A vertical rod of mass m is kept on a wedge of mass M. If a horizontal force F acts on the wedge and the rod is constrained to move vertically, after releasing the rod-wedge system, (a) find their speeds when the wedge moves through a distance x. (b) What is the power delivered by the rod on the wedge aftre a time t measured from the instant of release?

Explanation

a. Let the displacement of rod is y when wedge moves a distance x,
`xsintheta=ycosthetaimpliesx=ytantheta`
Also `v_m=v_Mtantheta` and `a_m=a_Mtantheta`
Apply work-energy theorem:
`W=DeltaKE`
`impliesFx-mgy=1/2Mv_M^2+1/2mv_m^2`
`implies(F-mg tan theta)x=1/2Mv_M^2+1/2mv_M^2tan^2theta`

`impliesv_M=sqrt((2[F-mg tan theta]x)/(M+mtan^2theta))`
and
`v_m=v_M tan theta=sqrt((2[F-mg tan theta]x)/(M cot^2 theta+m))`
b. From FBD of wedge and rod,

`F-Nsin theta=Ma_M` (ii)
`N cos theta-mg= ma_m` (ii)
From (i), and (ii)
`a_M=(F-mg tan theta)/(M+m tan^2 theta)`
`N=((Ftan theta+Mg)m)/((M+mtan^2theta)cos theta)`
Velocity of M after time t,
`v_M=a_Mt`
Power given by rod by wedge
`P=-(N sin theta)v_M`
`=-Na_Msin thetat`
`=(-mtantheta[Ftantheta+Mg][F-mg tantheta])/([M+m tan^2theta]^2)`