ecat.montana.edu pm Homework 10 D2L Due Wed. Nov 1 of 2 ment of Inertia and Torq

Question

ecat.montana.edu pm Homework 10 D2L Due Wed. Nov 1 of 2 ment of Inertia and Torque Problem 1: Shown below are six cases in which signs suspended from equal length rigid rods attached to the side of a building. The mass of the rod compared to that of the sign is small and can be ignored The rod in cases B and D is horizontal; in the other cases, the angle that the rod makes with the vertical is given. The length of the rope attached to the signs is also indicated in each figure. Rank the situations based on the magnitude of the torque about the point in which the rod is attached to the side of the building from greatest to least cm 30° 50 cm 100 kg 50 kg 80 kg 90 kg 70 kg 60 kg

Explanation / Answer

here,

Torque is given by :

t = f*d*SinA

Case A:
d = 50 = 0.50 m
F = Tension in string = mg
F = 50*9.8
F = 490 N

t = 0.5 * 490 * sin30
t = 245 * 0.5
t = 122.5 N

Case B:
d = 50 cm = 0.50 m
F = Tension in string = mg
F = 100*9.8
F = 980 N

t = 0.5 * 80 * sin90
t = 450 N

Case C:
d = 50 = 0.50 m
F = Tension in string = mg
F = 80*9.8
F = 784 N

t = 0.5 * 784 * sin60
t = 392 * 0.866
t = 339.48 N

Case D:
d = 1 m
F = Tension in string = mg
F = 90*9.8
F = 882 N

t = 1 * 882 * sin90
t = 882 N

Case E:
d = 1 m
F = Tension in string = mg
F = 70*9.8
F = 686 N

t = 1 * 686 * sin60
t = 686 * 0.866
t = 594.076 N

Case F:
d = 1 m
F = Tension in string = mg
F = 60*9.8
F = 588 N

t = 1 * 588 * sin30
t = 588 * .5
t = 294 N

Case D > Case E > Case B > Case C > Case F > Case A

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