Atwood machine rope tension problem

[Freeq]

Homework Statement

Smaller mass on an Atwood machine is 5.2 kg (m₁). If the mass accelerates at 4.6 m/s²what is the mass of the object? What is the tension of the rope?

Homework Equations

a = g(m₂-m₁)/(m₂+m₁)

The Attempt at a Solution

I can't approach this questions, because we always worked with two masses and had to find acceleration and tension of the string, but we have never worked without the second mass.

 

[gneill]

Draw a free body diagram of the forces acting on the smaller mass. What forces are operating on it? What does Newton's second law have to say about the net force and resulting acceleration on a given mass?

 

[Freeq]

That's what I did. According to Newton's law the forces need to be equal on boths sides.


(Left side)
Ft=m₁g+m₁a
Ft=(5.2x9.8)+(5.2x4.6)
Ft=74.88N

Then I can sub in Ft into the second equation.

(Right side)
Ft=m₂g-m₂a
Ft=m₂(g-a)


When I want to check it, it works perfectly!
Ft=m₂g-m₂a
Ft=(14.4x9.8)-(14.4x4.6)
Ft=74.88N

Thanks!

 

[gneill]

I see that in your diagram, on the left hand side, you have an equation relating the forces acting on the smaller mass.

Your equation is FT = mg + ma. Since FT must be acting upwards on m, and mg must be operating downwards, and since the acceleration of the mass must be upwards (it's the smaller mass), then the equation should probably look like: FT = ma - mg for this smaller mass.

Now, for which of the variables in that equation do you have values?

EDIT: Silly me. ma should be the reaction force. That does indeed make it FT = mg + ma as you wrote. Apologies.

 

[Freeq]

Your equation is FT = mg + ma. Since FT must be acting upwards on m, and mg must be operating downwards, and since the acceleration of the mass must be upwards (it's the smaller mass), then the equation should probably look like: FT = ma - mg for this smaller mass.

Not really, because when you think about tension of the rope, you need to add two opposite forces (or substract two forces in the same direction), and even if it should be mg-ma, and this is equation for the bigger mass. Which grade are you in? :D

 

[gneill]

If the smaller mass is accelerating upwards, then the force on that mass due to the tension in the cord exceeds the force due to gravity on that mass. Thus I write ma = T - mg, or T = ma - mg. This allows you to solve for the tension given values that you have in hand, namely the mass and acceleration of the small block. Try it.

EDIT: Doh! ma = T - mg, so T = ma + mg.

Sorry about that. I've been up to long...

 

[Freeq]

ok I am going to teach you a bit of physics, the greater the downward force of gravity on the object, the greater the upward tension in the rope. In other words, the more the material is stretched, the greater the tension in the material. Now let's apply this to my assigmnent question : the smaller mass is stretched downward by gravity and upward by acceleration where, the bigger mass is only stretched downward by gravity and that stretch is reduced by upward acceleration, so the tension of the smaller object needs to be added.

 

[gneill]

Well, Freeq, if you have all the answers, you should be able to solve the problem without help! Write your equation and solve for the tension.

As for the heavier mass, you can solve the equation that you already wrote:

a = g(m₂-m₁)/(m₂+m₁)

for m₂ and plug in the known values of a and m₁to find m₂.

Good luck!

 

[Freeq]

(Right side)
Ft=m₂g-m₂a
Ft=m₂(g-a)

I forgot to fiinish this so,
m₂= Ft/(g-a)
m₂=74.88/(9.8-4.6)
m₂=14.4kg


BTW: I goofed on the picture - I wrote the mass of smaller object 5.3kg which it should be m₁ 5.2kg
Then I want to prove you AGAIN that this is right, by using the formula you mentionied in your post above.

a= g(m₂-m₁)/(m₂+m₁)
a= 9.8(14.4-5.2)/(14.4+5.2)
a= 4.6 m/s²

Which is right, because the acceleration given in the question is 4.6 m/s²


Dear gneill, I'm glad that I helped you understand physics a litte bit better :)

For more info/refference visit:
http://en.wikipedia.org/wiki/Atwood_machine
http://en.wikipedia.org/wiki/Tension_(physics )

Have fun!