box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

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Sheetal kumari · Answer

For downward motion F+f = mgsin theta we know f=mu N and N=mgcos thetaSo f=mumgcos thetaF= mgsintheta-mumgcosthetaThen for upward motion2F-f = mgsinthetaf=mumgcostheta2mgsintheta-2mumgcostheta=mumgcostheta-mgsinthetamgsintheta=3mumgcosthetaThen mu= mgsintheta/mgcostheta&times;3mu=tantheta/3 please kindly approved me

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box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

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box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

box of mass 8 kg is placed on a roughinclined plane of inclination theta. Its downwardmotion can be prevented by applying anupward pull F and it can be made to slideupwards by applying a force 2F. What is thecoefficient of friction between the box andthe inclined plane

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