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June 2009 p4 q4
4010
A block of mass 8 kg is at rest on a plane inclined at 20° to the horizontal. The block is connected to a vertical wall at the top of the plane by a string. The string is taut and parallel to a line of greatest slope of the plane (see diagram).
(i) Given that the tension in the string is 13 N, find the frictional and normal components of the force exerted on the block by the plane.
The string is cut; the block remains at rest, but is on the point of slipping down the plane.
(ii) Find the coefficient of friction between the block and the plane.
Solution
(i) Resolve forces parallel to the plane:
\(F + T = 8 \times 10 \sin 20^{\circ}\)
\(F + 13 = 8 \times 10 \times 0.342\)
\(F = 27.36 - 13 = 14.36 \text{ N}\)
Frictional component is 14.4 N (rounded).
Resolve forces normal to the plane:
\(R = 8 \times 10 \cos 20^{\circ}\)
\(R = 80 \times 0.94 = 75.2 \text{ N}\)
Normal component is 75.2 N.
(ii) When the string is cut, the block is on the point of slipping, so frictional force \(F = 8 \times 10 \sin 20^{\circ}\).
Coefficient of friction \(\mu = \tan 20^{\circ}\).