Two straight ladders, AC and BC, have equal weights $W$. They stand in equilibrium with their feet, A and B, on rough horizontal ground and their tops freely hinged at C. Angle CAB = 60$ ^o$, angle CBA = 30$^o$ and AB = $l$ (see diagram). Find the vertical reactions at A and B.
Resolving moments about A:
$$\frac12l\sin30^o\cdot\sin30^o\cdot W + (l-\frac12l\sin60^o\cdot\sin60^o)W = lN$$
$$\implies N = \frac{3W}4$$
Resolving moments about B:
$$\frac12l\sin60^o\cdot\sin60^o\cdot W + (l-\frac12l\sin30^o\cdot\sin30^o)W = lR$$
$$\implies R = \frac{5W}4$$
Find the magnitude and direction of the force exerted on BC by AC.
I couldn't work out how to solve this?
The required answer is $\frac12W$, 30$^o$ above horizontal.
Best Answer
You'll need to include a vertical and horizontal force at point C due to ladder AC (since we shouldn't assume a direction for the total force -- even though our intuition may prove to be correct). The vertical forces on ladder BC are then related by $F_v + N - W = 0$ and you have already established that $N = \frac{3W}{4}$.
There are now five forces acting on ladder BC, but if we look at the force moments about B, we can neglect the torque due to N and the static friction acting on the foot of the ladder. We can then sum the torques due to W , $F_v$, and $F_h$ to zero. You should indeed find that the total force F at C (using the Pythagorean Theorem) comes to $\frac{W}{2}$ and the direction is 30ยบ above the horizontal (I find it pointing to the right).