Some solutions!
Problems:
6. (I will use g = 10m/s^2 to make things easier to calculate.)
a)weight = 200 N, normal force 200 N
b) Table exerts 300 N, bottom box exerts a normal force ON the top box of 100 N.
28.
a)FBD should include F at 45 degrees below the horizontal to0 the right, a friction force directed to the left, and the weight (mg) directed downwards.
b) Friction force F = 88 N cos(45) = 62.2 N
c) Fn = 88 N sin 45 + (14.5 kg)(9.8 m/s^2) = 204.3 N
78.
page 267:
12. T1 (diagonal cord) = 3920 N, T2 = 3395 N
13. right cord = 176.9 N, left cord = 234.8 N
16. tension = 48.2 N. The tension is so great because of the small angle. 2T sin(theta) = mg. Since theta is small, T is large compared with the weight mg.
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