Help with an Atwood's Machine problem

[jcranford]

Homework Statement

Consider the ideal Atwood's machine in Figure 4-65. When N washers are transferred from the left side to the right side, the right side descends 47.1cm in 0.40s. Find N.

Homework Equations

a = (m₂-m₁/m₁+m₂)g

The Attempt at a Solution

I'm honestly not sure where to start with this one. We did one Atwood example in class and it wasn't anything like this. I'm not sure if that equation is relevant either, it's just the one that was given when we did the example in class.

The first thing I did was convert the given distance and time into an acceleration of m/s², so I converted the 47.1cm -> .471m / .4s and got 1.17m/s, then divided by .4s again to get 2.94m/s², which I'm not sure if that's the right way to do that or not. I think I have to solve for the masses, but I'm just really lost without a clue on this one. A push in the right direction would be greatly appreciated.Also, if anyone wants to take a look at this one while they're at it. This is what I've done so far.

For (a), I basically found the max force of static friction and said that was the most horizontal force it could have exerted on it for the top block not to slip. (b), I used the net force it gave me to calculate m(a) for 1.5m/s² and then calculated the force of friction by subtracting the net force of the smaller block from static friction for the smaller block. I'm not really sure that's a legal maneuver. I'm just lost on (c). At first I just used the new force to calculate an acceleration, but then I realized they would have two different accelerations due to this force being greater than the max static friction force. It's probably because it's late but I'm just blanking on what I'm supposed to do next. Any suggestions would be great. Thank you all.

 

[Simon Bridge]

I'm honestly not sure where to start with this one ... I converted the 47.1cm -> .471m / .4s and got 1.17m/s, then divided by .4s again to get 2.94m/s2, which I'm not sure if that's the right way to do that or not

Hint: kinematic equations. You have the distance and the time, and you know it is constant acceleration.
Note:
- if the machine is ideal - what does that say about the friction?
- the initial condition (in the reference figure) has 5 washers on each side for $m_1=m_2=5m$- what would the acceleration be in this situation?
- can you write an expression for $m_1$ and $m_2$ after $N$ washers have been moved from one side to the other?

 

[jcranford]

Hint: kinematic equations. You have the distance and the time, and you know it is constant acceleration.
Note:
- if the machine is ideal - what does that say about the friction?
- the initial condition (in the reference figure) has 5 washers on each side for $m_1=m_2=5m$- what would the acceleration be in this situation?
- can you write an expression for $m_1$ and $m_2$ after $N$ washers have been moved from one side to the other?

Ideal would mean frictionless, correct? Also, that the cables have no mass.

In the initial condition, acceleration would be zero. (5-5)9.8 / (5+5) would be 0/10 and therefore zero. I kind of figured out that with a = 2.94, m₁ must equal 3.5 and m₂ must equal 6.5. This balances out the equation for acceleration, however I have no method by which I did this. I basically plugged in some numbers and went with it when it added up. I'd love to know what it looks like to solve for this algebraically and how to go about it. I think that's what you're getting at in your third point, but I don't think I'm getting it.

 

[Simon Bridge]

That's good so far - however, I cannot help you f you don't follow the hints (you only addressed two of them):

Hint: kinematic equations. You have the distance and the time, and you know it is constant acceleration.
Note:
- can you write an expression for $m_1$ and $m_2$ after $N$ washers have been moved from one side to the other?

... presumably an ideal machine would have a zero-mass pulley too.

 

[azizlwl]

2nd. Question.
a.) F=(M+m)a
ma=mgμ_s

b) F/2=(M+m)a.

c) (1.) 2F-mgμ_k=Ma_M
(2.) ma_m=mgμ_k