a) Explanation Given information: The given force acts at the point ( 3 , − 1 , 0 ) . Given line is r = ( 2 i − k ) + ( 3 j − 4 k ) t . Calculation: Given that the force is F = 2 i − 5 k acts at the point ( 3 , − 1 , 0 ) and the vector is r = ( 2 i − k ) + ( 3 j − 4 k ) t First find the vector torque about the point on the line say ( 2 , 0 , − 1 ) in r × F . In r , is the vector from the point about which we want torque, to the point at which F acts From ( 2 , 0 , − 1 ) to ( 3 , − 1 , 0 ) the vector is given by r = ( 3 , − 1 , 0 ) − ( 2 , 0 , − 1 ) r = ( 1 , − 1 , 1 ) Therefore the vector torque is r × F = | i j k 1 − 1 1 2 0 − 5 | = i ( 5 − 0 ) − j b) To determine To find: the torque of F about the line r = i + 4 j + 2 k + ( 2 i + j − 2 k ) t .

Mathematical Methods in the Physical Sciences

3rd Edition

ISBN: 9780471198260

Author: Mary L. Boas

Publisher: Wiley, John & Sons, Incorporated

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Question

Chapter 6.3, Problem 20P

To determine

To find: the torque of F .

To determine

To find: the torque of F about the line r=i+4j+2k+(2i+j−2k)t .

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Chapter 6.3, Problem 19P

Chapter 6.4, Problem 1P

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Chapter 6 Solutions

Mathematical Methods in the Physical Sciences

Ch. 6.3 - If A=2ijk,B=2i3j+k,C=j+k, find (AB)C,A(BC),(AB)C,...Ch. 6.3 - For Problems 2 to 6, given A=i+j2k,B=2ij+3k,C=j5k:...Ch. 6.3 - For Problems 2 to 6, given A=i+j2k,B=2ij+3k,C=j5k:...Ch. 6.3 - For Problems 2 to 6, given A=i+j2k,B=2ij+3k,C=j5k:...Ch. 6.3 - For Problems 2 to 6, given A=i+j2k,B=2ij+3k,C=j5k:...Ch. 6.3 - For Problems 2 to 6, given A=i+j2k,B=2ij+3k,C=j5k:...Ch. 6.3 - A force F=2i3j+k acts at the point (1,5,2). Find...Ch. 6.3 - Prob. 8P Ch. 6.3 - Prob. 9P Ch. 6.3 - In Figure 3.5, let r be another vector from O to...

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the torque of F .

Solution Summary: The author calculates the torque of F based on ncdot (rtimes F)=135

To find: the torque of F .

To find: the torque of F about the line r=i+4j+2k+(2i+j−2k)t .

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Chapter 6 Solutions