Lab 9: Torque & Equilibrium I. To start, draw a free-body diagram of the forces

Question

Lab 9: Torque & Equilibrium I. To start, draw a free-body diagram of the forces acting on the meter stick. Including gravity, your diagram should have five forces! (Omit the weight due to the 300 gram mass because you don't know where to hang it, yet). Choose the rotation axis for all torque calculations to be located at the 30 cm mark of the ruler (i.e. this also the location of the pivot). 2ocm 50gm ogm 2. For the meter stick to be in equilibrium, what must be the net torque about the axis? Explain. x- 2233c 30p 3. How much torque does the force of the pivot support produce about the axis? Explain Toryvt bt(#6fof the f.at of piv.I /5 mo , br@wSt +4 (

Explanation / Answer

7. given mass of ruler, m1 = 56 gram

mass m2 = 200 g

m3 = 500 g

pivot location at 30 cm mark of ruler

m4 = 300 g

let the 300 g mass be located at distance x to the right of the pivot

then from moment balance about the pivot

m1*20 + m2*10 = m3*10 + m4*x

56*20 + 200*10 = 500*10 + 300*x

x = -6.2666

hence the 300 gram mass has to be hung at 30 + 6.2666 = 36.266 cm mark

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