Note the following relationships:
tan(θ)=FvFh→Fh=Fvtan(θ)
F1=μN1=μm1g
F2=μN2wherem2g=N2+Fv
Fh=Fd+F1+F2
Making the appropriate substitutions give:
Fvtan(θ)=(m1+m2)⋅¯a+μm1g+μ(m2g−Fv)
After some algebraic rearrangement, we arrive at:
Fv=tan(θ)1+μtan(θ)[m1+m2(μg+¯a)]
Hence, P=Fvsin(θ)=tan(θ)sin(θ)[1+μtan(θ)][m1+m2(μg+¯a)]
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