Asked by Ammar Hafez
a meter stick of wight 0.8 n is pivoted at 40 cm mark at which mark 1n load should be located to balence the sticke
Answers
Answered by
Writeacher
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Answered by
drwls
Since the stick by itself is unbalanced with the fulcrum off center, the weight must be placed on the short (40 cm) side. Let x be the distance of the 1 N weight from the fulcrum, measured toward the short end, in cm. The stick's weight acts at the center of mass at x = -10 cm
0.8*(-10) + 1*x = 0 at equilbrium
x = 8 cm
0.8*(-10) + 1*x = 0 at equilbrium
x = 8 cm
Answered by
Anonymous
0.8*40-1x=0
x=32cm
x=32cm
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