A classical mechanics problem by Ajit Athle
Two right circular rollers (radii, r & R) rest on a rough horizontal plane as shown in the diagram. The larger roller (R) has a string around it with which it can pulled with a force P. Assuming that the coeff. of friction (f) is the same everywhere, determine the condition that the larger roller may be pulled over the smaller roller i.e. show that f>=(r/R)^(1/2). I have tried to draw the FBD but my diagram may not be correct.
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1 solution
Consider horizontal and vertical resolution of forces just after the bigger cylinder has lost contact with the ground i.e. reaction from the ground onto the bigger roller = 0. Let θ = Angle between the line joining the centres & the vertical, W1 = Weight of the bigger roller and N3 = the reaction from the smaller roller on to the bigger one. We've, cos(θ) = (R-r)/(R+r) and sin(θ)=2 √(Rr)/(R+r) W1 = N3cos(θ) + fN3sin(θ) ------------- (1) N3sin(θ) = fN3cos(θ) + P ------------- (2) W1 R sin(θ)=P(1+cos(θ))R ------------ (3) (Taking moments about the point of contact) or W1 = P(1+cos(θ))/sin(θ) & P = N3sin(θ) - fN3cos(θ) Plug these values in (1) above, (N3sin(θ) - fN3cos(θ)(1+cos(θ))/sin(θ) = N3cos(θ) + fN3sin(θ) or (sin(θ) - f cos(θ)(1+cos(θ)) = (cos(θ) + fsin(θ))sin(θ) sin(θ) + sin(θ)cos(θ)-f cos(θ)-f cos²(θ) = sin(θ)cos(θ)+f sin²(θ) or f(sin²(θ) +cos²(θ)) + f*cos(θ) = sin(θ) or f = sin(θ)/(1+cos(θ)) = (r/R)^(1/2)