CHAPTER 4 - 8

# CHAPTER 4 - 8 - 37 There is no acceleration From the...

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Unformatted text preview: 37. There is no acceleration. From the symmetry of the force diagram, we see that the tension is the same on each side of Arlene. We write ?F = ma from the force diagram: ?F = ma : 2F sin θ – mg = 0, which gives y y T F Tmin = mg/2 sin θ 3 max 2 = (50.0 kg)(9.80 m/s )/2 sin 10° = 1.4 × 10 N. 38. There is no acceleration. From symmetry we have 3 pairs of wires, with the vertical components equal to one­sixth the weight. We write ?F = ma from the force diagram: ?F = ma ; y y F = mg /6 = (70.0 kg)(9.80 m/s )/6 = 114 N; T1 T2 2 F cos 30° = mg/6 = (70.0 kg)(9.80 m/s )/6, which gives F = 132 N. T2 2 2 F cos 60° = mg/6 = (70.0 kg)(9.80 m/s )/6, which gives F = 229 N. T3 T3 L FA y θA θB FB x 39. There is no acceleration perpendicular to the line L. We write ?F = ma from the force diagram: ?F = ma ; x x F sin θ – F sin θ = 0; B B B A A F sin 30° – (4500 N) sin 50° = 0, which gives F = 6890 N. B The resultant force is in the y­direction: ?F = F cos θ + F cos θ = (6890 N) cos 30° + (4500 N) cos 50° = 8860 N. y B B A A FN θ y θ x mg 40. (a) From the force diagram for the block, we have ?F = ma: x­component: mg sin θ = ma; y­component: F – mg cos θ = 0. N From the x­equation we find the acceleration: a = g sin θ = (9.80 m/s ) sin 22.0° = 3.67 m/s . (b) For the motion of the block, we find the speed from v = v + 2a(x – x ); v = 0 + 2(3.67 m/s )(12.0 m – 0), which gives v = 9.39 m/s. 2 0 0 2 2 2 2 2 ...
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CHAPTER 4 - 8 - 37 There is no acceleration From the...

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