The magnitude and angle of displacement vector.

Question

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Answer 1

Question

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

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Answer 2

Question

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

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Answer 3

Question

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

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Answer 4

Question

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

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Answer 5

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

Answer 6

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Answer 7

Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

Answer 8

(a)

Expert Solution

Answer to Problem 54P

The magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The x and y components of a displacement vector are +8.5 m and −5.5 m .

Formula used:

Write the expression for x component of the displacement vector.

dx=dcosθ

Here, dx is the x component of the displacement vector, d is the magnitude of displacement vector and θ is the angle between the vector and horizontal component.

Rearrange above equation for d .

d=dxcosθ ……. (1)

Write the expression for y component of the displacement vector.

dy=dsinθ

Here, dy is the y component of the displacement vector.

Write the expression for the angle between two vectors.

tanθ=dydx ……. (2)

Calculation:

Substitute −5.5 m for dy and +8.5 m for dx in equation (2).

tanθ=−5.5 m+8.5 mθ=tan−1( −5.5 m +8.5 m)=−32.9°

Substitute −32.9° for θ and +8.5 m for dx in equation (1).

d=8.5 mcos( −32.9°)=10.12 m

Conclusion:

Thus, the magnitude of vector is 10.12 m and the angle taken counterclockwise from the +x direction is −32.9° , it is negative because it is taken in counterclockwise direction.

Answer 13

(b)

To determine

The magnitude and angle of velocity vector.

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Answer 14

(b)

To determine

The magnitude and angle of velocity vector.

Answer 15

(b)

Expert Solution

Answer to Problem 54P

The magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The x and y components of a velocity vector are −75 m/s and +35 m /s .

Formula used:

Write the expression for x component of the velocity vector.

vx=vcosθ

Here, vx is the x component of the velocity vector, v is the magnitude of the velocity vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for v .

v=vxcosθ ……. (3)

Write the expression for the angle between two vectors.

tanθ=vyvx ……. (4)

Here, vy is the y component of the velocity vector.

Calculation:

Substitute +35 m /s for vy and −75 m/s for vx in equation (4).

tanθ=+35 m /s−75 m/sθ=tan−1( +35 m /s −75 m/s)=−25.01°

Substitute −25.01° for θ and −75 m/s for vx in equation (3).

v=−75 m/scos( −25.01°)=−82.76 m/s

Conclusion:

Thus, the magnitude of vector is −82.76 m/s and the angle taken counterclockwise from the +x direction is −25.01° , it is negative because it is taken in counterclockwise direction.

Answer 20

To determine

The magnitude and angle of force vector.

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant

Answer 21

To determine

The magnitude and angle of force vector.

Answer 22

Expert Solution

Answer to Problem 54P

The magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in third quadrant.

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The x component of force vector is 40 lb and magnitude of force vector 50 lb .

Formula used:

Write the expression for x component of the force vector.

Fx=Fcosθ

Here, Fx is the x component of the force vector, F is the magnitude of the force vector and θ is the angle between the vector and horizontal component.

Rearrange the above equation for θ .

cosθ=FxFθ=cos−1( F x F) ……. (5)

Calculation:

Substitute 40 lb for Fx and 50 lb for F in equation (5).

θ=cos−1( F x F)θ=cos−1( 40 lb 50 lb)=−36.87°

Conclusion:

Thus, the magnitude of vector is 50 lb and the angle taken counterclockwise from the +x direction is −36.87° , it is negative because it is in the third quadrant