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Physics - Static Equilibrium?
A 155g stick is supported at the 43cm mark and has a 19g mass on the 92cm mark. Where would a 48g mass need to be located in order to balance the meter stick? (cm).
1 Answer
- igorotboyLv 71 decade agoFavorite Answer
If the fulcrum of this system is at 43 cm, then you have two torques on one side - from the 19 g mass and the 155 g center of gravity of the stick. We need to place the 48 g mass somewhere on the other side to balance it out:
f1d1 = f2d2 + f3d3
(0.048 kg)(9.81 m/s^2)d2 = (0.155 kg)(9.81 m/s^2)(0.07 m) + (0.019 g)(9.81 m/s^2)(0.49 m)
(0.471 N)d1 = 0.106 Nm + 0.091 Nm
(0.471 N)d1 = 0.197 Nm
d1 = 0.418 m
So the weight has to be placed 41.8 cm away from the fulcrum, or pretty close to the 1 cm mark.
