Lab 10 PreLab Questions 1 An equal arm balance is made by ba

Lab 10: Pre-Lab Questions 1. An equal arm balance is made by balancing a meter stick at its center, so it looks like a see-saw. The first mass, m 3.0 kg, is hung at 0.25 m to the left of the center. A second mass, m 1.1 r2 0.4 m to the right of center. At what location with respect to the center of the meter stick must a third mass, m3 0.8 kg, be hung for the meter stick to be in equilibrium? Hint t the angles are all 90°.(5 pts.)

Solution

m1 = 3 kg is at r1 = 0.25 m

m2 = 1.4 kg is at r2 = 0.4 m

m3 = 0.8 kg

let the distance of m3 be r3 from the center

m1 * r1 - m2 * r2 - m3 * r3 = 0

3 * 0.25 - 1.4 * 0.4 - 0.8 * r3 = 0

solving for r3

r3 = 0.24 m

m3 is 0.24 m to the right of center