question_answer 81)

                  A rectangular box lies on a rough incline surface. The co-efficient of friction between the surface and the box is \[\mu \]. Let the mass of the box be m.                 (a) At what angle of inclination \[\theta \] of the plane to the horizontal will the box just start to slide down the plane?                 (b) What is the force acting oil the box down the plane, if the angle of inclination the plane is increased to \[a>0\]?                 (c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?                 (d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?

Answer:

                  (a) The box will remain at rest it force of friction \[(f)>mg\,\sin \theta \]                 The box will first start to slide down the plane if                 \[mg\,\,\sin \,\theta \,=f=\,R=\,mg\,\cos \theta \]                 or \[\tan \,\theta \,=\mu \] or \[\theta \,=\,{{\tan }^{-1}}\,(\mu )\]                 (b) Force acting oil die box down the plane                 \[=\,mg\,\sin \,\alpha -\,f=\,mg\,\sin \alpha -\,\mu R\]                 \[=\,mg\,\,\sin \,\alpha -\,\mu \,mg\,\cos \alpha \]                 \[=\,mg\,\,(\sin \,\alpha -\,\mu \,\cos \alpha )\]                 (c) \[F=mg\text{ }\sin \,\theta +f=mg\sin \,\theta +\mu \text{ }mg\cos \,\theta \]                 \[=mg\cos \theta =mg\,(\sin \theta +\mu \,\cos \theta )\]                 (d) Force oil the box, when moves with an acceleration \[\alpha ,\,\,{{f}_{1}}=ma\]                 \Force applied upward                 \[=F+{{f}_{1}}=mg(\sin \theta +\mu \cos \theta )+ma\]