Added captions to figures and numbering for equations

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2024-03-13 14:53:52 -05:00
parent 6726565ab8
commit d5538d5761
2 changed files with 54 additions and 35 deletions

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@@ -18,17 +18,36 @@
justify: true,
)
#show par: set block(spacing: 2em)
#set heading(numbering: "1.")
#set math.equation(numbering: it => {
locate(loc => {
let count = counter(heading).at(loc).last()
numbering("1.1", count, it)
})
})
#show heading: it => {
underline(stroke: 0.05em, it)
v(0.3em)
let count = locate(loc => [
#counter(heading).at(loc).last()
#text(")")
])
set text(size: 16pt)
block()[
#count
#underline(it.body)
]
v(0.5em)
}
#show figure.caption: it => {
set par(leading: 0.8em)
it
}
#metro-setup(inter-unit-product: $dot.c$)
#box(height: 100%, width: 100%)[
#set align(horizon + center)
#set par(leading: 1em)
@@ -57,18 +76,21 @@ The masses of both weights can be measured and the forces can be calculated from
By varying the weights and measuring acceleration, the relationship between forces and acceleration can be calculated.
= Theory
#grid(
columns: (50%, 50%),
rows: (auto),
box(width: 100%)[An Atwood machine consists of two weights ($m_1$ and $m_2$) connected by a string ($S$).
box(width: 100%)[
= T#h(0.02em)heory
An Atwood machine consists of two weights ($m_1$ and $m_2$) connected by a string ($S$).
The string is placed on a wheel that allows the weights to move up and down.
The system can be modeled with the two free body diagrams:],
The system can be modeled with the two free body diagrams:
],
figure(
image("./001.png", width: 80%),
caption: [A model of the Atwood machine used in the procedure]
)
)
)
#grid(
columns: (50%, 50%),
@@ -110,53 +132,52 @@ By varying the weights and measuring acceleration, the relationship between forc
The sum of forces in the y direction for each weight can be found by adding
the two forces in each diagram. Since these are the only forces acting on the weights
$ sum F_(y,m_1) = F^T_(S,m_1) + F^G_(g,m_1) $
$ sum F_(y,m_1) = F^T_(S,m_1) + F^G_(g,m_1) $
$ sum F_(y,m_1) = F^T_(S,m_1) + F^G_(g,m_1) $ <eq-1>
$ sum F_(y,m_1) = F^T_(S,m_1) + F^G_(g,m_1) $ <eq-2>
Taking the downward direction to be positive, $F^G_(g,m_1)$ and
$F^G_(g,m_2)$ can be found with the equation
t
$ F = m a $
$ F^G_(g,m_1) = m_1 g $
$ F^G_(g,m_2) = m_2 g $
$ F = m a $ <eq-3>
$ F^G_(g,m_1) = m_1 g $ <eq-4>
$ F^G_(g,m_2) = m_2 g $ <eq-5>
Assuming that the string is not stretching, $m_1$ and $m_2$ are each
exerting equal forces on each of the weights
$ F^T_(S,m_1) = F^T_(S,m_2) = F^T $
$ F^T_(S,m_1) = F^T_(S,m_2) = F^T $ <eq-6>
Since the rope is not stretching, the objects are
accelerating with the same magnitude but in
opposite directions. Using this fact,
the values for $F^G$ and $F^T$ found previously,
the values found in @eq-4, @eq-5, and @eq-6,
and that $sum F_y = m a_y$ the equations can be simplified to:
$ m_1 a = F^T + m_1 g $
$ -m_2 a = F^T + m_2 g $
$ m_1 a = F^T + m_1 g $ <eq-7>
$ -m_2 a = F^T + m_2 g $ <eq-8>
The first equation can now be solved for $F^T$ and can be plugged into
the second equation
$ F^T = m_1 a - m_1 g $
$ -m_2 a = m_2 g - (m_1 a - m_1 g) $
$ -m_2 a = m_2 g - m_1 a + m_1 g $
$ F^T = m_1 a - m_1 g $ <eq-9>
$ -m_2 a = m_2 g - (m_1 a - m_1 g) $ <eq-10>
$ -m_2 a = m_2 g - m_1 a + m_1 g $ <eq-11>
Isolating $a$ then gives us an equation for acceleration in terms of $m_1$ and $m_2$
Isolating $a$ then gives an equation for acceleration in terms of $m_1$ and $m_2$
$ m_1 a - m_2 a = m_2 g + m_1 g $
$ a = (m_2 g + m_1 g)/(m_1 - m_2) $
$ m_1 a - m_2 a = m_2 g + m_1 g $ <eq-12>
$ a = (m_2 g + m_1 g)/(m_1 - m_2) $ <eq-13>
Pulling $g$ out of the right side of the equation gives
$ a = g ((m_2 + m_1)/(m_1 - m_2)) $
$ a = g ((m_2 + m_1)/(m_1 - m_2)) $ <eq-14>
Using $M$ to represent $(m_2 + m_1)/(m_1 - m_2)$, the equation used for this procedure is found:
$ a = g M $
$ a = g M $ <eq-15>
= Procedure
An Atwood Machine was created by suspending a string from a wheel attached to a lab support.
A photogate so that it was blocked multiple times while the wheel spun.
On each end of the string, weights were attached of varying masses.
@@ -178,8 +199,6 @@ any equipment due to the increased acceleration.
#align(center)[
#box(width: 85%)[
#set par(leading: 0.5em)
#figure(
caption: [A table containing all of the values collected during the experiment]
)[
@@ -210,19 +229,19 @@ any equipment due to the increased acceleration.
]
]
= Data Analysis
= Data Analysis
Since the function should yield a linear function with slope $g$,
an experimental value for $g$ can be found by finding the line of
best fit of the function.
$ g_"experimental" = 9.13 unit(meter/(second^2)) $
$ g_"experimental" = 9.13 unit(meter/(second^2)) $ <eq-16>
Comparing the calculated $g$ to the accepted $g = 9.81 unit(meter/(second^2))$ the percent
deviation can be calculated
$ "% deviation" = abs((T - E)/T) dot 100 $
$ "% deviation" = abs((9.81 - 9.13)/9.81) dot 100 $
$ "% deviation" = 6.9% $
$ "% deviation" = abs((T - E)/T) dot 100 $ <eq-17>
$ "% deviation" = abs((9.81 - 9.13)/9.81) dot 100 $ <eq-18>
$ "% deviation" = 6.9% $ <eq-19>
This error is somewhat high but is still a decent result.
The two dominant errors causing this error are systematic and random error.
@@ -243,8 +262,8 @@ minor source of error in the experiment. Both errors included here are somewhat
difficult to reduce given that they would require upgraded or new equipment.
However, the influence of the random error could be reduced by performing more trials.
= Conclusion
= Conclusion
An Atwood Machine can be a good method for determining acceleration due to gravity.
Although experimental errors caused a rather large error of 6.9%, it is still a relatively
good approximation. The results could likely be improved by running more trials to