CAIE FP2 2014 November — Question 4

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionNovember
TopicMoments

4
\includegraphics[max width=\textwidth, alt={}, center]{5d40f5b4-e3d4-482c-8d8d-05a01bd3b43f-3_513_643_260_749} A uniform rod \(A B\), of length \(l\) and mass \(m\), rests in equilibrium with its lower end \(A\) on a rough horizontal floor and the end \(B\) against a smooth vertical wall. The rod is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\), and is in a vertical plane perpendicular to the wall. The rod is supported by a light spring \(C D\) which is in compression in a vertical line with its lower end \(D\) fixed on the floor. The upper end \(C\) is attached to the rod at a distance \(\frac { 1 } { 4 } l\) from \(B\) (see diagram). The coefficient of friction at \(A\) between the rod and the floor is \(\frac { 1 } { 3 }\) and the system is in limiting equilibrium.
  1. Show that the normal reaction of the floor at \(A\) has magnitude \(\frac { 1 } { 2 } m g\) and find the force in the spring.
  2. Given that the modulus of elasticity of the spring is \(2 m g\), find the natural length of the spring.