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.
130 exam-style questions on Edexcel A Level Maths Applications of Forces, covering 7.1 Static Particles, 7.2 Modelling with Statics, 7.3 Friction and Static Particles, 7.4 Static Rigid Bodies, 7.5 Dynamics and Inclined Planes, and 7.6 Connected Particles 2. Each one has a worked solution and a mark scheme showing where the marks go.