Question

A horizontal meter stick has a mass of 211 g. Three weights ride on the meter stick: 253 g at 48.9 cm, 183 g at 77.7 cm, and 201 g at 94.1 cm. At what location on the meter stick would the system be in balance if it were suspended there?

Answer 1

We have to find center of mass of the system:

2 Center of mass cm m2 mxm2.x, m, + m For two masses: Xcm-

Thus, x_cm =( 211 x 50 + 253 x 48.9 + 183 x 77.7 + 201 x 94.1 ) / ( 211 +253 +183+201)

or, x_cm = 56054.9 / 848

or, x_cm = 66.1 cm

At 66.1 cm on the meter stick the system will be in balance if it were suspended there.

Answer 2

The total weight = 211 + 253 + 183+ 201 = 841g (strictly speaking this is a mass and should be converted to newtons to be a force, but this is unnecessary here)

When balanced, there will be an upwards force of 841g at the balance point at a distance x from the end of the metre stick.

When balanced, we can take moments (torques) about any convenient point. Use the end of the ruler (0.0cm) as the point:

841x = (211*50.0) + (253*48.9) + (183*77.7) + (201*94.1) = 56054.9

x = 66.65 cm

You can complete the calculation to find x.