COLLEGE PHYSICS
COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Chapter 8, Problem 101QAP
To determine

  (a)

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 0cm point.

Expert Solution
Check Mark

Answer to Problem 101QAP

Clockwise torque= τclockwise=1.75Nm

Counter-clockwise torquey1 = 3 τcounterclockwise=1.75Nm

Explanation of Solution

Given:

  COLLEGE PHYSICS, Chapter 8, Problem 101QAP , additional homework tip  1

Formula used:

Torque can be interpreted as,

  τ=Fr ( τ= Torque, F= Force, r= distance from rotational axis)

Calculation:

Consider the clockwise torque,

  τclockwise=0.1kg*10ms2*10*102m+50*103kg*10ms2*50*102m+0.2kg*10ms2*70*102m

  τclockwise=(1+2.5+14)*101Nm

  τclockwise=1.75Nm

Consider the vertical forces of the system,

  Fpivot=(0.1kg+0.2kg+0.050kg)10ms2

  Fpivot=(0.350kg)10ms2

  Fpivot=3.5N

Consider the counter-clockwise torque,

  τcounterclockwise=3.5N*50*102m

  τcounterclockwise=1.75Nm

Conclusion:

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 0cm point are;

  τclockwise=1.75Nm

  τcounterclockwise=1.75Nm

To determine

  (b)

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 50cm point.

Expert Solution
Check Mark

Answer to Problem 101QAP

Clockwise torque= τclockwise=0.4Nm

Counter-clockwise torque= τcounterclockwise=0.4Nm

Explanation of Solution

Given:

  COLLEGE PHYSICS, Chapter 8, Problem 101QAP , additional homework tip  2

Formula used:

Torque can be interpreted as,

  τ=Fr ( τ= Torque, F= Force, r= distance from rotational axis)

Calculation:

Consider the clockwise torque,

  τclockwise=0.2kg*10ms2*20*102m

  τclockwise=0.4Nm

Consider the counter-clockwise torque,

  τcounterclockwise=0.1kg*10ms2*(5010)*102m

  τcounterclockwise=0.4Nm

Conclusion:

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 50cm point are;

  τclockwise=0.4Nm

  τcounterclockwise=0.4Nm

To determine

  (c)

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 100cm point.

Expert Solution
Check Mark

Answer to Problem 101QAP

Clockwise torque= τclockwise=1.75Nm

Counter-clockwise torque= τcounterclockwise=1.75Nm

Explanation of Solution

Given:

  COLLEGE PHYSICS, Chapter 8, Problem 101QAP , additional homework tip  3

Formula used:

Torque can be interpreted as,

  τ=Fr ( τ= Torque, F= Force, r= distance from rotational axis)

Calculation:

Consider the clockwise torque,

  τclockwise=Fpivot*50*102m

  τclockwise=0.35N*50*102m

  τclockwise=1.75Nm

Consider the counter-clockwise torque,

  τcounterclockwise=0.2kg*10ms2*(10070)*102m+0.050kg*10ms2*(10050)*102m+0.100kg*10ms2*(10010)*102m

  τcounterclockwise=(0.6Nm+0.25Nm+0.9Nm)

  τcounterclockwise=1.75Nm

Conclusion:

Clockwise and counter-clockwise torques acting on the board due to the four forces shown about an axis pointing out of the page at the 100cm point are;

  τclockwise=1.75Nm

  τcounterclockwise=1.75Nm

To determine

  (d)

Verify the stability of the stick after the two masses have been added.

Expert Solution
Check Mark

Explanation of Solution

Given:

  COLLEGE PHYSICS, Chapter 8, Problem 101QAP , additional homework tip  4

Formula used:

Torque can be interpreted as,

  τ=Fr ( τ= Torque, F= Force, r= distance from rotational axis)

Calculation:

Consider the conclusions of (a),(b),(c) parts,

According to conclusion (a),

  τclockwise=1.75Nm

  τcounterclockwise=1.75Nm

So,

  τclockwise=τcounterclockwise...(1)

According to conclusion (b),

  τclockwise=0.4Nm

  τcounterclockwise=0.4Nm

So,

  τclockwise=τcounterclockwise...(2)

According to conclusion (c),

  τclockwise=1.75Nm

  τcounterclockwise=1.75Nm

So,

  τclockwise=τcounterclockwise...(3)

Conclusion:

According to (1),(2),(3) expressions,

Stability of the stick after the two masses have been added is still strong. In other words system has stabilized.

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Chapter 8 Solutions

COLLEGE PHYSICS

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