In the figure below, the light, taut, unstretchable cord B joins block 1 and the
ID: 2228149 • Letter: I
Question
In the figure below, the light, taut, unstretchable cord B joins block 1 and the larger-mass block 2. Cord A exerts a force on block 1 to make it accelerate forward. http://www.webassign.net/sercp9/4-p-034.gif (a) How does the magnitude of the force exerted by cord A on block 1 compare with the magnitude of the force exerted by cord B on block 2? Is it larger, smaller, or equal? equal smaller larger (b) How does the acceleration of block 1 compare with the acceleration (if any) of block 2? smaller larger equal (c) Does cord B exert a force on block 1? If so, is it forward or backward? Is it larger, smaller, or equal in magnitude to the force exerted by cord B on block 2? Yes; forward and greater than Yes; forward and equal Yes; forward and less than Yes; backward and greater than Yes; backward and equal Yes; backward and less than NoExplanation / Answer
Forces on block 1: Pull force P Tension T, in cord B Weight (w1) Normal force (N1) Forces on block 2: Tension T, in cord B Weight (W) Normal force (N2) The normal forces are unknowns, but we don't care about their values. Had friction been involved, we would, but it isn't. Define acceleration rightward as positive. Because cord B is inextensible, the blocks share a common acceleration. Because cord B has negligible mass, tension is the same on both sides, because of Newton's 3rd law. Newton's 2nd law equation for block 1: P - T = m1*a Newton's 2nd law equation for block 2: T = m2*a Solve our system for unknowns T and a: Substitute tension: P - m2*a = m1*a Gather a's: P = (m1 + m2)*a Thus, our concluding formulas are: a = P/(m1 + m2) T = P*m2/(m1 + m2) (A) Tension in cord B (T) is less than pull force P, because block 1 must have a sum of forces acting on it toward the right. (B) The accelerations are common to both blocks, because cord B is inextensible. (C) Yes, because of Newton's 3rd law. If block A pulls block B, then block B pulls block A.