a) A 0.101 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0.609 kg object hangs vertically from the 7.39 cm mark. A second mass is attached at another mark to keep it horizontal and in rotational and translational equilibrium. If the tension in the string attached to the ceiling is 16.7 N, find the value of the second mass. The acceleration due to gravity is 9.8 m/s2 . Answer in units of kg.
b) Find the mark at which the second mass is attached. Answer in units of cm.
A 0.101 kg meter stick is supported at its 40 cm mark by a string attached to the ceiling. A 0.609 kg object hangs vertically from ?
2020-07-11 4:29 pm
回答 (2)
2020-07-11 6:26 pm
✔ 最佳答案
Refer to the figure below:https://live.staticflickr.com/65535/50100332162_4538f88033_o.png
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a)
Consider translational equilibrium:
Total upward forces = Total downward forces
16.7 = (0.609 + 0.101 + m) × 9.8
m = 0.994 kg
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b)
Take moment about the center of the meter stick:
Total clockwise moments = Total anticlockwise moments
16.7 × (50 - 40) + 0.994 × 9.8 × (ℓ - 50) = 0.609 × 9.8 × (50 - 7.39)
9.7412 × (ℓ - 50) = 87.305
ℓ - 50 = 9.0
The mark at which the second mass is attached, ℓ cm = 59.0 cm
2020-07-11 6:11 pm
Let a mass of m kg is attached at a distance of a cm distance from the string on the opposite side of 0.609 kg mass, by balancing the forces vertically:
=>(m + 0.609 + 0.101)g = 16.7
=>m + 0.71 = 1.7
=>m = 0.99 kg ---------------------Answer (a)
by taking the torque at the string:
=>(40 - 7.39) x 0.609 + (40/2) x {0.101 x 40/100} = (60/2) x {0.101 x 60/100} + 0.99 x a
=> a = 19.04 cm
Thus the mark is ≈ (40 + 19) = 59 cm
=>(m + 0.609 + 0.101)g = 16.7
=>m + 0.71 = 1.7
=>m = 0.99 kg ---------------------Answer (a)
by taking the torque at the string:
=>(40 - 7.39) x 0.609 + (40/2) x {0.101 x 40/100} = (60/2) x {0.101 x 60/100} + 0.99 x a
=> a = 19.04 cm
Thus the mark is ≈ (40 + 19) = 59 cm
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