I am looking to find the torque produced by the weight of the meterstick at its

Question

I am looking to find the torque produced by the weight of the meterstick at its center of gravity?

Here is the information I have:

Mass of meterstick is .1265 kg

Weight of the meterstick is 1.2397 N (mass in kg X 9.8 (m/s^2)

Center of gravity is 50.3 cm

The position of the pivot point is 55 cm and the position of the ceter of gravity is 50.3 cm. The lever arm of center of gravity is 4.7 cm. The position of a 1 N weight is 30.8 cm, the lever arm of 1 N weight is 24.2 cm, the torque of the 1 N weight is.242 Nm. The position of a 2 N weight is 70 cm, the lever arm of 2 N weight is 15 cm and the torque of the 2 N weight is .30 Nm.

I need to find the torque produced by the weight of meterstick + torque of 1 N weight.

Explanation / Answer

the torque produced by the weight of the meterstick = F * R

                                                                                  = (0.1265 * 9.8) * 0.503 + 1 * 0.55

                                                                                  =   1.173 N-m

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