There is no friction anywhere. The cart system is let go with everything initially at rest. Find the acceleration for each of the three masses relative to ground. (they are all different)

- anonymous

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- anonymous

How should I approach this? I have no idea where to start

##### 1 Attachment

- anonymous

you can start drawing free body diagrams for each block, to do that you will need to identify the forces acting on each of them first. for example for block 3:
|dw:1434560772793:dw|
only the tension force \(T\) and the wight \(W_3\) are acting on this block.
try to do the same for the other blocks, and then write down the force sum for each of them.
If you get the diagrams done right you can get the equations directly from them.

- Michele_Laino

since there is no friction, the total momentum along the x-axis has to be conserved:
|dw:1434562048822:dw|

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## More answers

- Michele_Laino

here we have to compute the acceleration of the object #1

- Michele_Laino

|dw:1434562306655:dw|

- Michele_Laino

in order to find the acceleration of the object#1, we have to solve this vector system:
\[\left\{ \begin{gathered}
{m_1}{\mathbf{a}} = {\mathbf{\tau }} \hfill \\
{m_3}{\mathbf{g}} + {\mathbf{\tau }} = {m_3}{\mathbf{a}} \hfill \\
\end{gathered} \right.\]

- Michele_Laino

|dw:1434562777709:dw|
where \tau is the tension of the string connecting objects #1 and #3

- anonymous

so from the equation you have provided, is solved for the acceleration of M1 and I got:
\[a _{1}=\frac{ M _{3}g }{M _{3}-M _{1}}\]
assuming this is right, what should I do next to find the acceleration of M2?

- Vincent-Lyon.Fr

This can't be right.
If M1=M3 (which is possible), the acceleration would be infinite (which is physically impossible).

- anonymous

hmm. how do I solve for a1 then? :/

- IrishBoy123

.

- Vincent-Lyon.Fr

Do you have numerical values for M1, M2 and M3?

- anonymous

no

- anonymous

I have to state the answer in terms of M1 M2 M3 and g

- IrishBoy123

|dw:1434662031754:dw|
it is important for the solution that you go back to the original drawing and constrain m3 in the vertical plane. otherwise this is a real world problem that needs a computerised numerical solution.

- anonymous

what do you mean by constraining m3 in the vertical plane?

- anonymous

Ok. So this is what I did:
|dw:1434691193424:dw|
\[T = m _{1}a _{1}\
\\W-T = m _{3}a _{3}\]
\[m _{3}g - m _{1}a _{1} = m _{3}a _{3}\\m _{1}a _{1} = m _{3}g - m _{3}a _{3}\\a _{1} = m _{3}g - m _{3}a _{3}\]
am I do this right?

- anonymous

almost. you have to divide the right hand side of your last equation by m1

- anonymous

oh. oops, forgot about that.
So then it would be:
\[a _{1} = \frac{ m _{3}g - m _{3}a _{3} }{ m _{1}}\]

- anonymous

How can I then get rid of a3 so the answer is only in terms of masses and gravity?

- anonymous

so the way its drawn, I would constrain mass 2 and 3 to have the same horizontal velocity, and mass 1's horizontal velocity to equal mass 3s vertical velocity. for them to have equal velocities like that, they must also have equal accelerations (such that they all start from rest and have the same velocities throughout).
That gives us the following relations:
a1x = -a3y
a2x = a3x
we also know that m1 and m2 are constrained to horizontal motion:
a1y = 0
a2y = 0
treating m1,m2, m3 and g as independent variables, we have 6 unknowns that we need to solve for (a1x, a1y, a2x, a2y, a3x and a3y) but only 4 equations so far.
We can add equations using the free body diagrams made earlier for m1 and m2:
we know a1x = T/m1 and a3y = (T-g)/m3
Unfortunately, we just introduced another unknown (T)
These new relations bring us to:
T/m1 = -(T-g)/m3
Solving for T gives us T=m1*g/(m1+m3)
knowing T, we can solve for a1x and a3y
a1x = T/m1 = g/(m1+m3) <-- I just realized that my way to get this result was very circular... but bear with me
a3y = -T/m3 = m1*g/(m3*(m1+m3))
Just to recap where we are (for my sake)
a1x = g/(m1+m3)
a1y = 0
a2x = ?
a2y = 0
a3x = a2x
a3y = m1*g/(m3*(m1+m3))
So just one unknown left..

