- #36
Erdi
- 37
- 6
Im sorry, I am lost. I really don't have a clue.erobz said:We are freeing the pulley from the other bodies. Not from the forces. What are the internal forces inside the rope that we expose when we "free" the pulley.
Im sorry, I am lost. I really don't have a clue.erobz said:We are freeing the pulley from the other bodies. Not from the forces. What are the internal forces inside the rope that we expose when we "free" the pulley.
Which forces are associated with the ropes?Erdi said:Im sorry, I am lost. I really don't have a clue.
Lets back up. What are the forces acting on ##m_2##?Erdi said:Originally, the pulley B and m1
##T_1 \neq Mg## That would be the case if ##M## or ##m_2## were NOT accelerating, but they are.Erdi said:That should be just the gravity and tension 1, tension 1= M*g
Maybe I'm using confusing wording. That is just the force of weight. What is the sum of the forces acting on ##m_2##. I think you have already written this in an earlier post.Erdi said:its m2g
Ok good:Erdi said:If that's not T - mg = ma or just T - mg , then i really don't understand at all
close. be careful about the coordinate system you chose. If ##m2## goes up is positive ##a## then what does that mean for the acceleration of ##M## here. Also, you used two different symbols for ##M##Erdi said:Thats also got to T1 - Mg = ma
You chose a coordinate system where up was positive in #46 whether you did so intentionally or not. That is upward accelerations are positive. Upward forces are positive, downward forces are negative, downward accelerations are negative. Try again. And use a capital ##M## leave the little ##m##Erdi said:T1 + mg = ma?
Good. Please use capital ##M## though so you don't get confused about which hanging mass.Erdi said:Well i think the formula is correct, so what you mean is:
T - mg = m*(-a)
I know i edited it, guess it didnt go throug..erobz said:Good. Please use capital ##M## though so you don't get confused about which hanging mass.
Also its ##T_1## you missed the subscript.Erdi said:I know i edited it, guess it didnt go throug..
The ##m_2## 's weight does not make ##m_2## go up. The other force i.e. ##T_1## (the tension) does. The tension is supplied by the rope.Erdi said:Back to stage 1 again. That is m2g isn't it??
Welcome, @Erdi !Erdi said:I know that the tension from pulley B (T1) has to be equal the m*g of m1 for m1 to have acceleration = 0. But i can't figure how this works because the m2 is already heavier. And so the block(M) has to be negative weight?
Just imagine that you are removing M and that you are grabing the rope while standing on the ground.Erdi said:Im sorry, I am lost. I really don't have a clue.
It may seem, with the angst it can bring,kuruman said:You might profit if you research "Atwood machine" on the web. Here is a compact treatment but there is plenty more including videos. Please read carefully and try to understand how one proceeds to solve such problems. The method is straightforward:
See how this plan is executed in the link I provided. The link does not provide the tension, but you can find it easily with a little algebra from either ##F_{\text{net,M}}=Ma## or ##F_{\text{net},\text{m}_{2}}=-m_2a## if you have an expression for the acceleration ##a##.
- Choose a system, in this case one of the masses, say ##M##.
- Find the net force (sum of all the forces acting on it) ##F_{\text{net,M}}.##
- Set this sum equal to the mass of the system times its acceleration. This gives you one equation, ##F_{\text{net,M}}=Ma##
- Repeat for the other mass, ##m_2##. Note that if ##M## accelerates up, ##m_2## must accelerate down. You should get a second equation ##F_{\text{net},\text{m}_{2}}=-m_2a.##
- Combine the two equations to find the acceleration.
So M will be traveling upwards, if I am correct.erobz said:In which direction was the mass ##M## traveling?
##g## shouldn't be in the final result. You also changed your variable names, and your coordinate direction from what we were working towards earlier... not making it easy to follow along.Erdi said:I can quickly go through my math here, firstly i got:
m1: m1g -T1 = m1a
m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T)
M: Mg - T2 = Ma
Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2
Then i got to the acceleration relative to each other
a(m1) - a(pulleyB) = 0, a(m1) = -a(B)
a(M)-a(B)+a(m2)-a(B) = 0 => a(M)+a(m2) = 2*a(B) = -2*a(m1)
So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))
So there was a lot of math between right there to my final answer, but i ended up with:
a(m1) = (g(m1*M+m2*m1-4*M*m2)) / (M*m1+m2*m1+4*M*m2) = 0
Solved for M:
M = (m2*m1)/(4*m2-g*m1) = 0,32kg
Do the units make sense? You are subtracting a force from a mass in the denominator. Thats generally an indication something has gone haywire.Erdi said:Hmm, well here's almost the rest of the math. Why shouldn't g be there?
I got the same expression without the g for the reason already explained. The error is between the two steps belowErdi said:M = (m2*m1)/(4*m2-g*m1) = 0,32kg
The right-hand side of the bottom equation has dimensions of force but the left hand side has dimensions of acceleration. check your substitution from the top to the bottom equation.Erdi said:So i solved for
2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2)
2*a(m1) =-(g-(T2/M) + g-(T2/m2))