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 Equatio
Plane Trusses
It is defined as, two or more elements like beams or any two or more force members, which when assembled together, behaves like a complete structure or as a single structure. They generally consist of two force member which means any component structure where the force is applied only at two points. The point of contact of joints of truss are known as nodes. They are generally made up of triangular patterns. Nodes are the points where all the external forces and the reactionary forces due to them act and shows whether the force is tensile or compressive. There are various characteristics of trusses and are characterized as Simple truss, planar truss or the Space Frame truss.
Equilibrium Equations
If a body is said to be at rest or moving with a uniform velocity, the body is in equilibrium condition. This means that all the forces are balanced in the body. It can be understood with the help of Newton's first law of motion which states that the resultant force on a system is null, where the system remains to be at rest or moves at uniform motion. It is when the rate of the forward reaction is equal to the rate of the backward reaction.
Force Systems
When a body comes in interaction with other bodies, they exert various forces on each other. Any system is under the influence of some kind of force. For example, laptop kept on table exerts force on the table and table exerts equal force on it, hence the system is in balance or equilibrium. When two or more materials interact then more than one force act at a time, hence it is called as force systems.
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
With (Figure) and (Figure) for reference, we begin by finding the lever arms of the five forces acting on the stick:
![\[\begin{array}{ccc}\hfill {r}_{1}& =\hfill & 30.0\,\text{cm}+40.0\,\text{cm}=70.0\,\text{cm}\hfill \\ \hfill {r}_{2}& =\hfill & 40.0\,\text{cm}\hfill \\ \hfill r& =\hfill & 50.0\,\text{cm}-30.0\,\text{cm}=20.0\,\text{cm}\hfill \\ \hfill {r}_{S}& =\hfill & 0.0\,\text{cm}\,\text{(because}\,{F}_{S}\,\text{is attached at the pivot)}\hfill \\ \hfill {r}_{3}& =\hfill & 30.0\,\text{cm.}\hfill \end{array}\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-6fa006941a351ca4843b822e2ae0b1b0_l3.svg "Rendered by QuickLaTeX.com")
Now we can find the five torques with respect to the chosen pivot:
![\[\begin{array}{ccccc}\hfill {\tau }_{1}& =\hfill & +{r}_{1}{w}_{1}\text{sin}\,90\text{°}=\text{+}{r}_{1}{m}_{1}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill {\tau }_{2}& =\hfill & +{r}_{2}{w}_{2}\text{sin}\,90\text{°}=\text{+}{r}_{2}{m}_{2}g\hfill & & \text{(counterclockwise rotation, positive sense)}\hfill \\ \hfill \tau & =\hfill & +rw\,\text{sin}\,90\text{°}=\text{+}rmg\hfill & & \text{(gravitational torque)}\hfill \\ \hfill {\tau }_{S}& =\hfill & {r}_{S}{F}_{S}\text{sin}\,{\theta }_{S}=0\hfill & & \text{(because}\,{r}_{S}=0\,\text{cm)}\hfill \\ \hfill {\tau }_{3}& =\hfill & \text{−}{r}_{3}{w}_{3}\text{sin}\,90\text{°}=\text{−}{r}_{3}{m}_{3}g\hfill & & \text{(clockwise rotation, negative sense)}\hfill \end{array}\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-57f09a81d8faa87e214afc906e6c61f8_l3.svg "Rendered by QuickLaTeX.com")
The second equilibrium condition (equation for the torques) for the meter stick is
![\[{\tau }_{1}+{\tau }_{2}+\tau +{\tau }_{S}+{\tau }_{3}=0.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-c567baa281fc3b9e6be925a98d9862ff_l3.svg "Rendered by QuickLaTeX.com")
When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is
![\[+{r}_{1}{m}_{1}g+{r}_{2}{m}_{2}g+rmg-{r}_{3}{m}_{3}g=0.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-858a7a598b778fa4ef0ab8d1606da388_l3.svg "Rendered by QuickLaTeX.com")
Selecting the
![\[+y\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-8adfd9b44c3b1d1ee12b2c7982472ac3_l3.svg "Rendered by QuickLaTeX.com")
-direction to be parallel to
![\[{\overset{\to }{F}}_{S},\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-47ed29e7ef5fe35ba1c46ebc705e549f_l3.svg "Rendered by QuickLaTeX.com")
the first equilibrium condition for the stick is
![\[\text{−}{w}_{1}-{w}_{2}-w+{F}_{S}-{w}_{3}=0.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-92710ef9bb3e2817911f546dea6d9a7f_l3.svg "Rendered by QuickLaTeX.com")
Substituting the forces, the first equilibrium condition becomes
![\[\text{−}{m}_{1}g-{m}_{2}g-mg+{F}_{S}-{m}_{3}g=0.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-e993a7e6884e448e1ddc650df62a4b18_l3.svg "Rendered by QuickLaTeX.com")
We solve these equations simultaneously for the unknown values
![\[{m}_{3}\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-8f9d2de7f189031d4b7e4027b841bcca_l3.svg "Rendered by QuickLaTeX.com")
and
![\[{F}_{S}.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-16092ce96be752ea38ee8e7ed05e708a_l3.svg "Rendered by QuickLaTeX.com")
In (Figure), we cancel the g factor and rearrange the terms to obtain
![\[{r}_{3}{m}_{3}={r}_{1}{m}_{1}+{r}_{2}{m}_{2}+rm.\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-4d58924fed0bd29a55c8cd595f54a3d7_l3.svg "Rendered by QuickLaTeX.com")
To obtain
we divide both sides by
![\[{r}_{3},\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-1de187208d32b33b9d152ba8341b5d82_l3.svg "Rendered by QuickLaTeX.com")
so we have
![\[\begin{array}{cc} \hfill {m}_{3}& =\frac{{r}_{1}}{{r}_{3}}\,{m}_{1}+\frac{{r}_{2}}{{r}_{3}}\,{m}_{2}+\frac{r}{{r}_{3}}\,m\hfill \\ & =\frac{70}{30}\,(50.0\,\text{g})+\frac{40}{30}\,(75.0\,\text{g})+\frac{20}{30}\,(150.0\,\text{g})=316.0\frac{2}{3}\,\text{g}\simeq 317\,\text{g.}\hfill \end{array}\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-263ca8b0274763ab570818dfa59ca23e_l3.svg "Rendered by QuickLaTeX.com")
To find the normal reaction force, we rearrange the terms in (Figure), converting grams to kilograms:
![\[\begin{array}{cc} \hfill {F}_{S}& =({m}_{1}+{m}_{2}+m+{m}_{3})g\hfill \\ & =(50.0+75.0+150.0+316.7)\,×\,{10}^{-3}\text{kg}\,×\,9.8\,\frac{\text{m}}{{\text{s}}^{2}}=5.8\,\text{N}.\hfill \end{array}\]](https://opentextbc.ca/universityphysicsv1openstax/wp-content/ql-cache/quicklatex.com-d64e86a2c56997893d6e7f3163e89c57_l3.svg "Rendered by QuickLaTeX.com")
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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