Finding the reaction force at the hinge

Question

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

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

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:

$$\tan^{-1}(\frac{\sin(20)}{\cos(20)})=20°$$

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?

Answer 1

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.

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EDIT>>>

A second problem is with the equation for moments

$$\tau=T\sin(20)*2a\sin(50)-mg*a\sin(50)=0$$

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

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

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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.

Answer 2

Something is not right. Please check the calculation below.