This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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 ...
View Full Document