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A roller of mass 300 kg rests against a step of height 20 cm. If radius of the roller is 50 cm find the minimum horizontal force to be applied passing through its center of mass to take the roller on to the step. |
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Answer» Solution :TAKING the moments of forces about .O. `h=0.2m, R=0.5m, x^(2)=R^(2)-(R-h)^(2)` `x=sqrt(R^(2)-(R-h)^(2))=0.4m` To CLIMB up the step, moment due to horizontal force `ge` moment due to WEIGHT `F xx (R-h)ge mgx` `F ge (mgx)/(R-h)` `F ge (300xx9.8xx0.4)/(0.3)=3920N`
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