We calculate the torque around the step corner. The second diagram indicates that the distance from the line of Fto the corner is r– h, where ris the radius of the wheel and his the height of the step.The distance from the line of mgto the corner is rrhrhh2222+− =−bg. Thus, Fr h mg rh h−−− =bg202.The solution for Fis 2222222(6.00 10 m)(3.00 10 m) (3.00 10 m)2=(0.800 kg)(9.80 m/s )(6.00 10 m) (3.00 10 m)13.6 N.rh hFmgrh−××−×−=−×−×=21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.