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A block of mass 2kg rests on a rough inclined plane making an angle of `30^@` with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block isA. 9.8 NB. `0.7xx9.8xx sqrt(3)N`C. `9.8xxsqrt(3)N`D. `0.7xx9.8N` |
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Answer» Correct Answer - A Angle of repose, `theta = tan^(-1)(mu_s)=tan^(-1)(0.7)` or `tan theta_r=0.7` Angle of plane is `theta = 30^@, tan theta= tan30^@ = 0.577` Since `tan theta lt tan theta, theta lt theta` Block will not slide or `f=mgsintheta ne mu mgcostheta` or `f=(2)(9.8)sin 30^@ = 9.8N` |
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