EPChap12-Solution - Chap. 12. HW Questions 2. Figure 12-16...

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1 Chap. 12. HW Questions 2. Figure 12-16 shows four overhead views of rotating uniform disks that are sliding across a frictionless floor. Three forces, of magnitude F , 2F , or 3F , act on each disk, either at the rim, at the center, or halfway between rim and center. The force vectors rotate along with the disks, and, in the “snapshots” of Figure 12-16, point left or right. Which disks are in equilibrium? Assume positive direction is to RIGHT, assume positive rotation is counterclockwise [ccw] a) sum of forces along horizontal = 0 sum of torques about center = FR – 2F R/2 +0 = 0 IN EQUILIBRIUM b) sum of forces along horizontal = 4F sum of torques about center = -FR +0+FR = 0 NOT IN EQUILIBRIUM c) sum of forces along horizontal = 0 sum of torques about center = = -FR +0+FR = 0 IN EQUILIBRIUM d) sum of forces along horizontal = 0 sum of torques about center = -FR – 0 - 2FR = -3FR NOT IN EQUILIBRIUM 3. Figure 12-17 shows three situations in which the same horizontal rod is supported by a hinge on a wall at one end and a cord at its other end. Without written calculation, rank the situations according to the magnitudes of (a) the force on the rod from the cord, (b) the vertical force on the rod from the hinge, and (c) the horizontal force on the rod from the hinge, greatest first. a) Take torque about hinge. The torque due to mg must balance the torque due T. So for all three cases the torque due to T must be the same. This means that T is smallest for (2). The ranking is: (1) and (3) tie for largest T then (2) is smallest. b) Taking torque about the right side eliminates T and only includes mg and F v . Since mg does not change F v is the same for all three cases. c) If the x component of T is made bigger (i.e.
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EPChap12-Solution - Chap. 12. HW Questions 2. Figure 12-16...

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