Masses of 100 and 500 grams are placed at 0 cm and 100 cm points of a meter stic

Question

Masses of 100 and 500 grams are placed at 0 cm and 100 cm points of a meter stick respectively. Where must a single vertical force be placed to achieve a balance? What is the value of F in Newtons? The meter stick is uniform and has a mass of 80 grams. Position of F __________cm, Magnitude of F _____________Newton A meter stick whose mass is 0.200 kg is supported at the zero cm mark by a knife edge and a force F at the 100 cm point. A mass of 700 grams is attached to the stick at the 40 cm mark. Find the magnitude of N and F in Newtons. Magnitude of N ____________Newton, Magnitude of F ________________Newton

Explanation / Answer

the single force will be = (100+500+80)1000 x9.8 = 6.664 N

position let be x from 100 g mass side

so, (100/1000)x + (80/1000)x = (500/1000)*(1-x) + (80/1000)*(1-x)

or, 0.1x + 0.08x = (0.5+0.08)(1-x)

or, 0.18x = 0.58 - 0.58x

or, x = 0.76315 m = 76.315 cm ........................................ans

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