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Angle of Friction

Find the maxi...

Question

Find the maximum mass $m_{2}$ that can be hanged from the pulley such that the system remains in equilibrium. Coefficient of friction between block $m_{1}$ & wedge is $m u$ .

$m_{2} = m_{1} (s i n θ + μ c o s θ)$

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$m_{2} = m_{1} (s i n θ - μ c o s θ)$

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$m_{2} = \frac{m_{1}}{(s i n θ + μ c o s θ)}$

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$m_{2} = \frac{m_{1}}{(s i n θ - μ c o s θ)}$

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Solution

The correct option is A
$m_{2} = m_{1} (s i n θ + μ c o s θ)$

So maximum mass is asked that means if I increase the mass even a little the system should start moving. So if I increase $m_{2}$ little bit more than maximum, then $m_{2}$ block will go down and $m_{1}$ will have a tendency to slide up the block so friction on block $m_{1}$ will be maximum and in downward direction.

Lets draw free body diagram of equilibrium condition

T= $m_{2} g$

N= $m_{1} g c o s θ$ ..........(i)

$f_{r} = μ N = μ m_{1} g c o s θ$

T= $m_{1} g s i n θ + μ m_{1} g c o s θ$ ........(ii)

solving (i) & (ii)

$m_{2} g = m_{1} g s i n θ + μ m_{1} g c o s θ$

$m_{2} = m_{1} (s i n θ + μ c o s θ)$

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Find the maximum mass m2 that can be hanged from the pulley such that the system remains in equilibrium. Coefficient of friction between block m1 & wedge is mu.

Find the maximum mass $m_{2}$ that can be hanged from the pulley such that the system remains in equilibrium. Coefficient of friction between block $m_{1}$ & wedge is $m u$ .