Question

A uniform ladder of 7m rests against a vertical wall with which it makes an angle of 45º, the coefficient of friction between the ladder and the wall is 0.4 and that between ladder and the floor is 0.5. If a man, whose weight is one half of that of the ladder, ascends it, how high will it be when the ladder slips?

Answer 1

X = Distance between A and the man, when the ladder is at the point of slipping. 

W = Weight of the ladder 

Weight of man = W/2 = 0.5W 

Fr₁ = 0.5R₁ ...(i) 

Fr₂ = 0.4R₂ ...(ii) 

Resolving the forces vertically 

R₁ + Fr₂ – W – 0.5W = 0 

R₁ + 0.4R₂ = 1.5W ...(iii) 

Resolving the forces Horizontally 

R₂ – Fr₁ = 0; R₂ = 0.5R₁ ...(iv) 

Solving equation (iii) and (iv), we get 

R₂ = 0.625W, Fr₂ = 0.25W 

Now taking moment about point A, 

W × 3.5cos45° + 0.5W × xcos45° – R₂ × 7sin45° – Fr₂ × 7cos45° 

Putting the value of R₂ and Fr_2, we get X 

= 5.25m.

Comments

What is W