10_InstSolManual_PDF_Part21 - Dynamics of Rotational Motion...

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Figure 10.60 (b) for the lower rod: and for the lower rod: for the middle rod: and for the middle rod: for the upper rod: and for the upper rod: 10.61. Set Up: Apply to the roller, with counterclockwise torque positive and the axis at the point of contact between the roller and the brick. When the roller is on the verge of rolling over the brick, the normal force on it from the ground becomes zero. Solve: (a) The free-body diagram for the roller is given in Figure 10.61a. and are the componets of the force exerted by the brick. Figure 10.61 The moment arm for w is The moment arm for F is 0.400 m. gives and (b) The free-body diagram for the roller is given in Figure 10.61b. Now the moment arm for F is gives and Reflect: Less force is required when the force is applied at the top of the roller. F 5 1 0.300 m
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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