Question
Two blocks are connected and pulled up along an incline at alpha . The pulling force T2 is angled at the angle of beta with respect to the surface of the incline. Using the notation, axis and terms in the figure answer the questions below. Let, mu be the coefficient of kinetic friction and T1 be the tension in the connecting rope as shown in the figure. Assume the pulling force T2 is just enough to maintain constant uphill speed. What is the equation of motion for object 1 in the x direction ? Simplify as much as you can. Final answer should be only in terms of T1,m1,alpha,mu and g. Do the same for object 2 and simplify as much as you can. Final answer should be only in terms of T1, m2,alpha,mu, T2,beta and g.
Explanation / Answer
Make a free body diagram of each block, and set up Newton's 2nd law in direction down the incline. First block is pulled down the incline by gravity and the string with tension T while frictional force acts up the plane. So m1·a1 = m1·g·sin(?) + T - µ1·m1g·cos(?) So the acceleration of the 1st block is given by a1 = g·sin(?) - µ1·g·cos(?) + T/m1 Second block is pulled down the incline by gravity while string tension and frictional pulls it up: m2·a2 = m2·g·sin(?) - T - µ2·m2·g·cos(?) So the acceleration of the 1st block is given by a2 = g·sin(?) - µ2·g·cos(?) - T/m2 The blocks connected by a stretched string. Therefore the move with the same displacement , velocity and acceleration, i.e. a1 = a2 So you can equate the two accelerations equations and solve for the tension: g·sin(?) - µ1·g·cos(?) + T/m1 = g·sin(?) - µ2·g·cos(?) - T/m2 T/m1 + T/m2 = µ1·g·cos(?) - µ2·g·cos(?) T = g·cos(?)·(µ1 - µ2) / (1/m1 + 1/m2)