question_answer

A cubical block of mass m rests on a rough horizontal surface. \[\mu \] is the coefficient of static friction between the block and the surface. A force mg acting on the cube at an angle \[\theta \] with the vertical side of the cube pulls the block. If the block is to be pulled along the surface, then the value cot (\[\theta \]/2) is:

A)  less than \[\mu \]                          

B)  greater than \[\mu \]

C)  equal to \[\mu \]                           

D)  not dependent on \[\mu \]

Correct Answer: B

Solution :

                 A force mg acts on a cube of mass m at angle \[\theta \] with the vertical side of cube and pulls the block. The forces acting on the block are: (i) Applied force mg at an angle \[\theta \] with vertical side of cube. (ii) Weight mg of cube vertically downward. (iii) Reaction of surface vertically upward (iv) Friction force\[f\] Resoling the components of mg along horizontal and vertical i.e., \[mg\,\sin \,\theta \]and \[mg\,\cos \,\theta .\] Component \[mg\sin \theta \] moves the block in forward direction for this we have \[mg\sin \theta >f\]                                       ?(i) Also    \[mg\cos \theta +R=mg\] or      \[R=mg(1-cos\theta )\]                       ...(ii) and  \[f=\mu R=\mu mg(1-cos\theta )\]                ...(iii) From Eqs. (i) and (iii), we have \[mg\sin \theta >\mu mg(1-\cos \theta )\] \[\sin \theta >\mu (1-cos\theta )\] or \[2\sin \theta /2\cos \theta /2>\mu 2{{\sin }^{2}}\theta /2\] or            \[\frac{\cos \theta /2}{\sin \theta /2}>\mu \] \[\cot \theta /2>\mu \]