How Do You Calculate the Second Mass in an Atwood Machine Problem?

[Kamisama]

Homework Statement

atwood machine: pulley with negligible mass and frictionless with 2 blocks hanging at rest.
The first block "m" has a mass of 3.66kg
The second Block "M" is unknown
The blocks are released; block M accelerates at .255*g m/s2 downward
Find the mass of block "M"
Hint: There are 2 solutions.

Homework Equations

F=ma
a= (Mg-mg)/(m+M)

The Attempt at a Solution

I rearranged the equation to M=m(a+g)/(g-a) and solved getting 6.17 as my first answer but I can't figure out how to get a second solution after trying a few different ways that which all lead me to 6.17 (pretty much different forms of the acceleration formula, kinda lost). How else can I get the mass of M with what I'm given? Also, was my first answer correct?

 

[Paul Colby]

There is one equation and one unknown (and you've solved it). The equation is linear so there would be a unique answer. Possible the hint is wrong?

 

[Kamisama]

I don't think the hint is wrong. There are two answer boxes. I tried answering it with only one solution and it told me it was completely wrong with no partial credit which makes me think I'm completely off.

 

[TSny]

I also don't see two solutions coming out of this. Just to make sure, does the statement of the problem definitely say that the unknown mass accelerates downward?

 

[Paul Colby]

I could well be wrong but I'm fairly certain the physics isn't determined by the number of solution boxes. A pulley (massless) a rope and 2 masses. The acceleration of the mass M is known (both magnitude and direction) and fixing the acceleration of m. Taking up as positive I get,
T - Mg = Ma,
where a = -0.255g for mass M and,
T - mg = -ma,
solving for M,
(m - M)g = (M+m)a
M = m(g - a)/(g+a)
one and only one solution

 

[Kamisama]

I also don't see two solutions coming out of this. Just to make sure, does the statement of the problem definitely say that the unknown mass accelerates downward?

Yes, well rather the 2 blocks on the frictionless, massless pulley both accelerate at 0.255g m/s2 once they're released with M>m

M = m(g - a)/(g+a)
one and only one solution

M= (3.66)(9.81 -(-0.255*9.81)) / (9.81- 0.255*9.81) = 6.1655 = 6.17kg..

I did calculate this correctly, right?

Unfortunately it wants 2 answers. Is there another equation I could use?

 

[TSny]

Yes, well rather the 2 blocks on the frictionless, massless pulley both accelerate at 0.255g m/s2 once they're released with M>m

I'm a little confused. In the original statement of the problem there was no mention of M being greater than m, but it did state that M accelerates downward.

Did you state the problem in post #1 exactly as given to you (word for word)?

I'm just trying to make sure that the problem statement states or implies that M accelerates downward.

 

[haruspex]

Yes, well rather the 2 blocks on the frictionless, massless pulley both accelerate at 0.255g m/s2 once they're released with M>m

We need to be completely clear on this. Are you saying that it does not specify the direction of acceleration, but that it does specify M>m? Maybe it would be best if you were to quote the complete question word for word. (Or is it a translation?)

 

[Kamisama]

It specifies that M > m which is why I said downward. Sorry for any confusion.

 

[TSny]

It specifies that M > m which is why I said downward. Sorry for any confusion.

OK. Thanks. I don't see how there can be two answers. It still might be good if you quoted the problem exactly.

 

[Kamisama]

2 blocks of masses m1 and m2 are suspended by a massless string over a frictionless pulley with negligible mass, as in an atwood machine . the blockss are held motionless and the released. if m1 = 3.66kg, what value does m2 have to have in order for the system to experience and acceleration a= .255g? (hint: there are two solutions to this problem)

word for word

 

[TSny]

There is no mention of M > m or M accelerating downward. So, see if you can see why there are in fact two solutions.

 

[Kamisama]

ah I see now, my original thinking is that if m2 is to accelerate, it would only accelerate downward. How would i go about finding the other acceleration.

 

[TSny]

ah I see now, my original thinking is that if m2 is to accelerate, it would only accelerate downward. How would i go about finding the other acceleration.

Assume the acceleration of M is upward?

 

[Kamisama]

Assume the acceleration of M is upward?

Well, that would make sense wouldn't it...haha

So for one,

M= (3.66)(9.81 -(-0.255*9.81)) / (9.81- 0.255*9.81) = 6.1655 = 6.17kg

and for the second one,

M= (3.66)(9.81-0.255*9.81) / (9.81+0.255*9.81) = 2.17kg

if my thought process is correct..

 

[TSny]

Looks good.

 

[Kamisama]

Thanks again for all your help as well as everyone else! I finally got it!

 

[TSny]

Good work and welcome to PF.