Solving for Masses in an Atwood Machine

  • #1
Venturi365
12
3
Homework Statement
An Atwood's machine uses two masses ##m_{1}## and ##m_{2}##. Starting from rest, the velocity of the two masses is ##v=4\,\mathrm{\frac{m}{s}}## after ##3\,\mathrm{s}##. In that instant, the kinetic energy of the system is ##E_{k}=80\,\mathrm{J}## and each one of the masses has moved ##6\,\mathrm{m}##. Calculate the values of ##m_{1}## and ##m_{2}##
Relevant Equations
##P=\frac{E_{k}}{t}##
##P=F\cdot v##
##\sum F=ma##
##E_{k}=\frac{1}{2}mv^{2}##
I've tried to solve this exercise but I haven't used one of the properties of the system (the displacement of the masses) so I don't know if I'm wrong about my procedure.

First of all, we (obviously) know that

$$
P=P
$$

And since we can express the power of a force in two different ways, we can say that:

$$
\frac{E_{k}}{t}=F\cdot v \quad \text{or} \quad \frac{mv^{2}}{2t}=F\cdot v
$$

Since the energy of the system is known and the mass of the system is ##(m_{1}+m_{2})## then:

$$
\begin{split}
\frac{E_{k}}{t}&=\frac{mv^2}{2t}\\
\frac{80}{3}&=\frac{(m_{1}+m_{2})4^{2}}{2\cdot 3}\\
(m_{1}+m_{2})&=\frac{80\cdot 2\cdot 3}{16\cdot 3}\\
m_{1}+m_{2}&=10\,\mathrm{kg}
\end{split}
$$

Applying Newton's Second law we can express the forces on each mass like this:

$$
\begin{split}
w_{1}-T&=m_{1}a\\
T-w_{2}&=m_{2}a
\end{split}
$$

Adding up both equations we get:

$$
\begin{split}
m_{1}g-m_{2}g&=m_{1}a+m_{2}a\\
g(m_{1}-m_{2})&=a(m_{1}+m_{2}\\
(m_{1}+m_{2})a&=g(m_{1}-m_{2})\\
(m_{1}-m_{2}&=\frac{a(m_{1}+m_{2})}{g}\\
m_{1}-m_{2}&=\frac{\frac{4}{3}\cdot10}{9.81}\\
m_{1}-m_{2}&\approx =1.36\,\mathrm{kg}
\end{split}
$$

Now we have a simple system of equations:

$$
\begin{cases}
m_{1}+m_{2}=10\\
m_{1}-m_{2}=1.36
\end{cases}
$$

Solving we get:

$$
\begin{array}{ll}
m_{1}=1.36+m_{2} & m_{1}+4.32=10\\
1.36+m_{2}+m_{2}=10 & m_{1}=10-4.32\\
m_{2}=\frac{10-1.36}{2} & m_{1}=5.68\,\mathrm{kg}\\
m_{2}=4.32\,\mathrm{kg}
\end{array}
$$

So, am I right or did I do something wrong? The only thing that I really doubt about my procedure is the fact that I assume that the net force is ##F=a(m_{1}+m_{2})## but still I'm quite insecure.
 
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  • #2
Here's the diagram btw

imagen_2023-11-08_144406652.png
 
Last edited by a moderator:
  • #3
You cannot use P = E/t since the energy put into the system varies with time. In particular, at t=0 the velocity is zero so P = Fv is zero for both masses.
 
  • #4
You have already found that ##(m_1+m_2)a=(m_1-m_2)g## where ##a=\frac{4}{3}~##m/s2. That's one equation relating the masses.
Conservation of mechanical energy is a second equation relating the masses.
You have a system of two equations and two unknowns.
 
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  • #5
Orodruin said:
You cannot use P = E/t since the energy put into the system varies with time. In particular, at t=0 the velocity is zero so P = Fv is zero for both masses.
While that is true, and is certainly worth pointing out for future reference, it does not appear that @Venturi365 used ##P=Fv##. Rather, ##\frac 12mv^2=E_k## was used to find the sum of the masses and ##F=ma, a=v/t## were used to find the difference.
kuruman said:
You have already found that ##(m_1+m_2)a=(m_1-m_2)g## where ##a=\frac{4}{3}~##m/s2. That's one equation relating the masses.
Conservation of mechanical energy is a second equation relating the masses.
You have a system of two equations and two unknowns.
Seems to me that is what @Venturi365 did. The question is why the info re the distance moved was not needed. The solution to that is that, assuming constant acceleration, it was redundant: average velocity ##=\frac{u+v}2=\frac st##, ##\frac{0+4}2+\frac 63##.
 
  • #6
haruspex said:
Seems to me that is what @Venturi365 did.
I agree. The two relevant equations that OP already has are
##m_{1}g-m_{2}g=m_{1}a+m_{2}a##
##m_1+m_2=\dfrac{2\Delta K}{v^2}##.
In this form one can solve the second equation for one of the masses and substitute in the first. The displacements would be needed if the second equation were written in terms of ##h## and ##\Delta U = -80## J. In that case, the speed would not be needed.
 
  • #7
haruspex said:
While that is true, and is certainly worth pointing out for future reference, it does not appear that @Venturi365 used ##P=Fv##. Rather, ##\frac 12mv^2=E_k## was used to find the sum of the masses and ##F=ma, a=v/t## were used to find the difference.

Seems to me that is what @Venturi365 did. The question is why the info re the distance moved was not needed. The solution to that is that, assuming constant acceleration, it was redundant: average velocity ##=\frac{u+v}2=\frac st##, ##\frac{0+4}2+\frac 63##.
It's right that I started with one statement and ended up forgetting it, I apologise for that. However, this means that I didn't commit any misconception, right?
 
  • #8
Venturi365 said:
It's right that I started with one statement and ended up forgetting it, I apologise for that. However, this means that I didn't commit any misconception, right?
It appears that you had a misconception, but did not rely on it in your solution.
 
  • #9
haruspex said:
It appears that you had a misconception

Yeah, you're right about that.

Thank you so much!
 
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