Question 49
A uniform rod of length 2L metres and mass m m\,m kilograms rests horizontally in equilibrium on two fixed supports at the points A A\,A and BBB.
A A\,A lies at a distance x x\,x metres from one end of the rod. B B\,B lies at a distance (x+0.15)(x + 0.15)(x+0.15) metres from the other end of the rod.
The reaction force of the support on the rod at B B\,B is three times the reaction force of the support on the rod at AAA.
Show that L−x=kL - x = kL−x=k, where k k\,k is a constant to be found.