Torque Worksheet
Torque Worksheet
Do the following on a separate sheet of paper. Show your work! No work, No credit! Watch your units and significant figures! HINT: Draw a picture for each problem!
1. A force of 60 Newtons is applied to the end of a wrench 12 centimeters long. How much torque is produced?
2. 30 Newton•meters of torque is required to close a door 1.5 meters wide. What force is needed to cause this torque?
3. A 90.0 gram mass is 75 centimeters from the pivot point on a balance scale. What mass must be placed 45 centimeters from the pivot to create rotational equilibrium?
4. Two students sit on a see-saw. Archie is a hulking football player with a mass of 120 kg. Clementine is a dainty cheerleader with a mass of 40 kg. The see-saw is 3.5 m in total length with the fulcrum at the center. If Clementine sits at the end on one side, where must Archie sit relative to the center to keep the see-saw balanced?
5. Where should a third 25.0 gram mass be placed on the mobile drawn below so that the mobile will hang motionless? The mobile is 125 centimeters long and the support is at the 50.0-centimeter mark. The first mass, m1 is 25.0 centimeters from the left end of the mobile and the third mass, m3 is at the right end of the mobile (125.0 centimeters).
[pic]
Answer Key
1. Torque = Perpendicular Force x distance = (60 N) x ( 0.12 meters) = 7.2 Nm
2. Torque = Force x perpendicular distance so
[pic]
3. Torque on Left Side of Pivot = Torque on Right Side of Pivot
(90 grams) x ( 75 cm) = ( ? grams) x ( 45 cm) ( 150 grams = 0.15 kilograms
Why was it ok to use grams and Why didn’t I include the acceleration of gravity,
g, in this calculation???
4. Torque on Left Side of Pivot = Torque on Right Side of Pivot
(40 grams) x ( 1.75 m) = ( 120kg) x ( ? m) ( 0.58m
Still ok to leave “g” out of calculation???
5. Torque on Left Side of Pivot = Torque on Right Side of Pivot
(25 grams) x ( 25 cm) + (25 grams) x ( ? cm) = ( 25 grams) x ( 75 cm)
25 cm + ? cm = 75 cm => 50 cm = 0.50 meters
Note that this is 50 cm FROM THE PIVOT, so the mass would be at the very end
of the mobile.
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