How to find the reaction force from the joint from a bar supported to a wall where a block is resting?

Jun 2017
399
6
Lima, Peru
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.



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.
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
about the pivot, $\tau_{ccw} = \tau_{cw}$

$T\cos(53) \cdot 6a = 14(3a) + (28-T) \cdot 2a \implies T = 17.5 \text{ N}$

$\sum F_x = 0 \implies T\sin(53) = R_x \implies R_x = 14 \text{ N}$

$\sum F_y = 0 \implies T\cos(53) + R_y - (28-T) - 14 \implies R_y = 14 \text{ N}$

$R = 14\sqrt{2} \text{ N}$
 
Jun 2017
399
6
Lima, Peru
about the pivot, $\tau_{ccw} = \tau_{cw}$

$T\cos(53) \cdot 6a = 14(3a) + (28-T) \cdot 2a \implies T = 17.5 \text{ N}$

$\sum F_x = 0 \implies T\sin(53) = R_x \implies R_x = 14 \text{ N}$

$\sum F_y = 0 \implies T\cos(53) + R_y - (28-T) - 14 \implies R_y = 14 \text{ N}$

$R = 14\sqrt{2} \text{ N}$
Can you please include a FBD for this thing?. I'm getting tangled with the direction of the vectors acting in the object.

My major confusion is what's exactly this problem asking?. Is it a reaction caused by the pushing against the wall and experienced by the joint. Or is it the reaction vertically experienced by the joint?. Btw your answer is correct. I think when the question is put without giving any specification it is meant about the total reaction which is the vector sum of both. Am I right with this?

The confusion is exactly which forces are acting vertically?

In other words why is there is a negative sign in front of $-(28-T)$?

The reaction force over the bar where the block is standing isn't going upwards?. Can you help me with this?

$N+T=28$

hence $N=28-T$

When an object is supported over another generates a reaction but the one which is on top experiences the reaction. But what happens to the one which is on bottom. Is it the force with the same sign?. Can you please explain this part please?. :rolleyes:
 
Jun 2017
399
6
Lima, Peru
You try one first ...
Okay here's what I got. It took me some time to figure out where are the forces acting. Please revise my work if I'm getting the idea correctly.



I'm assuming that the reaction created by the block standing over the bar will not produce any torque, but the reaction going downwards it will. Thus is is why it appears of the right side of your equation. Am I right with this?. :)
 

skeeter

Math Team
Jul 2011
3,363
1,854
Texas
The magnitude of the reaction force of the block on the beam, $(28-T)$, exerts a clock-wise torque, $\tau = (28-T) \cdot 2a$, about the wall pivot. Note this is a scalar equation. I'm not going to put a negative sign in front to confuse the issue.

Beam_equilibrium.jpg