question_answer

A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. The magnitude of the force acting upwards at an angle of \[60{}^\circ \] from the horizontal that will just start the block moving is

A) \[5\,N\]            

B) \[\frac{20}{2+\sqrt{3}}N\]

C) \[\frac{20}{2-\sqrt{3}}N\]         

D) \[10\,N\]

Correct Answer: B

Solution :

[b] \[R+F\sin 60{}^\circ =mg\] \[R=mg-\frac{\sqrt{3}F}{2}\] If block just starts moving \[F\cos 60{}^\circ =f=\mu R\] Or \[F+\frac{\sqrt{3}F}{2}=10\]or \[F=\frac{20}{2+\sqrt{3}}\]