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Figure 4 shows a uniform metre rule of weight I N with two weights of 0.18 N and 0.12 N suspended from its ends.

Figure 4 shows a uniform metre rule of weight I N with two weights of 0.18 N and 0.12 N suspended from its ends.
balance4182017319.jpg
Determine how far from the 0.18 N weight a pivot should be placed in order to balance the meter rule

Answers


John
Let the pivot be at point X, x cm from the 50 cm mark on the left hand side,

clockwise moment = Anticlockwise moment

(x x 1) + 0.12 (50 + x) = (50 - x)0.18

x = 2.307 cm, Answer = 50 cm - 2.307 cm = 47.69 cm (4 s.f)
johnmulu answered the question on April 18, 2017 at 12:25

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