Question

a beam LO weighing 50N rests against a wall, the lower part of the beam extending 15 m from the base of the wall. Calculate the minimum value of the force F that can be applied to keep the beam in equilibrium. The coefficient of friction at L and O is 0.30

Answer 1

Answer

Let's assume that beam is making angle 45° with horizontal then

Moment about point O

$50(\frac{15}{2}) (\sqrt{2})sin45° +0.3N_2(15) (\sqrt{2}) cos45°=N_2(15) \\N_2=35.7N$

Force requires to keep the beam equilibrium

$35.7(0.3) +P=50cos45°\\P=24.6N$

So required force is 24.6N .