A meter stick with a mass of 143 grams is allowed to pivot about its 45 cm mark. How much mass
must be placed at the 13 cm mark in order to balance the meter stick?
Physics problem. Rotational Equilibrium.?
2018-03-02 12:41 pm
回答 (3)
2018-03-02 1:16 pm
✔ 最佳答案
Assume the meter stick is of uniform constructionSum moments about the pivot point to zero
Let m be the unknown mass
143(g)[45 - 50] + m(g)[45 - 13] = 0
143(g)[- 5] + m(g)[32] = 0
m(g)[32] = 143(g)[5]
32m = 715
m = 22.34375
m = 22.3 grams
I hope this helps.
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2018-03-02 1:21 pm
Refer to the diagram below.
Take moment about the pivot.
Total clockwise moment = Total anticlockwise moment
(45 - 13)m = 143 × (50 - 45)
m = 22.3
Mass must placed at the 13-cm mark = 22.3 g
Take moment about the pivot.
Total clockwise moment = Total anticlockwise moment
(45 - 13)m = 143 × (50 - 45)
m = 22.3
Mass must placed at the 13-cm mark = 22.3 g
2018-03-02 1:45 pm
143/100 = 1.43g/cm
1.43 X 45/2 + m(45-13) = 1.43 X 5/2 <<< solve for m
1.43 X 45/2 + m(45-13) = 1.43 X 5/2 <<< solve for m
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