Question

As shown in Fig.7.40, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. (Take g = 9.8 m/s²)

(Hint: Consider the equilibrium of each side of the ladder separately.)

Answer 1

The given situation can be shown as:

N_B = Force exerted on the ladder by the floor point B 

N_C = Force exerted on the ladder by the floor point C

Draw a perpendicular from A on the floor BC. This intersects DE at mid-point H

Hence, F is the mid-point of AD.

FG||DH and F is the mid-point of AD. Hence, G will also be the mid-point of AH.

For translational equilibrium of the ladder, the upward force should be equal to the downward force.

N_c + N_B = mg = 392 … (iii)

For rotational equilibrium of the ladder, the net moment about A is:

Adding equations (iii) and (iv), we get:

Comments

Why do u not consider the hinge foces offered by hinge to the ladders

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It is considered as the pivot point and here hinge force is meaningless