Question

A young lady (m=45kg) ascends a ladder 12m (M=30kg) positioned against a wall. The ladder makes an angle of θ=70 with the ground as seen in the diagram. The coefficient of static friction between the ladder and ground is 0.25 and b/t the ladder and wall is smooth. Determine how high up the ladder she can ascend before the ladder slips.

Answer 1

Let she can ascend a distance x up the ladder from the ladder base. The weight of the lady acts at this distance vertically downwards.

The ladder's weight acts at 6m vertically downwards at its centre of mass.

The static friction acts along the x direction.

The normal reaction 'N' from contact of ladder and wall acts vertically upwards at that point.

The normal reaction 'F' acts towards left horizontally due to the contact of ladder and wall.

Summing the forces along the vertical direction,

N - mg - Mg = 0

N = g(m+M) = 9.8(45+30) = 735N

Hence maximum frictional force will be Ffr = 0.25*735 = 183.75N

Along the horizontal direction, F = Ffr = 183.75N

Let us take sum of moments about this point.

mg*xcos70 + Mg*6cos70 = Fsin70*12

45*9.8*x*0.342 + 30*9.8*6*0.342 = 183.75*0.939*12

=> 150.8x + 603.3 = 2070.5

x = 9.7 m

Hence answer is 9.7 m.