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 \[\frac{\cos \theta /2}{\sin \theta /2}>\mu \] \[\cot \theta /2>\mu \]You need to login to perform this action.
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