Question

1.  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. Height from the bottom of the ground up to the top of the ladder is 30m. F is not at the middle of the ladder. The componenets acting on the top the ladder are R_L and N_L. The componenets acting on the top the ladder are R₀ and N₀. The distance of force F from the ground is 10m.

Answer 1

Answer

Answer

Let's assume that beam is making angle 45° with horizontal then moment about point O

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

Force required to keep the beam in equilibrium

$35.7(0.3) +P=50cos45°\\P=24.6N$