Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2012 p42 q4
4085
A ring of mass 4 kg is attached to one end of a light string. The ring is threaded on a fixed horizontal rod and the string is pulled at an angle of 25° below the horizontal (see diagram). With a tension in the string of \(T\) N the ring is in equilibrium.
(i) Find, in terms of \(T\), the horizontal and vertical components of the force exerted on the ring by the rod.
The coefficient of friction between the ring and the rod is 0.4.
(ii) Given that the equilibrium is limiting, find the value of \(T\).
Solution
(i) To find the horizontal component of the force exerted by the rod, resolve the tension \(T\) horizontally:
\(T \cos 25^{\circ} = 0.906T\).
To find the vertical component of the force exerted by the rod, resolve the forces vertically:
\(4g + T \sin 25^{\circ} = 40 + 0.423T\).
(ii) Given that the equilibrium is limiting, the frictional force \(F\) is equal to the maximum static friction, which is \(0.4R\), where \(R\) is the normal reaction force. Using the horizontal component: