EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100654428
Author: Jewett
Publisher: Cengage Learning US
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Chapter 11, Problem 11.6P

(a)

To determine

The quantity cos1(ABAB).

(a)

Expert Solution
Check Mark

Answer to Problem 11.6P

The quantity cos1(ABAB) is 168°.

Explanation of Solution

Given info: The vector B is 6i^10j^+9k^ and vector A is 3i^+7j^4k^

The formula to calculate the dot product of two vector is,

AB=(x1i^+x2j^+x3k^)(y1i^+y2j^+y3k^)=(x1y1+x2y2+x3y3)

Here,

x1 is the x-component of first vector.

x2 is the y-component of first vector.

x3 is the z-component of first vector.

y1 is the x component of second vector.

y2 is the y-component of second vector.

y3 is the z- component of second vector.

Substitute 3 for x1,7 for x2,4 for x3,6 for y1,10 for y2,9 for y3 in the above formula to find AB.

AB=(3)(6)+(7)(10)+(4)(9)=187036=124

Thus, the value of dot product of vectors is 124.

The formula to calculate the magnitude of vector A is,

|A|=(x1)2+(x2)2+(x3)2

Here,

x1 is the x-component of first vector.

x2 is the y-component of first vector.

x3 is the z-component of first vector.

Substitute 3 for x1,7 for x2,4 for x3 in the above formula to find |A|

|A|=(3)2+(7)2+(4)2=9+49+16=74=8.60

Thus, the value of magnitude of vector A is 8.60.

The formula to calculate the magnitude of vector B is,

|B|=(y1)2+(y2)2+(y3)2

y1 is the x component of second vector.

y2 is the y-component of second vector.

y3 is the z- component of second vector.

Substitute 6 for y1,10 for y2,9 for y3 in the above formula to find |B|

|B|=(6)2+(10)2+(9)2=36+100+81=217=14.73

Thus, the value of magnitude of vector B is 14.7.

The formula to calculate the angle between the vectors is,

cosθ=AB|A||B|θ=cos1(AB|A||B|)

Here,

AB is the dot product of  two vectors.

|A||B| is the magnitude of two vectors.

Substitute 124 for AB,8.60 for |A|,14.7 for |B| in the above formula to find θ.

θ=cos1(AB|A||B|)=cos1(124(8.60)(14.7))=168.52°168°

Conclusion:

Therefore, the quantity cos1(ABAB) is 168°

(b)

To determine

The magnitude of the total angular momentum of the system of the system about the axle of the pulley.

(b)

Expert Solution
Check Mark

Answer to Problem 11.6P

The magnitude of the total angular momentum of the system of the system about the axle of the pulley is 11.88°.

Explanation of Solution

Given info: The mass of counterweight is 4.00kg,mass of pulley is 2.00kg,radius of the pulley is 8.00cm.

The formula to calculate the cross product of two vector is,

A×B=|i^j^k^x1x2x3y1y2y3|

Substitute 3 for x1,7 for x2,4 for x3,6 for y1,10 for y2,9 for y3 in the above formula to find A×B.

A×B=|i^j^k^3746109|=i^(7×9(4×9))j^(3×9(4×6))+k^(3×10)(7×6)=23i^+3j^12k^

Thus, the value of cross product of vectors is 23i^+3j^12k^.

The formula to calculate the magnitude of cross product of two vectors is

|A×B|=(a1)2+(a2)2+(a3)2

Here,

a1 is the x-component of |A×B|

a2 is the y-component of |A×B|

a3 s the z-component of |A×B|

Substitute 23 for a1, 3 for a2 and 12 for a3 in the above formula to find |A×B|.

|A×B|=(23)2+(3)2+(12)2=529+9+144=682=26.115

Thus, the value of magnitude of |A×B| is 26.115.

The formula to calculate the magnitude of vector A is,

|A|=(x1)2+(x2)2+(x3)2

Here,

x1 is the x-component of first vector.

x2 is the y-component of first vector.

x3 is the z-component of first vector.

Substitute 3 for x1,7 for x2,4 for x3 in the above formula to find |A|

|A|=(3)2+(7)2+(4)2=9+49+16=74=8.60

Thus, the value of magnitude of vector A is 8.60.

The formula to calculate the magnitude of vector B is,

|B|=(y1)2+(y2)2+(y3)2

y1 is the x component of second vector.

y2 is the y-component of second vector.

y3 is the z- component of second vector.

Substitute 6 for y1,10 for y2,9 for y3 in the above formula to find |B|

|B|=(6)2+(10)2+(9)2=36+100+81=217=14.73

Thus, the value of magnitude of vector B is 14.7.

The formula to calculate the angle between the vectors is,

sinθ=|A×B||A||B|θ=sin1(A×B|A||B|)

Here,

|A×B| is the magnitude of cross  product of  two vectors.

|A||B| is the magnitude of two vectors.

Substitute 26.115 for A×B,8.60 for |A|,14.7 for |B| in the above formula to find θ.

θ=sin1(26.115(8.60)(14.7))=sin1(26.115126.42)=sin1(0.206)=11.88°

Conclusion:

Therefore, the quantity sin1(A×B|A||B|) is 11.88°

(c)

To determine

Whether the angle between the vectors is given buy part (a) or part (b).

(c)

Expert Solution
Check Mark

Answer to Problem 11.6P

The angle between the vectors is given by part (a).

Explanation of Solution

Given info: The vector B is 6i^10j^+9k^ and vector A is 3i^+7j^4k^

From part (a), the value of the angle obtained is 168° and the value of the angle obtained from part (b) is 11.88°. The value of angle obtained in part (a) is same as obtained in part (b) but the difference is the value of angle obtained in part (a) is anticlockwise.

The value of angle obtained in part (a) is,

sin(168°)=sin(180°168°)=cos(11.88°)=cos(11.88°)

Conclusion:

Therefore, the angle between the vectors is given by part (a).

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Chapter 11 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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