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June 2012 p43 q6
4027
A block of weight 6.1 N is at rest on a plane inclined at angle \(\alpha\) to the horizontal, where \(\tan \alpha = \frac{11}{60}\). The coefficient of friction between the block and the plane is \(\mu\). A force of magnitude 5.9 N acting parallel to a line of greatest slope is applied to the block.
(i) When the force acts up the plane (see Fig. 1) the block remains at rest. Show that \(\mu \geq \frac{4}{5}\).
(ii) When the force acts down the plane (see Fig. 2) the block slides downwards. Show that \(\mu < \frac{7}{6}\).
(iii) Given that the acceleration of the block is 1.7 m s\(^{-2}\) when the force acts down the plane, find the value of \(\mu\).
Solution
(i) Resolve forces parallel to the plane: The force equation is \(F = 5.9 - 6.1 \sin \alpha\). The normal reaction is \(R = 6.1 \cos \alpha\). For equilibrium, \(5.9 - 6.1 \sin \alpha \leq \mu (6.1 \cos \alpha)\). Given \(\tan \alpha = \frac{11}{60}\), \(\sin \alpha = \frac{11}{61}\) and \(\cos \alpha = \frac{60}{61}\). Substituting, \(5.9 - 6.1 \times \frac{11}{61} \leq \mu \times 6.1 \times \frac{60}{61}\). Simplifying gives \(\mu \geq \frac{4}{5}\).
(ii) When the force acts down the plane, the net force is \(6.1 \times \frac{11}{61} + 5.9 - \mu \times 6.1 \times \frac{60}{61} > 0\). Simplifying gives \(\mu < \frac{7}{6}\).