Question

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far along the length of the ladder can a 69.1-kg painter climb before the ladder begins to slip?

Answer 1

Answer:

The weight of the ladder is given as well as the length ;  l = 11.9 m long and weighing W = 46.1 N rests against a smooth vertical wall.

how far along the length of the ladder = x = 4.14m

Explanation:

The detailed step and calculation is as shown in the attached file.

Answer 2

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=$\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip