sol_4.65 - 4.65. O a d G P Figure 1 We rst draw a free-body...

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4.65. θ φ G O P d a Figure 1 We first draw a free-body diagram of the half cylinder as shown in Figure 2. θ φ G O P d a W N F φ Figure 2 The normal reaction force N must be perpendicular to the plane and hence to the tangent at the point P . This force must therefore pass through the center of the half cylinder at O . Summing the forces along the direction of N , we have N = W cos θ 1
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and summing along the direction of the friction force F , F = W sin θ . For there to be no slip, we require F < μ N , with = 0 . 3, so W sin < 0 . 3 W cos , or tan < 0 . 3 and < 16 . 7 o . To find the corresponding value of φ we sum the moments about O , obtaining Wd sin - Fr = 0 (remember that N passes through
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This note was uploaded on 11/22/2011 for the course ME ME211 taught by Professor Barber during the Fall '10 term at University of Michigan.

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sol_4.65 - 4.65. O a d G P Figure 1 We rst draw a free-body...

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