The question given is

A uniform rod, of mass $m$ and length $2a$ smoothly hinged to a vertical wall is connected to a point on the wall above the hinge by a light inelastic string. Find the magnitude and direction of the force on the rod from the hinge.

The diagram given is

Diagram 1

I have constructed the forces on to the freebody diagram as such

Diagram 2

Using the vertical component of net force: $$F_x=T \sin(20)-mg+R_V=0$$ $$\implies T\sin(20)=mg-R_V$$

Using the horizontal component of net force: $$F_y=T \cos(20)-R_H=0$$ $$\implies T\cos(20)=R_H$$ $$\implies T=\frac{R_H}{\cos(20)}$$

Finding the moment around the hinge A: $$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$ $$\implies T\sin(20)=\frac{mg}{2}$$ $$\implies T=\frac{mg}{2\sin(20)}$$

Then, equating the two equations for $T\sin(20)$ I get: $$\frac{mg}{2}=mg-R_V$$ $$\implies 2R_V=mg$$

Finally equating both equations for $T$: $$\frac{mg}{2\sin(20)}=\frac{R_H}{\cos(20)}$$ $$\implies \frac{2R_V}{2\sin(20)}=\frac{R_H}{\cos(20)}$$ $$\implies \frac{R_V}{R_H}=\frac{\sin(20)}{\cos(20)}$$

And since $R_V,R_H$ are the vertical and horizontal components of the same force respectively, the $tan^{-1}$ of this ratio is equal to the angle between the resultant and the horizontal hence:


Which is wrong as the answer given in the book is that the angle is $82.7°$ and the reaction force is $0.474mg$, I am aware that this entire method is a hackjob but I have no clue on how to tackle this question properly, are the reaction forces I drew the problem or is it my working onwards?

How would I tackle this problem properly?


2 Answers 2


The problem is that the angle between T and the 20 degrees is not only 20 degrees but 20 + 40=60 deg. Look at the triangles that are formed.


A second problem is with the equation for moments


Although you have included the moment produced by $T_y = T\sin(20+40)$, you need to also include the momemt produced by the x-component i.e. $T_x = T\cos(20+40)$, so the above equation should read

$$\sum M_A = T\sin(60)*2a\sin(50)-T\cos(60)*2a\cos(50)-mg*a\sin(50)=0$$

<<< end of edit

Apart from that the basic equations seem fine. If you redo the calcs with the angle modification you should get the right result.

Also another thing is that I would have used $F_x$ for the horizontal component but that's not important for the calculation. Its more a matter of preference.


Something is not right. Please check the calculation below.

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