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Paper 4 (9164, Section B) · November 2004 · Connected Particles

A particle A of mass 6m6m kg rests on a rough plane inclined at 30∘30^{\circ} to the horizontal, with coefficient of friction 0.25. A light inextensible string runs from A over a smooth pulley at the top of the plane to a particle B of mass 2m2m kg hanging freely. The system is released from rest. Does A slide down the plane?

AYes: the limiting friction is 0.25×2mgcos⁡30∘=0.433mg0.25\times2mg\cos30^{\circ}=0.433mg, which is less than the driving force mgmg.
BNo: the driving force 6mgsin⁡30∘−2mg=mg6mg\sin30^{\circ}-2mg=mg is less than the limiting friction 1.299mg1.299mg.
CYes: 6mgsin⁡30∘=3mg6mg\sin30^{\circ}=3mg exceeds 2mg2mg, so A slides down whatever the friction.
DNo: B is lighter than A, so the string stays slack and nothing moves.

Explanation

The test for motion compares the net driving force with the limiting friction. Down the plane A is pulled by 6mgsin⁡30∘=3mg6mg\sin30^{\circ}=3mg and held back by the tension, which cannot exceed B's weight 2mg2mg while B is on the point of rising, leaving mgmg to overcome friction. The limiting friction uses A's own normal reaction, R=6mgcos⁡30∘R=6mg\cos30^{\circ}, so it is 0.25×6mgcos⁡30∘=1.299mg0.25\times6mg\cos30^{\circ}=1.299mg. Since mg<1.299mgmg<1.299mg, the system stays at rest. Comparing 3mg3mg with 2mg2mg ignores friction altogether, and the friction depends on the mass on the plane, not on the hanging mass.

Derived from ZIMSEC Mathematics 9164/4 Paper 4, November 2004, Q15

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