question_answer

Consider a block held stationary over a rough incline by applying a force F parallel to the incline as shown below.

Force applied is initially, \[{{F}_{i}}=mg\,(\sin \theta -{{\mu }_{s}}\cos \theta )\] and is increased finally to \[{{F}_{f}}=mg\,(\sin \theta -{{\mu }_{K}}\cos \theta ),\]

where \[{{\mu }_{s}}\] and \[{{\mu }_{K}}\] are coefficient of friction and \[\theta \] is angle of inclination. Now, correct graph of F (Force) versus f (friction) is

A)

B)

C)

D)

Correct Answer: D

Solution :

If applied force is just sufficient to hold the block over incline, friction acts in upwards direction (so treated positive).

\[F={{F}_{i}}=mg\,(\sin \theta -{{\mu }_{s}}\cos \theta )\]

\[\Rightarrow \]\[{{F}_{i}}mg\sin \theta -f\] or \[f=mg\sin \theta -F\]

When \[F=mg\sin \theta ,\]friction \[f\]is zero.

If applied force exceeds \[mg\sin \theta ,\] then block moves upwards and friction now acts downwards.

\[\therefore \]\[{{F}_{f}}=mg(\sin \theta +{{\mu }_{K}}\cos \theta )\]

Hence, friction in now negative.

As \[{{\mu }_{s}}>{{\mu }_{K}},\]graph must be [d].