Three blocks are placed on a smooth inclined plane with force acting on m1 parallel to the inclined plane. Find the contact force between m2 and m3.
A
(m1+m2+m3)Fm3
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B
m3Fm1+m2+m3
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C
F−(m1+m2)g
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D
None of these
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Solution
The correct option is Bm3Fm1+m2+m3 Analysing the motion of (m1+m2+m3) as a system,
Acceleration of the system, a=Net pushing forceTotal mass =F−(m1+m2+m3)gsinθm1+m2+m3−−(i)
Let N be the contact force between blocks m2 and m3.
Drawing FBD of m3 and analysing its motion,
Using the above figure, we can write N−m3gsinθ=m3a−−(ii)
Using equation (i) in (ii), we get N=m3gsinθ+m3[F−(m1+m2+m3)gsinθm1+m2+m3] ⟹N=m3Fm1+m2+m3