Determine the smallest counter-clockwise couple moment M that will cause cylinde

Question

Determine the smallest counter-clockwise couple moment M that will cause cylinder 2 to rotate. Clearly demonstrate if cylinder 1 also rotates or remains stationary.

The two cylinders in Figure 2 each have a mass of 75 kg. The coefficients of static friction on the surfaces are muA = muB = muC = 0.25 and muD = 0.4. Using appropriate free body diagrams Determine the smallest counter-clockwise couple moment M that will cause Cylinder 2 to rotate. Your answer must clearly demonstrate if Cylinder 1 also rotates or remains stationary.

Explanation / Answer

F = Friction force,

N = Normal force

For cyl2:

Fd + Nc = mg

Nd = Fc

M = (Fc + Fd)*r

Also,

Fa = 0.25*Na

Fb = 0.25*Nb

Fc = 0.25*Nc

Fd = 0.4*Nd

Therefore,

0.4*Nd + Nc = mg

Nd = 0.25*Nc

M = (0.25*Nc + 0.4*Nd)*r

0.4*0.25*Nc + Nc = mg

Nc = mg / (1 + 0.4*0.25)

M = (0.25*Nc + 0.4*0.25*Nc)*r

M = 1.4*0.25*Nc*r

M = 1.4*0.25*[mg / (1 + 0.4*0.25)]*r

M = 1.4*0.25*[75*9.81 / (1 + 0.4*0.25)]*0.5

M = 117.05 Nm

Nc = 75*9.81 / (1 + 0.4*0.25) = 668.86 N

Nd = 0.25*Nc = 167.2 N

For cyl1:

Fd - Fa + Nb = mg

Nd - Na = Fb

M1 = (Fa + Fb + Fd)*r

0.4*Nd - 0.25*Na + Nb = mg

Nd - Na = 0.25*Nb

So, 0.4*Nd - 0.25*(Nd - 0.25*Nb) + Nb = mg

0.4*167.2 - 0.25*(167.2 - 0.25*Nb) + Nb = 75*9.81

Nb = 668.86 N

Na = 167.2 - 0.25*668.86

Na = 0

M1 = (0 + 0.25*668.86 + 0.4*167.2)*0.5

M1 = 117.05 Nm

Since M1 = M, cyl1 will also rotate.

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