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 to Problem 101QAP

Clockwise torque= ${\tau }_{clockwise}=1.75Nm$

Counter-clockwise torquey1 $=$ 3 ${\tau }_{counter-clockwise}=1.75Nm$

Explanation of Solution

Given:

  

Formula used:

Torque can be interpreted as,

  $\tau =Fr$ ( $\tau =$ Torque, $F=$ Force, $r=$ distance from rotational axis)

Calculation:

Consider the clockwise torque,

  ${\tau }_{clockwise}=0.1kg*10m{s}^{-2}*10*{10}^{-2}m+50*{10}^{-3}kg*10m{s}^{-2}*50*{10}^{-2}m+0.2kg*10m{s}^{-2}*70*{10}^{-2}m$

  ${\tau }_{clockwise}=(1+2.5+14)*{10}^{-1}Nm$

  ${\tau }_{clockwise}=1.75Nm$

Consider the vertical forces of the system,

  $F_{pivot}=\left(0.1kg+0.2kg+0.050kg\right)10m{s}^{-2}$

  $F_{pivot}=\left(0.350kg\right)10m{s}^{-2}$

  $F_{pivot}=3.5N$

Consider the counter-clockwise torque,

  ${\tau }_{counter-clockwise}=3.5N*50*{10}^{-2}m$

  ${\tau }_{counter-clockwise}=1.75Nm$

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;

  ${\tau }_{clockwise}=1.75Nm$

  ${\tau }_{counter-clockwise}=1.75Nm$

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.

Answer to Problem 101QAP

Clockwise torque= ${\tau }_{clockwise}=0.4Nm$

Counter-clockwise torque= ${\tau }_{counter-clockwise}=0.4Nm$

Explanation of Solution

Given:

  

Formula used:

Torque can be interpreted as,

  $\tau =Fr$ ( $\tau =$ Torque, $F=$ Force, $r=$ distance from rotational axis)

Calculation:

Consider the clockwise torque,

  ${\tau }_{clockwise}=0.2kg*10m{s}^{-2}*20*{10}^{-2}m$

  ${\tau }_{clockwise}=0.4Nm$

Consider the counter-clockwise torque,

  ${\tau }_{counter-clockwise}=0.1kg*10m{s}^{-2}*(50-10)*{10}^{-2}m$

  ${\tau }_{counter-clockwise}=0.4Nm$

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;

  ${\tau }_{clockwise}=0.4Nm$

  ${\tau }_{counter-clockwise}=0.4Nm$

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.

Answer to Problem 101QAP

Clockwise torque= ${\tau }_{clockwise}=1.75Nm$

Counter-clockwise torque= ${\tau }_{counter-clockwise}=1.75Nm$

Explanation of Solution

Given:

  

Formula used:

Torque can be interpreted as,

  $\tau =Fr$ ( $\tau =$ Torque, $F=$ Force, $r=$ distance from rotational axis)

Calculation:

Consider the clockwise torque,

  ${\tau }_{clockwise}=F_{pivot}*50*{10}^{-2}m$

  ${\tau }_{clockwise}=0.35N*50*{10}^{-2}m$

  ${\tau }_{clockwise}=1.75Nm$

Consider the counter-clockwise torque,

  ${\tau }_{counter-clockwise}=0.2kg*10m{s}^{-2}*(100-70)*{10}^{-2}m+0.050kg*10m{s}^{-2}*(100-50)*{10}^{-2}m+0.100kg*10m{s}^{-2}*(100-10)*{10}^{-2}m$

  ${\tau }_{counter-clockwise}=(0.6Nm+0.25Nm+0.9Nm)$

  ${\tau }_{counter-clockwise}=1.75Nm$

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;

  ${\tau }_{clockwise}=1.75Nm$

  ${\tau }_{counter-clockwise}=1.75Nm$

To determine

  $(d)$

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

Explanation of Solution

Given:

  

Formula used:

Torque can be interpreted as,

  $\tau =Fr$ ( $\tau =$ Torque, $F=$ Force, $r=$ distance from rotational axis)

Calculation:

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

According to conclusion $(a)$,

  ${\tau }_{clockwise}=1.75Nm$

  ${\tau }_{counter-clockwise}=1.75Nm$

So,

  ${\tau }_{clockwise}={\tau }_{counter-clockwise}$$\mathrm{...}(1)$

According to conclusion $(b)$,

  ${\tau }_{clockwise}=0.4Nm$

  ${\tau }_{counter-clockwise}=0.4Nm$

So,

  ${\tau }_{clockwise}={\tau }_{counter-clockwise}$$\mathrm{...}(2)$

According to conclusion $(c)$,

  ${\tau }_{clockwise}=1.75Nm$

  ${\tau }_{counter-clockwise}=1.75Nm$

So,

  ${\tau }_{clockwise}={\tau }_{counter-clockwise}$$\mathrm{...}(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.