Identify what's going to rotate2. where is it going to rotate (place an X on it)3. Identify the forces4. Draw an Extended Free Body Diagram5. Write a Net Torque Equation What's gonna Rotate?This is aWhere's it gonna Rotate?floating BusStop at someheight, h=0.5m,above theWhy's it gonna Rotate?ground+1.5 m-4.5 m0.5 m

Answered: where is it going to rotate (place an X…

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.122P: The 12-ft wide quarter-circular gate AB is hinged at A. Determine the contact force between the gate...

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Question

Identify what's going to rotate

2. where is it going to rotate (place an X on it)

3. Identify the forces

4. Draw an Extended Free Body Diagram

5. Write a Net Torque Equation

What's gonna Rotate?
This is a
Where's it gonna Rotate?
floating Bus
Stop at some
height, h=0.5m,
above the
Why's it gonna Rotate?
ground
+1.5 m-
4.5 m
0.5 m

Expert Solution

Step 1

With (Figure) and (Figure) for reference, we begin by finding the lever arms of the five forces acting on the stick:

$\begin{array}{ccc}\hfill {r}_{1}& =\hfill & 30.0\,\text{cm}+40.0\,\text{cm}=70.0\,\text{cm}\hfill \\ \hfill {r}_{2}& =\hfill & 40.0\,\text{cm}\hfill \\ \hfill r& =\hfill & 50.0\,\text{cm}-30.0\,\text{cm}=20.0\,\text{cm}\hfill \\ \hfill {r}_{S}& =\hfill & 0.0\,\text{cm}\,\text{(because}\,{F}_{S}\,\text{is attached at the pivot)}\hfill \\ \hfill {r}_{3}& =\hfill & 30.0\,\text{cm.}\hfill \end{array}$

Now we can find the five torques with respect to the chosen pivot:

$\begin{array}{ccccc}\hfill {\tau }_{1}& =\hfill & +{r}_{1}{w}_{1}\text{sin}\,90\text{°}=\text{+}{r}_{1}{m}_{1}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill {\tau }_{2}& =\hfill & +{r}_{2}{w}_{2}\text{sin}\,90\text{°}=\text{+}{r}_{2}{m}_{2}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill \tau & =\hfill & +rw\,\text{sin}\,90\text{°}=\text{+}rmg\hfill & & \text{(gravitational torque)}\hfill \\ \hfill {\tau }_{S}& =\hfill & {r}_{S}{F}_{S}\text{sin}\,{\theta }_{S}=0\hfill & & \text{(because}\,{r}_{S}=0\,\text{cm)}\hfill \\ \hfill {\tau }_{3}& =\hfill & \text{−}{r}_{3}{w}_{3}\text{sin}\,90\text{°}=\text{−}{r}_{3}{m}_{3}g\hfill & & \text{(clockwise rotation, negative sense)}\hfill \end{array}$

The second equilibrium condition (equation for the torques) for the meter stick is

${\tau }_{1}+{\tau }_{2}+\tau +{\tau }_{S}+{\tau }_{3}=0.$

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

$+{r}_{1}{m}_{1}g+{r}_{2}{m}_{2}g+rmg-{r}_{3}{m}_{3}g=0.$

Selecting the

$+y$

-direction to be parallel to

${\overset{\to }{F}}_{S},$

the first equilibrium condition for the stick is

$\text{−}{w}_{1}-{w}_{2}-w+{F}_{S}-{w}_{3}=0.$

Substituting the forces, the first equilibrium condition becomes

$\text{−}{m}_{1}g-{m}_{2}g-mg+{F}_{S}-{m}_{3}g=0.$

We solve these equations simultaneously for the unknown values

${m}_{3}$

and

${F}_{S}.$

In (Figure), we cancel the g factor and rearrange the terms to obtain

${r}_{3}{m}_{3}={r}_{1}{m}_{1}+{r}_{2}{m}_{2}+rm.$

To obtain

${m}_{3}$

we divide both sides by

${r}_{3},$

so we have

$\begin{array}{cc} \hfill {m}_{3}& =\frac{{r}_{1}}{{r}_{3}}\,{m}_{1}+\frac{{r}_{2}}{{r}_{3}}\,{m}_{2}+\frac{r}{{r}_{3}}\,m\hfill \\ & =\frac{70}{30}\,(50.0\,\text{g})+\frac{40}{30}\,(75.0\,\text{g})+\frac{20}{30}\,(150.0\,\text{g})=316.0\frac{2}{3}\,\text{g}\simeq 317\,\text{g.}\hfill \end{array}$

To find the normal reaction force, we rearrange the terms in (Figure), converting grams to kilograms:

$\begin{array}{cc} \hfill {F}_{S}& =({m}_{1}+{m}_{2}+m+{m}_{3})g\hfill \\ & =(50.0+75.0+150.0+316.7)\,×\,{10}^{-3}\text{kg}\,×\,9.8\,\frac{\text{m}}{{\text{s}}^{2}}=5.8\,\text{N}.\hfill \end{array}$

Significance

Notice that (Figure) is independent of the value of g. The torque balance may therefore be used to measure mass, since variations in g-values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

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