Magnitude and direction of the force exerted by the wall on a rod smoothly hinged at its end on the wall.

Question

A uniform rigid rod $AB$ has length $2\text{m}$ and weight $60\text{N}$. The rod is smoothly hinged at its end $A$ to a vertical wall. The rod is held at an angle of $60^\circ$ downward from the wall by a force of magnitude $F$ $\text{N}$ acting at $B$. The force acts at an angle $\theta$ upwards from the horizontal.

(i) Show that $$F\cos(60^{\circ} - \theta)$$ is approx $26\text{N}.$

(ii) Find the magnitude and direction of the force exerted by the wall on the rod at $A$ when

a) $\theta = 30^\circ$ b) $\theta = 60^\circ$ c) $\theta = 90^\circ$

Answers: a)$52\text{N}$ at $30^\circ$ to the vertical, b) $39.7\text{N}$ at $19.1^\circ$ to the vertical, c) $30\text{N}$ vertical.

Some workings:

For the second part a) $$F\cos(60^\circ −30^\circ)=F\cos(30^\circ)=15\sqrt{3}$$ and let $P$ be the reaction force from the wall on the rod and $\alpha$ be the angle of $P$ to the vertical.

Resolving horizontally: $$P\sinα=F\cosθ$$ so $$P=\frac{F\cos\theta}{\sin\alpha} = \frac{30\cos30^\circ}{\sin30^\circ} ≈ 52\text{N}.$$

Answer 1

For the first part, taking moments about A, you should have$$60m\sin 60=F.2m\cos(60-\theta)$$ and thus $$F\cos(60-\theta)=15\sqrt{3}$$

Answer 2

Here is the way that I did it, for $θ=30^o$ - having $P$ point upwards and to the right ($NE$ direction) from the wall and calling the angle of $P$ to the vertical wall angle $α$:

Using the $Fcos(60-θ)=15√3$ relation to work out $F$:

$$Fcos(60-30)=15√3$$ giving $$F=30N$$

Resolving horizontally:

$$Psinα = Fcosθ$$

Resolving vertically:

$$Pcosα + Fsinθ = W$$

With $θ = 30^o$ we have, on dividing the horizontal by the vertical equation:

$$α =arctan(\frac{30cos30}{60-30sin30}) = 30^o$$

Using the horizontal components equation:

$$P=\frac{30cos30}{sin30}≈51.96N$$

This works for the other angles for $θ$, as well, except for $θ=90^o$ where you have to use the vertical components equation to work out $P$, as it is vertical (so $α=0^o$).