The figure shows a block of mass m placed on a smooth wedge of mass M calculate the value of M' and the tension in the string so that the block of mass m will move vertically downward with acceleration of 10 metre per second square (take G as 10 metre per second square)
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The value are M' = Mcotθ / 1 - cotθ and T = 2 N respectively.
Explanation:
For the value of M:
M'g - T = M'a ....(1)
T = Ma .....(2)
M'g / M' + M = a ....(3)
Now
mgcosθ - masinθ = 0
a = gcotθ
M'g / M' + M = a
M'g / M' + M = gcotθ
M' = Mcotθ / 1 - cotθ
Acceleration to the given problem, m is a freely falling body i.e Contact force between M and m is zero.
Under this condition acceleration of M leftwards will be
a = gcotθ = g × √3
tension T=Ma
= 1 / 5√3×√3g
= 2 N .
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