A recovery vessel is salvageing a piece of wreckage from the seabed by pulling it up a steady underwater slope. The slope is inclined at a constant angle β \beta\,β to the horizontal.
The wreckage has a mass of M M\,M kg. It is hauled up the slope by a cable that makes a constant angle θ \theta\,θ with the slope itself. The cable is modeled as light and inextensible, and it remains taut throughout the motion. The wreckage is modeled as a particle.
The coefficient of friction between the wreckage and the seabed is μ\muμ. The wreckage moves up the slope with a constant acceleration a a\,a m s−2^{-2}−2 under a constant tension T T\,T Newtons.
Show that:
T=M(a+gsinβ+μgcosβ)cosθ+μsinθ T = \frac{M(a + g\sin\beta + \mu g\cos\beta)}{\cos\theta + \mu\sin\theta} T=cosθ+μsinθM(a+gsinβ+μgcosβ)In a specific salvage operation, it is found that when M=1200M = 1200M=1200, β=12∘\beta = 12^{\circ}β=12∘, θ=20∘\theta = 20^{\circ}θ=20∘, and T=4500T = 4500T=4500, the wreckage remains stationary. The salvage team attempts to use these values in the formula from part (a), with a=0a = 0a=0, to determine the coefficient of friction μ\muμ. Explain why this method might lead to an incorrect value for μ\muμ.