A person of mass m2= 85.0 kg is standing on a rung, one third of the way up a ladder of length d= 3.0 m. The mass of the ladder is m1= 15.0 kg, uniformly distributed. The ladder is initially inclined at an angle θ= 35.0∘ with respect to the horizontal. Assume that there is no friction between the ladder and the wall but that there is friction between the base of the ladder and the floor with a coefficient of static friction μs .

Question

Start this problem by drawing a free-body force diagrams showing all the forces acting on the person and the ladder. Indicating a choice of unit vectors on your free-body diagrams may be helpful.

(a) Using the equations of static equilibrium for both forces and torque, find expressions for the normal and horizontal components of the contact force between the ladder and the floor, and the normal force between the ladder and the wall. Consider carefully which point to use for computing the torques. Determine the magnitude of the frictional force (in Newton) between the base of the ladder and the floor below.

fs=

(b) Find the magnitude for the minimum coefficient of friction between the ladder and the floor so that the person and ladder does not slip.

μs=

(c) Find the magnitude Cladder,ground (in Newton) of the contact force that the floor exerts on the ladder. Remember, the contact force is the vector sum of the normal force and friction.

Cladder,ground=

unanswered

Find the direction of the contact force that the floor exerts on the ladder. i.e. determine the angle α (in radians) that the contact force makes with the horizontal to indicate the direction.

α=

unanswered

Answer 1

b) Find Mu_s first
Mu_s= (m_p/3 + m_l/2) *cotan (theta)/(m_p+m_l)

a) Force = Mu_s*g(m_p+m_l)

I don't know the rest.

Answer 2

a) fs = gcotanè(mp/3+ml/2)