(III) A uniform rod AB of length 5.0 m and mass M = 3.8 kg is hinged at A and he
ID: 1610577 • Letter: #
Question
(III) A uniform rod AB of length 5.0 m and mass M = 3.8 kg is hinged at A and held in equilibrium by a light cord, as shown in Fig. 9-67. A load W = 22 N hangs from the rod at a distance d so that the tension in the cord is 85 N. (a) Draw a free-body diagram for the rod. (b) Determine the vertical and horizontal forces on the rod exerted by the hinge. (c) Determine d from the appropriate torque equation. (III) You are on a pirate ship and being forced to walk the plank (Fig. 9-68). You are standing at the point marked C. The plank is nailed onto the deck at point A, and rests on the support 0.75 m away from A. The center of mass of the uniform plank is located at point B. Your mass is 65 kg and mass of the plank is 45 kg. What is the minimum down ward force the nails must exert on the plank to hold it in place?Explanation / Answer
27.solution
The tension has components Tx = Tsin = 85N * sin37º = 51.15 N
and Ty = Tcos = 85N *cos37º = 67.88 N
A)
Sum the vertical forces: F + Ty = mg + W
F + 67.88 N = 3.8 kg * 9.8m/s² + 22N
F = - 8.64 N (up? not intuitive)
B)
Sum the horizontal forces: F = Tx = 51.15 N
28.solution
Let F = minimum downward force of nails
moment/torque is measured about plank PIVOT
moment/torque of nails* = F(0.75) CCW
moment/torque of U = mg(3.0 - 0.75) = 65*(9.8)*(2.25) = 1433.25 m-N CW
moment/torque of plank = mg(0.75) = 45(9.8)(0.75) 331 m-N CW
Condition of Equilibrium:
0.75F = 1433.25 + 331 = 1764.25 N
F = 1764.25 / 0.75 2352.33 N ANS
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