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=[tex]\frac{N2}{mg}*l[/tex] * 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= [tex]\frac{241.75}{69.1*9.8} *l[/tex] * tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
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.
