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Holding a meter stick with a 50 gram weight on the end. What is the difference in...

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MaxxVolume | eNotes Newbie

Posted October 8, 2013 at 12:59 AM via web

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Holding a meter stick with a 50 gram weight on the end. What is the difference in torque if I held it parallel to the ground ( - ), slanted upwards ( / ), and slanted downward ( \ )?

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ishpiro | Teacher | (Level 1) Associate Educator

Posted October 8, 2013 at 2:55 PM (Answer #1)

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Assume that you are holing the meter stick at one end and the weight is on the other end.

The torque, relative to the point where you are holding it, will be

`tau=l*mg*sin(theta)`

where l is the length of the stick and `theta` is the angle between the force of gravity (which is always vertical) and the direction of the stick.

If you hold the stick parallel to the ground, that is, the stick is horizontal, the angle between the stick and the gravity is 90 degrees. This means `sin(theta) = 1` , which results in maximum possible torque, `tau = lmg` .

For l = 1 meter and m = 50 gram = 0.05 kg,

`tau = 1*0.05*9.8 = 0.49 N*m`

If the stick is slanted upward, the torque will depend on the angle between the stick and the vertical:

`tau=0.49sin(theta) N*m` .

The larger the angle, which means, the flatter the stick, the greater the torque.

If the stick is slanted downward, the value and direction of torque will be the same if it slanted upward (as long as in both cases the second end is pointing always to the right or always to the left.)

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