# How do I solve for the reaction force at the hinge

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: $$F_y=T \sin(20)-mg+R_V=0$$ $$\implies T\sin(20)=mg-R_V$$

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

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?

• Why is there no tension in the rod? Your Tsin20 and Tcos20 makes me think that you think the rod is horizontal which it isn’t. Tcos20 could be a force parallel to the rod and Tsin20 perpendicular to the rod, if you want to do it that way. But they’re not vertical or horizontal forces - you seem to have completed ignored the $50 ^\circ.$ Something else you didn’t do is to take moments of the forces on the rod about $B$, the point where the string and rod meet. Doing so could simplify your task. Jan 23 '21 at 23:32
• Also, either draw the force $T$, or draw the components of $T$, but don’t draw both. Jan 23 '21 at 23:38

HINT for "How would I tackle this problem properly?"

There is a very clever method for tackling this type of problem. The key is not to split the forces up into components as you have done.

The point is that the rod is acted upon by just three forces.

These must add up to $$0$$ and so can be drawn head-to-tail to form a triangle and trig can then be used e.g. cosine rule/sine rule, ...

Even more usefully the three forces must pass through the same point! In your diagram, the forces of $$mg$$ and $$T$$ pass through a point, $$P$$ say, on the rope directly over the middle of the rod. Then these two forces have no moment about $$P$$. But the moment of all the forces about $$P$$ must be zero and so the reaction must also pass through $$P$$ in order to have zero moment about $$P$$.

The entire question is thus reduced to the geometry of triangles.

• Ah that’s clever and a nice answer! Jan 23 '21 at 23:40