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Nov 2007 p4 q5
4094
A ring of mass 4 kg is threaded on a fixed rough vertical rod. A light string is attached to the ring, and is pulled with a force of magnitude \(T\) N acting at an angle of \(60^\circ\) to the downward vertical (see diagram). The ring is in equilibrium.
(i) The normal and frictional components of the contact force exerted on the ring by the rod are \(R\) N and \(F\) N respectively. Find \(R\) and \(F\) in terms of \(T\).
(ii) The coefficient of friction between the rod and the ring is 0.7. Find the value of \(T\) for which the ring is about to slip.
Solution
(i) Resolve horizontally to find the normal force \(R\):
\(R = T \sin 60^\circ\)
Resolve vertically to find the frictional force \(F\):
\(F = W + T \cos 60^\circ\)
Given \(W = 4 \times 10 = 40\) N (weight of the ring),
\(F = 40 + T \cos 60^\circ\)
(ii) Use the equation for limiting friction:
\(F = \mu R\)
Substitute \(\mu = 0.7\), \(R = T \sin 60^\circ\), and \(F = 40 + T \cos 60^\circ\):