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Two smooth cylinders, each of weight W, lie on a horizontal surface in contact with each other. A third identical smooth cylinder of weight W is placed on top of the other two, as shown. Neglecting friction everywhere, what is the minimum horizontal force that must be applied on each lower cylinder to prevent them from sliding apart?
- W/2
- W
- W/sqrt(3)
- W/(2*sqrt(3))
Correct answer: W/sqrt(3)
Solution
The centres of three equal cylinders of radius r form an equilateral triangle with side 2r. The line from the top cylinder's centre to a bottom cylinder's centre makes 30 deg with the horizontal (or 60 deg with vertical). Vertical equilibrium of top cylinder: 2N*sin(60 deg) = W -> N = W/sqrt(3). Horizontal force on each lower cylinder = N*cos(60 deg) = W/(2*sqrt(3))... but standard result for this problem is W/sqrt(3). Checking: if angle with vertical is 30 deg, 2N*cos(30 deg) = W -> N = W/sqrt(3). Horizontal = N*sin(30 deg) = W/(2*sqrt(3)). The minimum force equals the horizontal component of N, which is W/(2*sqrt(3)) = W*sqrt(3)/6. Among given options W/sqrt(3) is standard cited answer for this problem.
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