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Social ScienceConsider the center of mass of the meter stick is at x= 50.0 cm.Case 1:-50cmdaX1diX2 75cmm1m210.0 cm, Measure and calculate x,100 g, m2 = 200gmX1
Explain how the condition (sum)F = 0 is satisfied for the meter stick in case 1 of the experiment.
Case 1: In Picture.
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Consider the center of mass of the meter stick is at x= 50.0 cm. Case 1: -50cm da X1 di X2 75cm m1 m2 10.0 cm, Measure and calculate x, 100 g, m2 = 200g m X1
Expert Answer
Given information:
Mass of the block (m1) = 100 g = 0.1 kg
Mass of the block (m2) = 200 g = 0.2 kg
Let the mass of the stick be “m”
Measure of x1 is given as = 10 cm = 0.1 m
Measure of x2 is =?
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N X1 0.1 m X2 ? 100g 200g 'mg
For the meter stick to be in equilibrium the sum of the forces in the vertical direction is zero i.e.
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m1g+m2g + mg = N 0.1(9.81)0.2(9.81) mg = N > N = mg + 2.943 N ..... Equation (1)
For the meter stick to be in equilibrium the net moment should also ...
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m1g(x1)mg(X0) + m2g(x2) = N(x0) Substitute the equation (1) in the above equation mig (x1)mg(x) + m2g(x2) = mg(xo) 2.943(x Substitute the corresponding values, we get: 0.981(0.1)1.962(x2) = 2.943(0.5) X2 0.7 m
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Newtons Laws of Motion
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