ladder slip:
A uniform ladder having length of 7m rests against vertical wall with which it makes an angle of 45º, coefficient of friction between ladder and wall is 0.4 and that between ladder and the floor is 0.5. If a man having weight is one half of that of ladder, ascends it, how high will it be when ladder slips?
Sol.: Let,
X = Distance between A and man, when ladder is at point of slipping.
W = Weight of ladder
Weight of man = W/2 = 0.5W
Fr_{1} = 0.5R_{1 } ...(i)
Fr_{2} = 0.4R_{2 } ...(ii)
Resolving forces vertically
R_{1} + Fr_{2} - W - 0.5W = 0
R_{1} + 0.4R2 = 1.5W ...(iii)
Resolving forces Horizontally
R_{2} - Fr_{1} = 0; R_{2} = 0.5R_{1} ...(iv)
By solving equation (iii) and (iv), we get
R_{2 }= 0.625W, Fr_{2} = 0.25W
Now taking moment about the point A,
W × 3.5cos45° + 0.5W × xcos45° - R_{2} × 7sin45° - Fr_{2} × 7cos45°
By putting the value of R_{2} and Fr_{2}, we get
X = 5.25m .......ANS