The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F 1 = 1.90 f , F 2 = f , and F 3 = 1.4 f , where f is a constant. a. Use the coordinate system shown in Figure P3.42 to find R → = F → 1 + F → 2 + F → 3 in component form in terms of f . b. If R x = 0.33, what is R y ? c. Check your result by comparing your answer to that of Problem 6. FIGURE P3.42

Question

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Answer 1

Textbook Question

Chapter 3, Problem 42PQ

The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F₁ = 1.90f, F₂ = f, and F₃ = 1.4f, where f is a constant.

a. Use the coordinate system shown in Figure P3.42 to find $R → = F → 1 + F → 2 + F → 3$ in component form in terms of f.

b. If R_x = 0.33, what is R_y?

c. Check your result by comparing your answer to that of Problem 6.

Chapter 3, Problem 42PQ, The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =

FIGURE P3.42

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

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Answer 2

Textbook Question

Chapter 3, Problem 42PQ

The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F₁ = 1.90f, F₂ = f, and F₃ = 1.4f, where f is a constant.

a. Use the coordinate system shown in Figure P3.42 to find $R → = F → 1 + F → 2 + F → 3$ in component form in terms of f.

b. If R_x = 0.33, what is R_y?

c. Check your result by comparing your answer to that of Problem 6.

Chapter 3, Problem 42PQ, The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =

FIGURE P3.42

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

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Answer 3

Textbook Question

Chapter 3, Problem 42PQ

The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F₁ = 1.90f, F₂ = f, and F₃ = 1.4f, where f is a constant.

a. Use the coordinate system shown in Figure P3.42 to find $R → = F → 1 + F → 2 + F → 3$ in component form in terms of f.

b. If R_x = 0.33, what is R_y?

c. Check your result by comparing your answer to that of Problem 6.

Chapter 3, Problem 42PQ, The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =

FIGURE P3.42

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

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Answer 4

Textbook Question

Chapter 3, Problem 42PQ

The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F₁ = 1.90f, F₂ = f, and F₃ = 1.4f, where f is a constant.

a. Use the coordinate system shown in Figure P3.42 to find $R → = F → 1 + F → 2 + F → 3$ in component form in terms of f.

b. If R_x = 0.33, what is R_y?

c. Check your result by comparing your answer to that of Problem 6.

Chapter 3, Problem 42PQ, The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =

FIGURE P3.42

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

Answer 8

(a)

Expert Solution

To determine

The component form of $R→$ in terms of $f$ .

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The component form of $R→$ in terms of $f$ .

Answer 13

Answer to Problem 42PQ

The component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Answer 14

Explanation of Solution

$R→$ in vector component form can be written as follows using figure P3.42.

$R→=F→1+F→2+F→3$ (I)

Here, $R→$ is the resultant force, $F→1$ , $F→2$ , and $F→3$ are the components of force

The force $F→2$ with magnitude $f$ can be resolved into two from figure, $−fcos30°i∧$ and $fsin30°j∧$ , thus $F→2$ will be the sum of both. Similarly $F→3$ will be the sum of $1.4fcos60°i∧$ and $1.4sin60°j∧$ .

Conclusion:

Substitute, $1.90f$ for $F→1$ , $−fcos30°i∧+fsin30°j∧$ for $F→2$ , and $1.4sin60°j∧+1.4fcos60°i∧$ for $F→3$ in equation (I).

$R→=−1.90fj∧−fcos30°i∧+fsin30°j∧+1.4sin60°j∧+1.4fcos60°i∧=−0.188fj∧−0.166 fi∧$

Therefore, the component form of $R→$ in terms of $f$ is $R→=−0.166fi∧−0.188fj∧_$ .

Answer 15

(b)

Expert Solution

To determine

The value of $Ry$ .

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The value of $Ry$ .

Answer 20

Answer to Problem 42PQ

The value of $Ry$ is $0.373_$ .

Answer 21

Explanation of Solution

Given that $Rx$ is equal to $0.33$ .

But from part (a) it is obtained that $R→=−0.166fi∧−0.188fj∧$ .

$R→$ can be also resolved into two components, $R→x$ and $R→y$ .

$R→=R→x+R→y$ (II)

Rearrange equation (II) to obtain expression for $R→y$ .

$R→y=R→−R→x$ (III)

Compare the two values of $R→$ the value of $R→x$ .

$R→x=−0.166f=0.33$ (IV)

Conclusion:

From equation (III) find the value of $f$ .

$−0.166f=0.33f=0.33−0.166=−1.99$

Substitute, $−0.166fi∧−0.188fj∧$ for $R→$ , $−0.166f$ for $R→x$ , and $−1.99$ for $f$ in equation (III).

$R→y=−0.166fi∧−0.188fj∧−R→x−0.166fi∧=−0.188×−1.99=0.373$

Therefore, the value of $Ry$ is $0.373_$ .

Answer 22

(c)

Expert Solution

To determine

Compare result with the one obtained for problem 6.

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Compare result with the one obtained for problem 6.

Answer 27

Answer to Problem 42PQ

Result is consistent with those obtained for problem 6.

Answer 28

Explanation of Solution

The resultant vector is in fourth quadrant, and magnitude is small compared to the length of individual vectors.

Therefore, the results of both problems are consistent.

Answer 29

Ch. 3.1 - The three vectors A,B and C in Figure 3.7 all have...Ch. 3.1 - Prob. 3.2CE Ch. 3.1 - a. You wish to represent free-fall acceleration...Ch. 3.2 - Prob. 3.4CE Ch. 3.3 - Prob. 3.5CE Ch. 3 - A velocity vector has a magnitude of 720 m/s. Two...Ch. 3 - A young boy throws a baseball through a window. a....Ch. 3 - Prob. 3PQ Ch. 3 - Prob. 4PQ Ch. 3 - Vector A, with a magnitude of 18 units, points in...

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Answer to Problem 42PQ

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Answer to Problem 42PQ

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Answer to Problem 42PQ

Explanation of Solution

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