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An equilateral prism of mass m rests on a rough horizontal surface (coefficient of static friction mu). A horizontal force F is applied on the prism (at the top edge). If friction is high enough to prevent sliding before toppling, find the minimum force required to topple the prism.

1. mg / sqrt(3)
2. mg / 4
3. mu*mg / sqrt(3)
4. mu*mg / 4

Correct answer: mg / sqrt(3)

Solution

Equilateral prism (cross-section is equilateral triangle) of side a. Height of CG from base = h/3 where h = a*sqrt(3)/2. So CG height = a*sqrt(3)/6 (from base... actually for equilateral triangle CG is at 1/3 of height from base, so height of CG from base = (1/3)*(a*sqrt(3)/2) = a*sqrt(3)/6. No: height h = a*sqrt(3)/2, centroid at h/3 = a*sqrt(3)/6 from base. Horizontal distance of CG from right bottom edge = a/2 (half base, since CG is at the horizontal centre). Force F is applied horizontally. If F is applied at the top vertex (height = h = a*sqrt(3)/2), torque of F about bottom-right edge = F * (a*sqrt(3)/2). Torque of mg (restoring) about bottom-right edge = mg * (a/2). For toppling: F*(a*sqrt(3)/2) >= mg*(a/2). F >= mg*(a/2)/(a*sqrt(3)/2) = mg/sqrt(3). So minimum F = mg/sqrt(3).

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