A 150-kg ladder leans against a smooth wall, making an angle of 30 degrees with the floor. The centre of gravity of the ladder is one-third the way up from the bottom. How large a horizontal force must the floor provide if the ladder is not to slip?

Question

anonymous · Answer

Check the attached file. It shows all the forces acting on the ladder. Wall is smooth so it can apply only normal force ,N1. Floor is applying both frictional force, F and normal force, N2.

Equating the vertical forces, we get

N2 = mg

Now apply rotational equilibrium condition. Take moments about A, we will get

N2 * (BC) = mg * (DE) + F * (AB)      .............................. (1)

We know that N2 = mg. Also you can find BC,DE,AB in terms of AC using trigonometry in triangles ADE and ABC. You also have to use AD = 2/3 AC as it is given that center of mass is at a one third distance of length of ladder from its foot. Apply trigonometry and find values of BC,DE,AB in terms of AC and put in eqn 1. Then AC will cancel out on both sides of the equation and you can easily solve for "F" which is the required horizontal force to be applied by the floor.