In Fig. 6-12, if the box is stationary and the angle θ between the horizontal and force F → is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) F x ; (b) ƒ s ; (c) F N ; (d) ƒ s , max ? (e) If, instead, the box is sliding and θ is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same? Figure 6-12 Question 1.

Question

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Answer 1

Textbook Question

Chapter 6, Problem 1Q

In Fig. 6-12, if the box is stationary and the angle θ between the horizontal and force F → is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) F_x; (b) ƒ_s; (c) F_N; (d) ƒ_{s, max}? (e) If, instead, the box is sliding and θ is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same?

Chapter 6, Problem 1Q, In Fig. 6-12, if the box is stationary and the angle between the horizontal and force F is

Figure 6-12 Question 1.

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

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05:30

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Answer 2

Textbook Question

Chapter 6, Problem 1Q

In Fig. 6-12, if the box is stationary and the angle θ between the horizontal and force F → is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) F_x; (b) ƒ_s; (c) F_N; (d) ƒ_{s, max}? (e) If, instead, the box is sliding and θ is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same?

Chapter 6, Problem 1Q, In Fig. 6-12, if the box is stationary and the angle between the horizontal and force F is

Figure 6-12 Question 1.

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

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05:30

Answer 3

Textbook Question

Chapter 6, Problem 1Q

In Fig. 6-12, if the box is stationary and the angle θ between the horizontal and force F → is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) F_x; (b) ƒ_s; (c) F_N; (d) ƒ_{s, max}? (e) If, instead, the box is sliding and θ is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same?

Chapter 6, Problem 1Q, In Fig. 6-12, if the box is stationary and the angle between the horizontal and force F is

Figure 6-12 Question 1.

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

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05:30

Answer 4

Textbook Question

Chapter 6, Problem 1Q

In Fig. 6-12, if the box is stationary and the angle θ between the horizontal and force F → is increased somewhat, do the following quantities increase, decrease, or remain the same: (a) F_x; (b) ƒ_s; (c) F_N; (d) ƒ_{s, max}? (e) If, instead, the box is sliding and θ is increased, does the magnitude of the frictional force on the box increase, decrease, or remain the same?

Chapter 6, Problem 1Q, In Fig. 6-12, if the box is stationary and the angle between the horizontal and force F is

Figure 6-12 Question 1.

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

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05:30

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

Answer 8

Expert Solution & Answer

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To find:

Whether the following quantities increase, decrease, or remain the same when an applied force is directed downward at an angle θ:

(a) Fx;

(b) fs;

(c) FN;

(d) fs,max

(e) And, whether the magnitude of the frictional force on the box increases, decreases, or remains the same if the angle θ is increased.

Answer 12

Answer to Problem 1Q

Solutions:

(a) The value of horizontal force Fx decreases.

(b) fs also decreases.

(c) The value of the normal force FN increases.

(d) fs,max also increases.

(e) The friction force increases when the box slides and the angle between the surface and box also increases.

Answer 13

Explanation of Solution

Concepts

If the block slides, then kinetic frictional force is given by fk= μk FN. When the block is stationary on an inclined plane, the kinetic frictional force balances the force component that attempts to slide the block along the surface.

Explanations:

Given Data:

In the problem (a) to (b): the box is stationary and the angle θ between the horizontal and force increases somewhat. In part (e), when the box slides and θ increases, the magnitude of the frictional force on the box may increase, decrease, or remain the same.

Note: It is clear from Fig. 6-12, the angle θ between the horizontal and force remains in the first quadrant.

Formula used:

The free-body depiction for the inclined slope is provided below.

From Fig. (6-19) of the textbook and the free body diagram, we can draw:

Fx=mgcosθ, and Fy=mgsinθ

FUNDAMENTALS OF PHYSICS (LLF)+WILEYPLUS, Chapter 6, Problem 1Q

No acceleration in the first case (from (a) to (d)) as the box is in the stationary position. Hence, acceleration is zero.

Applying Newton’s 2^nd law on the y-axis:

FN- Fsinθ-mg=0 (1)

If the block slides, the kinetic frictional force:

fk= μk FN (2)

If it does not slide, then the magnitude of maximum static friction:

fs,max= μs FN= μs ( Fsinθ+mg) (3)

Calculations: To find the magnitude of the different forces and their nature (increasing, decreasing, or no-change) when the angle (θ) at which force is applied is increased on the stationary box:

(a) The horizontal component of the force is Fx=mgcosθ. In this case, the horizontal force Fx decreases as the angle θ between the horizontal surface and force is increased due to the cosine factor in it. Thus, the value of horizontal force Fx decreases.

(b) If a body does not move, the static frictional force and the component parallel to the surface are equal in magnitude, and is directed opposite that component. If the component decreases, fs also decreases.

(c) The normal component of the force is given in Eq. (1).

FN= Fsinθ+mg

The normal component of the force is FN=Fsinθ+mg. In this case, the magnitude of the force FN increases as the angle θ between the horizontal surface and force is increased due to the sine (sin θ) factor in it. Thus, the value of the normal force FN increases.

(d) From Eq. (3), the magnitude of the maximum static friction will also increase as FN increases.

(e) In the sliding scenario, kinetic friction force can be explained by Eq. (2). This results in an increase in the kinetic frictional force. Thus, the friction force increases when the box slides and the angle between the surface and box increases.

Conclusion

If the value of the normal force FN increases, it will enhance both fs,max and fs. In a sliding scenario, kinetic friction force also increases.