In figure $m_{1} = 1$ kg and $m_{2} = 4$ kg. Find the mass $M$ (in nearest integer) of the hanging block which will prevent the smaller block from slipping over the triangular block (angle of wedge= $30^{o}$ ). All the surfaces are frictionless and the strings and the pulleys are light :

7 k g

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4 k g

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1 k g

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5 k g

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Solution

The correct option is A $7 k g$
The acceleration of whole system will be
$a = \frac{M g}{m_{1} + m_{2} + M}$ ...(1)
Since mass $m_{1}$ is not slipping
let the normal force be , so for mass $m_{1}$
$N c o s 30 = m_{1} g$ ..(2)
$N s i n 30 = m_{1} a$ ...(3)
on solving 2 and 3 we get
$a = \frac{g}{\sqrt{3}}$ ...(4)
putting the value of $a$ in equation 1
we get $M = 6.83 k g$

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In figure m1=1 kg and m2=4 kg. Find the mass M (in nearest integer) of the hanging block which will prevent the smaller block from slipping over the triangular block (angle of wedge= 30o). All the surfaces are frictionless and the strings and the pulleys are light :

The correct option is A 7kgThe acceleration of whole system will be a=Mgm1+m2+M ...(1) Since mass m1 is not slipping let the normal force be , so for mass m1Ncos30=m1g ..(2)Nsin30=m1a ...(3)on solving 2 and 3 we geta=g√3 ...(4)putting the value of a in equation 1we get M=6.83kg

In figure $m_{1} = 1$ kg and $m_{2} = 4$ kg. Find the mass $M$ (in nearest integer) of the hanging block which will prevent the smaller block from slipping over the triangular block (angle of wedge= $30^{o}$ ). All the surfaces are frictionless and the strings and the pulleys are light :