COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 4, Problem 62QAP
To determine

(a)

The magnitude of F2

Expert Solution
Check Mark

Answer to Problem 62QAP

The magnitude of F2=|F2|=10.7N

Explanation of Solution

Given info:

  COLLEGE PHYSICS, Chapter 4, Problem 62QAP , additional homework tip  1

Mass, M=3.00kg

Acceleration, a=7.00m/s2 in +x direction

And θ1=30.0

Formula Used:

  Fext,x=F1x+F2x

  Fext,y=F1y+F2y

  F2=F2x2+F2y2

  F=Ma

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope 1 applies a force with magnitude F1 at an angle of 30.0.

We need to determine the magnitude and direction of rope 2 's force, F2.

We are told that the box accelerates only in the x-direction at a rate of 7.00m/s2, which means the acceleration in the y-direction is equal to 0.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force 2.

  Fext,x=F1x+F2x=F1cos(30.0)+F2x=M(7.00m/s2)=>F2x=M(7.00m/s2)F1cos(30.0)

  Fext,y=F1y+F2y=F1sin(30.0)+F2y=0=>F2y=F1sin(30.0)

  F2=F2x2+F2y2=(M(7.00 m/s 2) F 1cos( 30.0 ))2+( F 1sin( 30.0 ))2=M2(49.0m2/s4)(14.0m/s2)MF1cos(30.0)+F12cos2(30.0)+F12sin2(30.0)=M2(49.0m2/s4)(14.0m/s2)MF1cos(30.0)+F12

  F2=(3.00kg)2(49.0m2/s4)(14.0m/s2)(3.00kg)(20.0N)cos(30.0)+(20.0N)2F2=10.7N

Conclusion:

The magnitude of F2 is 10.7N

To determine

(b)

A careful drawing to show direction of F2.

Expert Solution
Check Mark

Answer to Problem 62QAP

The below diagram shows the direction of F2.

  COLLEGE PHYSICS, Chapter 4, Problem 62QAP , additional homework tip  2

Explanation of Solution

Given info:

  COLLEGE PHYSICS, Chapter 4, Problem 62QAP , additional homework tip  3

Mass, M=3.00kg

Acceleration, a=7.00m/s2 in +x direction

And θ1=30.0

Formula Used:

  Fext,x=F1x+F2x

  Fext,y=F1y+F2y

  F2=F2x2+F2y2

  F=Ma

  θ2=tan1(F2yF2x)

Calculation:

Consider as two forces are attached to an object of mass M.

Let rope 1 applies a force with magnitude F1 at an angle of 30.0.

We need to determine the magnitude and direction of rope 2 's force, F2.

We are told that the box accelerates only in the x-direction at a rate of 7.00m/s2, which means the acceleration in the y-direction is equal to 0.

We can use Newton's second law in component form in order to calculate the components and, therefore, the magnitude and direction of force 2.

  Fext,x=F1x+F2x=F1cos(30.0)+F2x=M(7.00m/s2)=>F2x=M(7.00m/s2)F1cos(30.0)

  Fext,y=F1y+F2y=F1sin(30.0)+F2y=0=>F2y=F1sin(30.0)

On substituting given values we have: -

  F2x=(M(7.00m/s2)F1cos(30.0))=((3.00kg)(7.00m/s2)(20.0N)cos(30.0))F2x=3.68NF2y=F1sin(30.0)=(20.0N)sin(30.0)F2y=(10.0N)θ2=tan1(10.0N3.68N)=69.8=290.2

290.20

Conclusion:

Thus, we obtained the direction of F2 as below: -

  COLLEGE PHYSICS, Chapter 4, Problem 62QAP , additional homework tip  4

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Chapter 4 Solutions

COLLEGE PHYSICS

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