Question

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Answer 1

Solution

With Figure 12.9 and Figure 12.10 for reference, we begin by finding the lever arms of the five forces acting on the stick:

r1r2rrSr3=====30.0cm+40.0cm=70.0cm40.0cm50.0cm−30.0cm=20.0cm0.0cm(becauseFSis attached at the pivot)30.0cm.$$\begin{array}{ccc}\hfill r_1& =\hfill & 30.0\text{cm}+40.0\text{cm}=70.0\text{cm}\hfill \\ \hfill r_2& =\hfill & 40.0\text{cm}\hfill \\ \hfill r& =\hfill & 50.0\text{cm}-30.0\text{cm}=20.0\text{cm}\hfill \\ \hfill r_S& =\hfill & 0.0\text{cm}\text{(because}{F}_S\text{is attached at the pivot)}\hfill \\ \hfill r_3& =\hfill & 30.0\text{cm.}\hfill \end{array}$$

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

τ1τ2ττSτ3=====+r1w1sin90°=+r1m1g+r2w2sin90°=+r2m2g+rwsin90°=+rmgrSFSsinθS=0−r3w3sin90°=−r3m3g(counterclockwise rotation, positive sense)(counterclockwise rotation, positive sense)(gravitational torque)(becauserS=0cm)(clockwise rotation, negative sense)$$\begin{array}{ccccc}\hfill {\tau }_1& =\hfill & +r_1{w}_1\text{sin}90\text{°}=\text{+}{r}_1{m}_{1}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill {\tau }_2& =\hfill & +r_2{w}_2\text{sin}90\text{°}=\text{+}{r}_2{m}_{2}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill \tau & =\hfill & +rw\text{sin}90\text{°}=\text{+}rmg\hfill & & \text{(gravitational torque)}\hfill \\ \hfill {\tau }_S& =\hfill & r_S{F}_S\text{sin}{\theta }_S=0\hfill & & \text{(because}{r}_S=0\text{cm)}\hfill \\ \hfill {\tau }_3& =\hfill & \text{−}{r}_3{w}_3\text{sin}90\text{°}=\text{−}{r}_3{m}_{3}g\hfill & & \text{(clockwise rotation, negative sense)}\hfill \end{array}$$

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

τ1+τ2+τ+τS+τ3=0.$${\tau }_1+{\tau }_2+\tau +{\tau }_S+{\tau }_3=0.$$

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

+r1m1g+r2m2g+rmg−r3m3g=0.$$+r_1{m}_{1}g+r_2{m}_{2}g+rmg-r_3{m}_{3}g=0.$$

12.17

Selecting the +y$+y$-direction to be parallel to F⃗ S,${\overrightarrow{F}}_S,$ the first equilibrium condition for the stick is

−w1−w2−w+FS−w3=0.$$\text{−}{w}_1-w_2-w+F_S-w_3=0.$$

Substituting the forces, the first equilibrium condition becomes

−m1g−m2g−mg+FS−m3g=0.$$\text{−}{m}_{1}g-m_{2}g-mg+F_S-m_{3}g=0.$$

12.18

We solve these equations simultaneously for the unknown values m3$m_3$ and FS.$F_S.$ In Equation 12.17, we cancel the g factor and rearrange the terms to obtain

r3m3=r1m1+r2m2+rm.$$r_3{m}_3=r_1{m}_1+r_2{m}_2+rm.$$

To obtain m3$m_3$ we divide both sides by r3,$r_3,$ so we have

m3=r1r3m1+r2r3m2+rr3m=7030(50.0g)+4030(75.0g)+2030(150.0g)=316.023g≃317g.$$\begin{array}{}\\ \\ \hfill m_3& =\frac{r_1}{r_3}{m}_1+\frac{r_2}{r_3}{m}_2+\frac{r}{r_3}m\hfill \\ & =\frac{70}{30}(50.0\text{g})+\frac{40}{30}(75.0\text{g})+\frac{20}{30}(150.0\text{g})=316.0\frac{2}{3}\text{g}\simeq 317\text{g.}\hfill \end{array}$$

12.19

To find the normal reaction force, we rearrange the terms in Equation 12.18, converting grams to kilograms:

FS=(m1+m2+m+m3)g=(50.0+75.0+150.0+316.7)×10−3kg×9.8ms2=5.8N.$$\begin{array}{}\\ \\ \hfill F_S& =(m_1+m_2+m+m_3)g\hfill \\ & =(50.0+75.0+150.0+316.7)\times {10}^{\mathrm{-3}}\text{kg}\times 9.8\frac{\text{m}}{{\text{s}}^2}=5.8\text{N}.\hfill \end{array}$$

12.20

Significance

Notice that Equation 12.17 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.