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Nov 2020 p42 q6
4053
A block of mass 5 kg is placed on a plane inclined at 30° to the horizontal. The coefficient of friction between the block and the plane is \(\mu\).
(a) When a force of magnitude 40 N is applied to the block, acting up the plane parallel to a line of greatest slope, the block begins to slide up the plane (see Fig. 6.1). Show that \(\mu < \frac{1}{5} \sqrt{3}\).
(b) When a force of magnitude 40 N is applied horizontally, in a vertical plane containing a line of greatest slope, the block does not move (see Fig. 6.2). Show that, correct to 3 decimal places, the least possible value of \(\mu\) is 0.152.
Solution
(a) Resolve forces parallel and perpendicular to the plane. The normal reaction \(R = 5g \cos 30\). The force equation parallel to the plane is:
\(40 - 5g \sin 30 - F > 0\)
where \(F = \mu R = \mu \times 5g \cos 30\).
Substitute \(F\) into the inequality:
\(40 - 5g \sin 30 - \mu \times 5g \cos 30 > 0\)
Simplify to find \(\mu\):
\(\mu < \frac{1}{5} \sqrt{3}\)
(b) Resolve forces horizontally and vertically. The normal reaction \(R = 5g \cos 30 + 40 \sin 30\). The frictional force \(F = 40 \cos 30 - 5g \sin 30\).