( 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 0 − c m point.

Question

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Answer 1

Question

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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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Answer 2

Question

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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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Answer 3

Question

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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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Answer 4

Question

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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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Answer 5

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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.

Answer 6

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 $0−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

Answer 7

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 $0−cm$ point.

Answer 8

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torquey1 $=$ 3 $τcounter−clockwise=1.75Nm$

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.1kg*10ms−2*10*10−2m+50*10−3kg*10ms−2*50*10−2m+0.2kg*10ms−2*70*10−2m$

$τclockwise=(1+2.5+14)*10−1Nm$

$τclockwise=1.75Nm$

Consider the vertical forces of the system,

$Fpivot=(0.1kg+0.2kg+0.050kg)10ms−2$

$Fpivot=(0.350kg)10ms−2$

$Fpivot=3.5N$

Consider the counter-clockwise torque,

$τcounter−clockwise=3.5N*50*10−2m$

$τcounter−clockwise=1.75Nm$

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 $0−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

Answer 13

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 $50−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

Answer 14

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 $50−cm$ point.

Answer 15

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=0.4Nm$

Counter-clockwise torque= $τcounter−clockwise=0.4Nm$

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=0.2kg*10ms−2*20*10−2m$

$τclockwise=0.4Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.1kg*10ms−2*(50−10)*10−2m$

$τcounter−clockwise=0.4Nm$

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 $50−cm$ point are;

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

Answer 20

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 $100−cm$ point.

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

Answer 21

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 $100−cm$ point.

Answer 22

Expert Solution

Answer to Problem 101QAP

Clockwise torque= $τclockwise=1.75Nm$

Counter-clockwise torque= $τcounter−clockwise=1.75Nm$

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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,

$τclockwise=Fpivot*50*10−2m$

$τclockwise=0.35N*50*10−2m$

$τclockwise=1.75Nm$

Consider the counter-clockwise torque,

$τcounter−clockwise=0.2kg*10ms−2*(100−70)*10−2m+0.050kg*10ms−2*(100−50)*10−2m+0.100kg*10ms−2*(100−10)*10−2m$

$τcounter−clockwise=(0.6Nm+0.25Nm+0.9Nm)$

$τcounter−clockwise=1.75Nm$

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 $100−cm$ point are;

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

Answer 27

To determine

$(d)$

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

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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.

Answer 28

To determine

$(d)$

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

Answer 29

Expert Solution

Explanation of Solution

Given:

COLLEGE PHYSICS,VOLUME 1, 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 $(a)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(1)$

According to conclusion $(b)$ ,

$τclockwise=0.4Nm$

$τcounter−clockwise=0.4Nm$

So,

$τclockwise=τcounter−clockwise$ $...(2)$

According to conclusion $(c)$ ,

$τclockwise=1.75Nm$

$τcounter−clockwise=1.75Nm$

So,

$τclockwise=τcounter−clockwise$ $...(3)$

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.

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Ch. 8 - Prob. 1QAP Ch. 8 - Prob. 2QAP Ch. 8 - Prob. 3QAP Ch. 8 - Prob. 4QAP Ch. 8 - Prob. 5QAP Ch. 8 - Prob. 6QAP Ch. 8 - Prob. 7QAP Ch. 8 - Prob. 8QAP Ch. 8 - Prob. 9QAP Ch. 8 - Prob. 10QAP

( 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 0 − c m point.

Videos

$(a)$

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 $0−cm$ point.

Answer to Problem 101QAP

Explanation of Solution

$(b)$

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 $50−cm$ point.

Answer to Problem 101QAP

Explanation of Solution

$(c)$

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 $100−cm$ point.

Answer to Problem 101QAP

Explanation of Solution

$(d)$

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

Explanation of Solution

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Chapter 8 Solutions

( 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 0 − c m point.

Videos

Answer to Problem 101QAP

Explanation of Solution

Answer to Problem 101QAP

Explanation of Solution

Answer to Problem 101QAP

Explanation of Solution

(d)<m:math><m:mrow><m:mo stretchy="false">(</m:mo><m:mi>d</m:mi><m:mo stretchy="false">)</m:mo></m:mrow></m:math> Verify the stability of the stick after the two masses have been added.

Explanation of Solution

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Chapter 8 Solutions

$(d)$

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