When mass M is at the position shown, it is sliding down the inclined part of a

Question

When mass M is at the position shown, it is sliding down the inclined part of a slide at a speed of 2.45 m/s. The mass stops a distance S2 = 2.5 m along the level part of the slide. The distance S1 = 1.12 m and the angle ? = 28.10?. Calculate the coefficient of kinetic friction for the mass on the surface.

Explanation / Answer

As in any mechanics problem, you have to do a sum of force for every body. Then, Sum of forces on block b Tb - mb g = mb a Sum of forces on block c Tc - mc g = mc a Sum of forces on block a Tb - Tc -Fr = ma a In the case of constant velocity a=0 and thus Tb = mb g Tc = mc g Fr = Tb - Tc = (mb-mc) g and uk = Fr / (ma g) = (mb - mc)/ma where Tb and Tc are the tensions in the string at the respective points. In this case mb > mc. Since the system is symmetric then it doesn't matter which is mb and which is mc as long as mb > mc. To calculate the acceleration, simply go back to the original set of equations above Tb - mb g = mb a Tc - mc g = mc a Tb - Tc - uk ma g = ma a Now you have a system of three equations with three unknowns (Tb, Tc, and a). Go ahead and solve for a. _________A______ |...................6...............| C.................................B 1.5...............................3 Net force f = (B - C - kA)g = 0 as A is traveling at a constant velocity. So k = (B - C)/A = 1.5/6 = .25 Note that kAg = friction force on the table. The reason you can find k is exactly because there is no acceleration and, therefore, no net force on the system (the three blocks). Net force f = ma = (B + c + A)a = (B - c - kA)g; so that a = (B - c - kA)g/(B + c + A) = (3 - 1.2 -.25*6)9.81/(3 + 1.2 + 6) = .288 m/sec^2 to the right. Once you found k from the first question, finding the acceleration is just a matter of solving f = ma, where m = B + c + A is the sum of the three masses and c = C - .3 kg the reduced mass.

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