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
This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at University of California, Berkeley.
 Fall '08
 all
 Acceleration, Force

Click to edit the document details