Atwood's Machine with unknown masses

[Warmacblu]

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.

 

[method_man]

.786m₂ + m₂ = 1.612

2m₂ = 2.051

-------> Shouldn't this be 1.786m₂=1.612?

 

[Warmacblu]

-------> 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.

 

[method_man]

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?

 

[Warmacblu]

I got .786 from this:

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

 

[willem2]

the problem is: how do you get from

.786m_2 + m_2= 1.612

to

2m_2 = 2.051

 

[willem2]

1.176 = \frac {m_2 - m_1} {m_1 + m_2} g

10.976m_1 = 8.624m_2

this step is also wrong

 

[Warmacblu]

the problem is: how do you get from

.786m_2 + m_2= 1.612

to

2m_2 = 2.051

I divided 1.612 by .786 then I thought I could combine like terms.

1.176 = \frac {m_2 - m_1} {m_1 + m_2} g

10.976m_1 = 8.624m_2

this step is also wrong

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

 

[willem2]

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 m_2 + m_2/0.786 = 1.612/0.786 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 \frac {m_1 + m_2} {g} (you already know what m_1+m_2 is)

combine what you get with m_1 + m_2 = 1.612

 

[Warmacblu]

you have to divide ALL terms by .786. THat would give you m_2 + m_2/0.786 = 1.612/0.786 I don't think that helps.

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

combine what you get with m_1 + m_2 = 1.612[/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.

 

[willem2]

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

combine what you get with m_1 + m_2 = 1.612

Okay, so I did this:

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

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

not

a = \frac {m_2 - m_1} {(m_1 + m_2) g}

so you have to multiply both sides by m_1 + m_2 and then divide by g

 

[Warmacblu]

Okay, so I did this:

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

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

not

a = \frac {m_2 - m_1} {(m_1 + m_2) g}

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.

 

[willem2]

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

 

[Warmacblu]

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?

 

[willem2]

Looks OK. you can now add up the right and the left side of this equation.

 

[Warmacblu]

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?

 

[willem2]

.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.

 

[Warmacblu]

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.