2013: Problem 11

Topic: System of Masses
Concepts: Inclined plane, Newton’s laws


Solution:

Since the triangular block remains at rest, the net force on it is zero. In addition to its own weight, the block feels the normal forces from the two cubes and the ground. As the cubes are effectively sliding down an incline, the normal force on the left cube has magnitude

N1=mgcosαN_1=mg\cos\alpha

while the normal force on the right cube has magnitude

N2=mgcosβN_2=mg\cos\beta

By Newton’s 3rd law, the normal forces from the cubes on the block have the same magnitude in the opposite direction. Thus, the free-body diagram for MM is as follows:

Balancing forces in the vertical direction,

Ng=Mg+mgcos2α+mgcos2βN_g=Mg+mg\cos^2\alpha+mg\cos^2\beta

Since β=π2α\beta=\frac{\pi}{2}-\alpha,

cosβ=cos(π2α)=sinα\cos\beta=\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha

Since sin2θ+cos2θ=1\sin^2\theta+\cos^2\theta=1,

Ng=Mg+mgcos2α+mgsin2α=Mg+mgN_g=Mg+mg\cos^2\alpha+mg\sin^2\alpha=Mg+mg

so the answer is C.

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