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Condition for equilibrium: Torque and force satisfy; total torque and force are zero. Comment about this condition (statement-1): It shows that even if you apply just a tiny force, you can balance the torque due to a very large force, provided that you make your lever-arm sufficiently long. This fact led a well-known mathematician of long ago to claim that he could move the earth if given a long enough lever-arm. Case-1: forces F are applied upward at the ends of a stick of length L, and a force 2F is applied downward at the midpoint. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). If we wish, we may consider the stick to have a pivot at the left end. If we then erase the force F on the right end and replace it with a force 2F at the middle, then the two 2F forces in the middle will cancel, so the stick will remain at rest. Therefore, we see that a force Fapplied at a distance L from a pivot is equivalent to a force 2F applied at a distance L/2 from the pivot, in the sense that they both have the same effect in canceling out the rotational effect of the downwards 2F force. Case-2: forces F are applied upward at the ends, and forces F are applied downward at the L /3 and 2L/3 marks. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). Consider the stick to have a pivot at the left end. From the above paragraph, the force F at 2L/3 is equivalent to a force 2F at L/3. Making this replacement, we now have a total force of 3F at the L/3 mark. Therefore, we see that a force F applied at a distance L is equivalent to a force 3F applied at a distance L/3. Show that a force F applied at a distance L is equivalent to a force nF applied at a distance L/n, and to then argue why this demonstrates "statement-1"

Condition for equilibrium: Torque and force satisfy; total torque and force are zero. Comment about this condition (statement-1): It shows that even if you apply just a tiny force, you can balance the torque due to a very large force, provided that you make your lever-arm sufficiently long. This fact led a well-known mathematician of long ago to claim that he could move the earth if given a long enough lever-arm. Case-1: forces F are applied upward at the ends of a stick of length L, and a force 2F is applied downward at the midpoint. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). If we wish, we may consider the stick to have a pivot at the left end. If we then erase the force F on the right end and replace it with a force 2F at the middle, then the two 2F forces in the middle will cancel, so the stick will remain at rest. Therefore, we see that a force Fapplied at a distance L from a pivot is equivalent to a force 2F applied at a distance L/2 from the pivot, in the sense that they both have the same effect in canceling out the rotational effect of the downwards 2F force. Case-2: forces F are applied upward at the ends, and forces F are applied downward at the L /3 and 2L/3 marks. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). Consider the stick to have a pivot at the left end. From the above paragraph, the force F at 2L/3 is equivalent to a force 2F at L/3. Making this replacement, we now have a total force of 3F at the L/3 mark. Therefore, we see that a force F applied at a distance L is equivalent to a force 3F applied at a distance L/3. Show that a force F applied at a distance L is equivalent to a force nF applied at a distance L/n, and to then argue why this demonstrates "statement-1"

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter10: Fixed-axis Rotation
Section: Chapter Questions
Problem 74P: When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of...
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Condition for equilibrium: Torque and force satisfy; total torque and force are zero. Comment about this condition (statement-1): It shows that even if you apply just a tiny force, you can balance the torque due to a very large force, provided that you make your lever-arm sufficiently long. This fact led a well-known mathematician of long ago to claim that he could move the earth if given a long enough lever-arm. Case-1: forces F are applied upward at the ends of a stick of length L, and a force 2F is applied downward at the midpoint. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). If we wish, we may consider the stick to have a pivot at the left end. If we then erase the force F on the right end and replace it with a force 2F at the middle, then the two 2F forces in the middle will cancel, so the stick will remain at rest. Therefore, we see that a force F applied at a distance L from a pivot is equivalent to a force 2F applied at a distance L/2 from the pivot, in the sense that they both have the same effect in canceling out the rotational effect of the downwards 2F force. Case-2: forces F are applied upward at the ends, and forces F are applied downward at the L /3 and 2L/3 marks. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). Consider the stick to have a pivot at the left end. From the above paragraph, the force F at 2L/3 is equivalent to a force 2F at L/3. Making this replacement, we now have a total force of 3F at the L/3 mark. Therefore, we see that a force F applied at a distance L is equivalent to a force 3F applied at a distance L/3. Show that a force F applied at a distance L is equivalent to a force nF applied at a distance L /n, and to then argue why this demonstrates "statement-1"
Transcribed Image Text:Condition for equilibrium: Torque and force satisfy; total torque and force are zero. Comment about this condition (statement-1): It shows that even if you apply just a tiny force, you can balance the torque due to a very large force, provided that you make your lever-arm sufficiently long. This fact led a well-known mathematician of long ago to claim that he could move the earth if given a long enough lever-arm. Case-1: forces F are applied upward at the ends of a stick of length L, and a force 2F is applied downward at the midpoint. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). If we wish, we may consider the stick to have a pivot at the left end. If we then erase the force F on the right end and replace it with a force 2F at the middle, then the two 2F forces in the middle will cancel, so the stick will remain at rest. Therefore, we see that a force F applied at a distance L from a pivot is equivalent to a force 2F applied at a distance L/2 from the pivot, in the sense that they both have the same effect in canceling out the rotational effect of the downwards 2F force. Case-2: forces F are applied upward at the ends, and forces F are applied downward at the L /3 and 2L/3 marks. The stick will not rotate (by symmetry), and it will not translate (because the net force is zero). Consider the stick to have a pivot at the left end. From the above paragraph, the force F at 2L/3 is equivalent to a force 2F at L/3. Making this replacement, we now have a total force of 3F at the L/3 mark. Therefore, we see that a force F applied at a distance L is equivalent to a force 3F applied at a distance L/3. Show that a force F applied at a distance L is equivalent to a force nF applied at a distance L /n, and to then argue why this demonstrates "statement-1"
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