12.2 Examples of static equilibrium (Page 3/9)

University physics volume 1 Page 3 / 9

Figure is a schematic drawing of a force distribution for a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Force Fs at the point S is pointing upward. Force w3, to the right of point S and separated by distance r3 is pointing downward. Forces w, w2, and w1 are to the left of point S and are pointing downward. They are separated by distance r, r2, and r1, respectively. — Free-body diagram for the meter stick. The pivot is chosen at the support point S .

Solution

With [link] and [link] for reference, we begin by finding the lever arms of the five forces acting on the stick:

\begin{matrix} r_{1} & = & 30.0 cm + 40.0 cm = 70.0 cm \\ r_{2} & = & 40.0 cm \\ r & = & 50.0 cm - 30.0 cm = 20.0 cm \\ r_{S} & = & 0.0 cm (because F_{S} is attached at the pivot) \\ r_{3} & = & 30.0 cm. \end{matrix}

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

\begin{matrix} τ_{1} & = & + r_{1} w_{1} sin 90 ° = + r_{1} m_{1} g & (counterclockwise rotation, positive sense) \\ τ_{2} & = & + r_{2} w_{2} sin 90 ° = + r_{2} m_{2} g & (counterclockwise rotation, positive sense) \\ τ & = & + r w sin 90 ° = + r m g & (gravitational torque) \\ τ_{S} & = & r_{S} F_{S} sin θ_{S} = 0 & (because r_{S} = 0 cm) \\ τ_{3} & = & − r_{3} w_{3} sin 90 ° = − r_{3} m_{3} g & (clockwise rotation, negative sense) \end{matrix}

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

τ_{1} + τ_{2} + τ + τ_{S} + τ_{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 + r m g - r_{3} m_{3} g = 0 .

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

− w_{1} - w_{2} - w + F_{S} - w_{3} = 0 .

Substituting the forces, the first equilibrium condition becomes

− m_{1} g - m_{2} g - m g + F_{S} - m_{3} g = 0 .

We solve these equations simultaneously for the unknown values $m_{3}$ and $F_{S} .$ In [link] , we cancel the g factor and rearrange the terms to obtain

r_{3} m_{3} = r_{1} m_{1} + r_{2} m_{2} + r m .

To obtain $m_{3}$ we divide both sides by $r_{3},$ so we have

\begin{matrix} m_{3} & = \frac{r_{1}}{r_{3}} m_{1} + \frac{r_{2}}{r_{3}} m_{2} + \frac{r}{r_{3}} m \\ = \frac{70}{30} (50.0 g) + \frac{40}{30} (75.0 g) + \frac{20}{30} (150.0 g) = 316.0 \frac{2}{3} g ≃ 317 g. \end{matrix}

To find the normal reaction force, we rearrange the terms in [link] , converting grams to kilograms:

\begin{matrix} F_{S} & = (m_{1} + m_{2} + m + m_{3}) g \\ = (50.0 + 75.0 + 150.0 + 316.7) \times 10^{−3} kg \times 9.8 \frac{m}{s^{2}} = 5.8 N . \end{matrix}

Significance

Notice that [link] 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.

Check Your Understanding Repeat [link] using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

316.7 g; 5.8 N

Got questions? Get instant answers now!

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by [link] and [link] . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Forces in the forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in [link] . His forearm is positioned at

β = 60 °

with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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