Atwood's Machine with unknown masses

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  • #1
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Homework Statement



A simple Atwood's machine uses two masses m1 and m2. Starting from rest, the speed of the two masses is 8.7 m/s at the end of 7.4 s. At that time, the kinetic energy of the system is 61 J and each mass has moved a distance of 32.19 m.

Find the value of the heavier mass.

Find the value of the lighter mass.

Homework Equations



a = (m2 - m1) / (m1 + m2)g
Ke = 1/2mv2

The Attempt at a Solution



61 = 1/2m1v2 + 1/2m2v2
61 = 1/2(8.7)2(m1+m2)
61 = 37.845(m1+m2)
m1+m2 = 1.612

a = 8.7 / 7.4 = 1.176

1.176 = (m2 - m1) / (m1 + m2)g
10.976m1 = 8.624m2
m1 = .786m2

.786m2 + m2 = 1.612
2m2 = 2.051
m2 = 1.025

m1 + 1.025 = 1.612
m1 = .587

I was sure that I solved it correctly, perhaps my algebra for solving the acceleration equation is incorrect.

Thanks for any help.
 

Answers and Replies

  • #2
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.786m2 + m2 = 1.612
2m2 = 2.051
-------> Shouldn't this be 1.786m2=1.612?
 
  • #3
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-------> Shouldn't this be 1.786m2=1.612?
So you are saying my algebra for solving the acceleration equation is wrong? I got .786, not 1.786.
 
  • #4
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So you are saying my algebra for solving the acceleration equation is wrong? I got .786, not 1.786.
It seems to me so.
0.786m2 + m2 = 1.612
(0.786+1)m2=1.612
1.786m2=1.612
Could you write down exact solution?
 
  • #5
103
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I got .786 from this:

1.176 = (m2 - m1) / (m1 + m2)g
10.976m1 = 8.624m2
m1 = .786m2
 
  • #6
1,956
252
the problem is: how do you get from

[tex].786m_2 + m_2= 1.612 [/tex]

to

[tex] 2m_2 = 2.051 [/tex]
 
  • #7
1,956
252
[tex] 1.176 = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

[tex] 10.976m_1 = 8.624m_2 [/tex]

this step is also wrong
 
Last edited:
  • #8
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the problem is: how do you get from

[tex].786m_2 + m_2= 1.612 [/tex]

to

[tex] 2m_2 = 2.051 [/tex]
I divided 1.612 by .786 then I thought I could combine like terms.

[tex] 1.176 = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

[tex] 10.976m_1 = 8.624m_2 [/tex]

this step is also wrong
I was kinda finicky on this step. Could you give me a hint on how to solve it algebraically?
 
  • #9
1,956
252
I divided 1.612 by .786 then I thought I could combine like terms.
you have to divide ALL terms by .786. THat would give you [itex] m_2 + m_2/0.786 = 1.612/0.786 [/itex] I don't think that helps.

I was kinda finicky on this step. Could you give me a hint on how to solve it algebraically?
[/QUOTE]

multiply both sides by [itex] \frac {m_1 + m_2} {g} [/itex] (you already know what m_1+m_2 is)

combine what you get with [itex] m_1 + m_2 = 1.612 [/itex]
 
  • #10
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you have to divide ALL terms by .786. THat would give you [itex] m_2 + m_2/0.786 = 1.612/0.786 [/itex] I don't think that helps.
multiply both sides by [itex] \frac {m_1 + m_2} {g} [/itex] (you already know what m_1+m_2 is)

combine what you get with [itex] m_1 + m_2 = 1.612 [/itex][/QUOTE]

Okay, so I did this:

1.176 = (m2 - m1) / (m1 + m2) g
1.176 (m1 + m2) g = m2 - m1
1.176 (15.7976) = m2 - m1
18.578 = m2 - m1

I don't know if this is correct because I multiplied m1 + m2 and g. I wasn't sure if I should divide by g or multiply since it is in the denominator.
 
  • #11
1,956
252
multiply both sides by [itex] \frac {m_1 + m_2} {g} [/itex] (you already know what m_1+m_2 is)

combine what you get with [itex] m_1 + m_2 = 1.612 [/itex]
Okay, so I did this:

1.176 = (m2 - m1) / (m1 + m2) g
[/QUOTE]

note that it's [tex] a = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

not

[tex] a = \frac {m_2 - m_1} {(m_1 + m_2) g} [/tex]

so you have to multiply both sides by m_1 + m_2 and then divide by g
 
  • #12
103
0
Okay, so I did this:

1.176 = (m2 - m1) / (m1 + m2) g
note that it's [tex] a = \frac {m_2 - m_1} {m_1 + m_2} g [/tex]

not

[tex] a = \frac {m_2 - m_1} {(m_1 + m_2) g} [/tex]

so you have to multiply both sides by m_1 + m_2 and then divide by g
Okay, here's what I came up with.

1.176 = ((m2 - m1) / (m1 + m2))g

(1.176 * 1.162) / g = .19344 = m2 - m1

How does that look? I do not know what you mean when you say add it to m1 + m2 = 1.612.

Thanks for the help.
 
  • #13
1,956
252
adding 2 equations:

if you have A = B and C=D then A+C = B+D

now you have .19344 = m2 - m1

and m1 + m2 = 1.162
 
  • #14
103
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adding 2 equations:

if you have A = B and C=D then A+C = B+D

now you have .19344 = m2 - m1

and m1 + m2 = 1.162
A = .19344
B = m2 - m1
C = 1.162
D = m1 + m2

.19344 + 1.162 = (m2 - m1) + (m1 + m2)

How does that set-up look?
 
  • #15
1,956
252
Looks OK. you can now add up the right and the left side of this equation.
 
  • #16
103
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Looks OK. you can now add up the right and the left side of this equation.
.19344 + 1.162 = (m2 - m1) + (m1 + m2)

1.35544 = m2 + m1

m1 = m2 - 1.35544


Using the equation from Ke and Pe:

m1 + m2 = 1.612

m2 - 1.35544 + m2 = 1.612

m2 + m2 = 2.96744

2m2 = 2.96744

m2 = 1.48372

So m1:

m1 + 1.48372 = 1.612

m1 = .12828

Is my math okay?
 
  • #17
1,956
252
.19344 + 1.162 = (m2 - m1) + (m1 + m2)

1.35544 = m2 + m1

m1 = m2 - 1.35544
the only reason for adding the 2 equations was that m1 would disappear.

m1 - m2 + m1 + m2 = ?



if you have 2 equations like 2x - y = 7 and 3x + y = 4 it's often easy to eliminate one
of the variables by adding or subtracting them.
if you add them you get 2x - y + 3x + y = 7 + 4 so you get 5x = 11 and x = 11/5

then you substitute x = 11/5 in 2x - y = 7 to get y: 2 (11/5) - y = 7 => y = 22/5 -7 = 22/5 - 35/5 = - 13/5 = -2.6

adding the equations works because y appears in one of them and -y in the other, so you
know y will disappear.
 
  • #18
103
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I understand now, thanks. I also went to my professor to discuss this question and he explained an easier way to do it, but similar.

Thanks for all the help.
 

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