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Torque about a Point

Two identical...

Question

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$ . The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

A

f = (\frac{M + m}{2}) g tan θ

horizontally outwards

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B

f = (\frac{M + m}{2}) g cot θ

horizontally inwards

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C

f = (\frac{M + m}{2}) g cos θ

horizontally outwards

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D

None of these.

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Solution

The correct option is B $f = (\frac{M + m}{2}) g cot θ$ horizontally inwards
Drawing the FBD of both rods
As the rods are in equilibrium

$Σ F_{X} = 0$

$Σ F_{y} = 0$

$τ_{n e t} = 0$
For right rod

$N_{2} + \frac{m g}{2} + M g = N$ (1)

$N_{1} = f$ (2)
For left rod

$N_{2} + N = M g + \frac{m g}{2}$ (3)

From $(1) a n d (3) N_{1} = f$

$= N_{2} = 0 = N = M g + \frac{m g}{2}$

Balancing torque about $O$ ,

$M g \frac{L}{2} cos θ + f L sin θ = N L cos θ$

$\frac{M g cos θ}{2} + f sin θ = (M g + \frac{m g}{2}) cos θ$

$f = (\frac{M g}{2} + \frac{m g}{2}) cot θ = f = (\frac{M + m}{2}) g cot θ$

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