A uniform stationary ladder of length L = 4.5 m and mass M = 11 kg leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. The coefficient of static friction between floor and ladder is μ = 0.49. The ladder makes an angle θ = 54° with respect to the floor. A painter of mass 8M stands on the ladder a distance d from its base.
a. Find an expression for the magnitude of the normal force N exerted by the floor on the ladder.
b. Find an expression for the magnitude of the normal force NW exerted by the wall on the ladder.
c. Find an expression for the largest value of dmax for which the ladder does not slip.
d. What is the largest value for d, in centimeters, such that the ladder will not slip? Assume that ladder is 4.5 m long, the coefficient of friction is 0.59, the ladder is at an angle of 43.5°, and the ladder has a mass of 55 kg.

a)  N = 9 Mg

, b)N_w =  μ 9M

, c)  

Explanation:

a) For this part we write the equations of trslacinal equilibrium

Axis y

       N - Mg - 8M g = 0

       N = 9 Mg

        N = 9 11 9.8

        N = 970.2 N

b) the force on the horizontal axis (x) som

        fr -N_w = 0

        fr = N_w

friction force is

       fr = μ N

      N_w =  μ 9M

g

      fr = 0.59 970.2

      fr = N_w = 572,418 N

c) For this part we must use rotational equilibrium.

         Στ = 0

We set a frame of reference at the bottom of the ladder and assume that the counterclockwise acceleration is positive

the weight of it is at its midpoint (L / 2)

      - W L /2 cos 54 - 8M d_max cos 54+ NW L sin 54 = 0

        8M d_max cos 54 = - W L / 2 cos 54 + NW L sin 54

       d_max = L (-Mg 1/2 cos 54 + NW sin 54) / (8M cos 54)

       d_max = L (-g / 16 + μ 9Mg / 8M tan 54)

       d_max = L ( 9/8 μ g tan 54- g/16)

explanation:

as the height from which the ruler was dropped increased, the speed of the ruler at ryan's hand increased.

d eras and periods