In this question use g=9.8g = 9.8g=9.8 m s−2^{-2}−2.
A stagehand moves a 15 kg exhibition crate across a rough horizontal floor by pulling a light rope. The rope has constant tension 50 N and makes an angle θ \theta\,θ radians above the horizontal, where 0≤θ<π2\displaystyle 0 \le \theta < \frac{\pi}{2}0≤θ<2π.
The crate is modelled as a particle. The coefficient of friction between the crate and the floor is 0.25, and the frictional force is modelled as 0.25R 0.25R\,0.25R while the crate is moving, where R R\,R is the normal reaction.
Starting from the vertical and horizontal force equations, obtain a=103cosθ+56sinθ−2.45\displaystyle a = \frac{10}{3}\cos\theta + \frac{5}{6}\sin\theta - 2.45a=310cosθ+65sinθ−2.45, where a a\,a m s−2^{-2}−2 is the acceleration of the crate.
Determine the value of θ \theta\,θ for which the acceleration is greatest, and find this greatest acceleration.
Check that the crate remains in contact with the floor at the angle found in part (b).