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Question

In the arrangement shown in the figure [sin37=35]


A
direction of force of friction is up the plane.
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B
the magnitude of force of friction is zero.
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C
the tension in the string is 20 N.
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D
magnitude of force of friction is 56 N
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Solution

The correct option is A direction of force of friction is up the plane.

By balancing force on 10 kg mass in perpendicular direction to the incline we get
N=mgcos37
N=100×45N=80 N
Now we know friction is given by
f=μN=0.7×80f=56 N
By balancing force on 4 kg mass we get
T40=4a .....(i)
where a is the acceleration with which it is going upward
Now balancing force on 10 kg block in direction along the incline we get
100sin37Tf=10a ....(ii)
Acceleration of both the blocks will be same.
From (i) and (ii)
f=2014a .....(iii)

From (iii) we can conclude that
as we increase the magnitude of a, the value of friction is decreasing.
So
f20
Hence maximum value of friction can be 20
And T can be calculated from force balancing of 4kg block
T=40N
So only option (a) is correct as friction acts in the direction opposite to the tendency of slipping

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