Block B of mass 100 kg rests on a rough surface of friction coefficient μ = 1/3 . A rope is tied to

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4-2.Friction

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Block $B$ of mass $100 kg$ rests on a rough surface of friction coefficient $\mu = 1/3$. $A$ rope is tied to block $B$ as shown in figure. The maximum acceleration with which boy $A$ of $25 kg$ can climbs on rope without making block move is:

A$\frac{{4g}}{3}$

B$\frac{g}{3}$

C$\frac{g}{2}$

D$\frac{{3g}}{4}$

Solution

Let $a$ be the maximum acceleration of man.
The block will remain fixed if horizontal component of tension in the rope is equal to the friction force of block against the ground. i.e.
$T \cos 37^{\circ}=f=\mu\left(m_{B} g-T \sin 37^{\circ}\right)$
$\frac{4 T}{5}=\frac{100 g}{3}-\frac{T}{5}$
$\therefore T=\frac{100 g}{3}$
Now, $m_{A} a=T-m_{A} g$
or, $25 a=\frac{100}{3} g-25 g$
or, $a=\frac{4}{3} g-g=\frac{g}{3}$

Standard 11

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(AIPMT-2011)

Block $B$ of mass $100 kg$ rests on a rough surface of friction coefficient $\mu = 1/3$. $A$ rope is tied to block $B$ as shown in figure. The maximum acceleration with which boy $A$ of $25 kg$ can climbs on rope without making block move is:

Solution

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