Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 1, Problem 54P

(a)

To determine

The magnitude and angle of displacement vector.

(a)

Expert Solution
Check Mark

Answer to Problem 54P

The magnitude of vector is 10.12m 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 +8.5m and 5.5m .

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

  tanθ=5.5m+8.5mθ=tan1( 5.5m +8.5m)=32.9°

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

  d=8.5mcos( 32.9°)=10.12m

Conclusion:

Thus, the magnitude of vector is 10.12m 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
Check Mark

Answer to Problem 54P

The magnitude of vector is 82.76m/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 75m/s and +35m /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 +35m /s for vy and 75m/s for vx in equation (4).

  tanθ=+35m /s75m/sθ=tan1( +35m /s 75m/s)=25.01°

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

  v=75m/scos( 25.01°)=82.76m/s

Conclusion:

Thus, the magnitude of vector is 82.76m/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
Check Mark

Answer to Problem 54P

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

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=cos1( F x F) ……. (5)

Calculation:

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

  θ=cos1( F x F)θ=cos1( 40lb 50lb)=36.87°

Conclusion:

Thus, the magnitude of vector is 50lb 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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