JEE Main 2017 · Physics · Rotational Motion · medium

A uniform ladder of mass 10 kg rests against a smooth vertical wall, making an angle of 60^° with the rough horizontal floor. A man of mass 50 kg…

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A uniform ladder of mass

10\ \mathrm{kg}

rests against a smooth vertical wall, making an angle of

60^\circ

with the rough horizontal floor. A man of mass

50\ \mathrm{kg}

stands on the ladder at a point

\tfrac34

of its length from the bottom. Taking

g = 10\ \mathrm{m\,s^{-2}}

, the friction force exerted by the floor on the ladder (ladder still in equilibrium) is nearly

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Answer: (A) $245\ \mathrm{N}$

Take torques about the foot of the ladder (length

L

). The wall is smooth, so its reaction

N_w

is horizontal, with lever arm

L\sin60^\circ

. The weights act with lever arms

\tfrac L2\cos60^\circ

and

\tfrac{3L}{4}\cos60^\circ

. Thus

N_wL\sin60^\circ = \left(100\cdot\tfrac12 + 500\cdot\tfrac34\right)L\cos60^\circ

, giving

N_w = 425\cot60^\circ = \dfrac{425}{\sqrt3} \approx 245\ \mathrm{N}

. Horizontal equilibrium makes the floor friction

f = N_w \approx 245\ \mathrm{N}

. Using

\tan60^\circ

instead of

\cot60^\circ

gives

736\ \mathrm{N}

, and

425\ \mathrm{N}

forgets the angle altogether.

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake