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Nov 2009 p41 q4
4082
A particle P of weight 5 N is attached to one end of each of two light inextensible strings of lengths 30 cm and 40 cm. The other end of the shorter string is attached to a fixed point A of a rough rod which is fixed horizontally. A small ring S of weight W N is attached to the other end of the longer string and is threaded on to the rod. The system is in equilibrium with the strings taut and AS = 50 cm (see diagram).
By resolving the forces acting on P in the direction of PS, or otherwise, find the tension in the longer string. [3]
Find the magnitude of the frictional force acting on S. [2]
Given that the coefficient of friction between S and the rod is 0.75, and that S is in limiting equilibrium, find the value of W. [3]
Solution
(i) To find the tension in the longer string, resolve the forces acting on P in the direction of PS. The angle between PS and the vertical is 36.9° (since \\sin^{-1}(0.8) = 36.9°\\). The tension in the longer string T_{PS} is given by:
\(T_{PS} = 5 \sin 36.9^{\circ}\)
Calculating gives \(T_{PS} = 3 \text{ N}\).
(ii) The frictional force F acting on S is found by resolving horizontally. The horizontal component of the tension is:
\(F = T \cos(\sin^{-1}(0.6))\)
\(F = 2.4 \text{ N}\).
(iii) Given the coefficient of friction \(\mu = 0.75\), and S is in limiting equilibrium, the normal reaction R is: