sin (\theta )}{2 (3 \cos (\theta )-1)}\\ v= \frac{w (3 \cos (\theta )+1)}{2 (3 \cos (\theta )-1)}\\ t=\frac{3 w}{2 (3 \cos (\theta )-1)}\\ r= \frac{6 q \cos (\theta )-2 q-3 w}{2 (3 \cos (\theta )-1)} } $$

and with $q = 28N, w = 14N, \theta = \frac{53}{180}\pi$ we get at

$$ h = 20.8225N\approx 21N $$

which is the horizontal wall reaction force component.

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The correct answer is $14\sqrt{2}\,N$.

The free body diagram of the 28 N weight $Q$ is as follows, where

  • T is the tension in the string
  • Fc is the force of contact between the weight and the bar.

Free body diagram of Q

From this it can be seen that $Fc = 28-T$.

The free body diagram of the bar is as follows. The key here is the left end of the bar is pinned. That is, the "blue blob" in the original sketch is not a weld; it allows rotation of the bar.

free body diagram of bar

Since the bar is in equilibrium, the sum of the moments about the left end is zero.

$$\begin{align} \sum Moments=0&=Fc(2a)+14(3a)-T\cos(53)(6a)\\ &=(28-T)(2a)+14(3a)-T\cos(53)(6a) \end{align}$$ from which $T=17.466\,N$.

The sum of horizontal forces equal zero, so $Rh=T\sin(53)=13.95\,N$.

The sum of vertical forces equal zero, $$\begin{align} Rv&=Fc+14-T\cos(53)\\ &=(28-T)+14-T\cos(53)\\ &=14.02\,N \end{align}$$

Since the question refers to "the" reaction, that refers to the resultant of Rh and Rv. $$\begin{align} R&=\sqrt{Rh^2+Rv^2}\\ &\approx\sqrt{14^2+14^2}\\ &=\sqrt{14^2\times2}\\ &=14\sqrt{2}\,N \end{align}$$

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