- anonymous

Since we know (or at least I have assumed) that 2 and 3 move together in the x direction, lets consider them together in a free body diagram for object 2:
|dw:1434697943399:dw|

- anonymous

(also there should be a vertical normal force where block 1 is)... we don't really care about the vertical forces because we know they all cancel to 0. That just leaves us the horizontal force balance:
Since there is no friction, the only horizontal force is the tension force. We also know that m2 and m3 move together therefore:
a2x=a3x ---> (m2a2x+m3a3x)=T
we'll just combine a2 and a3 here to say a2x=a3x=a
a(m2+m3)=T
a2x = a3x = T/(m2+m3) = m1*g/[(m1+m3)*(m2+m3)]

- anonymous

not sure if you're still here, but that should be everything you need :)

- anonymous

yes im still am. im writing them down cuz i cant really see it in text form lol

- anonymous

if so, thank you! :)

- anonymous

Yeah, definitely write it out and make sure you understand the process! thats the most important part!

- anonymous

will do :)

- anonymous

i have a question. How did a3y become equal to -T/m3
I thought its either
a3y = -a1x = -T/m1 or
a3y = (T-g)/m3

- Michele_Laino

hint:
using the indicated reference system, those vector equations can be rewritten as below:
\[\left\{ \begin{gathered}
{m_1}a = \tau \hfill \\
- {m_3}g + \tau = - {m_3}a \hfill \\
\end{gathered} \right.\]
Solving that system for a and \tau, we get:
\[\left\{ \begin{gathered}
a = \frac{{{m_3}}}{{{m_1} + {m_3}}}g \hfill \\
\tau = \frac{{{m_1}{m_3}}}{{{m_1} + {m_3}}}g \hfill \\
\end{gathered} \right.\]

- Michele_Laino

\[\large \left\{ \begin{gathered}
a = \frac{{{m_3}}}{{{m_1} + {m_3}}}g \hfill \\
\hfill \\
\tau = \frac{{{m_1}{m_3}}}{{{m_1} + {m_3}}}g \hfill \\
\end{gathered} \right.\]

- Michele_Laino

Now using the fact that the total momentum is conserved along the horizontal direction, we can write:
\[\Large{m_2}A = {m_1}a\]
where A is the magnitude of the acceleration of the mass m_2

- IrishBoy123

3 is moving horizontally also!
that is why the constraint is important. if 3 were not constrained to sliding up and down that tube, you could not really model it....

- anonymous

so from that, I solved for A and got:
\[A = \frac{ m _{1}m _{3}g }{ m _{2} (m _{1}+m _{3})}\]

- anonymous

hmmm. two different answers....

- anonymous

and i see what you mean now @IrishBoy123

- anonymous

this problem makes me want to roll into a cocoon and cry. i thought i had it, but i dont, then i thought i had it, but then again i dont ._.

- IrishBoy123

well i happen to think the asnwer to this, **if calculated properly** is horrendously complicated and this is one reason why
from @LifeEngineer: " mass 1's horizontal velocity to equal mass 3s vertical velocity. " [which is otherwise great]
not true.
mass 3s vertical velocity is mass 1's horizontal velocity - mass 2's horizontal velocity because the pully point is moving too
the second one is conservation of momentum -- has to be is for all three particles as they are all moving in x direction
do you have a solution? if so i can scan what i came up with but i am really short on time....but the answer involves terms with combinations like m1+m2+m3.....
don't beat yourself up, this is tricky :p

- Michele_Laino

oops.. my preceding equation is wrong, namely the right equation is:
\[\Large \left( {{m_2} + {m_3}} \right)A = {m_1}a\]
so we get:
\[\Large A = \frac{{{m_1}{m_3}}}{{\left( {{m_1} + {m_3}} \right)\left( {{m_2} + {m_3}} \right)}}g\]

- IrishBoy123

@MicheleLaino
that adjustment sorts out momentum but i reckon you also have to factor in the motion of the pulley. it will be accelerating to the left as it pulls to the right.
and, in some more detail, conserving momentum in x direction is a consequence of Newton's 2nd, but the force that actually bring this about are the interaction between m2 and the force of the pulley and also the interaction in the x direction between m2 and m3 which will be in contact. m3 will move to the left due to it being pushed by m2. that is why the constraint is necessary.
i have tried an energy approach and gotten something really smelly but that is what this needs. a proper Lagrangian and not my Lagrangian-lite and at least we'd have verified the answer in several different ways.
i'd like to see it done but under kosh for next week or so and shouldn't really be here. but this is interesting.

- anonymous

Well, according to a yahoo source, the answer for a2 is a2 = -(M1M3g)/(M2M1+M2M3+2M1M3+M3^2)
I dont know if the source is reliable or not, but you can check it yourself
https://answers.yahoo.com/question/index?qid=20090722173918AARvrHi

- IrishBoy123

that looks more like it!
this is what i scribbled out. looks messy and not much fun to finish but the starting equations/conditions should be in right ball park.
doing this using forces is totally wrong way IMHO. look at energy.

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- anonymous

so the answer from yahoo answer is right?

- IrishBoy123

no idea, haven't seen it
do you understand the scribbles i posted? that is the way to do this with minimum fuss, but i think it is still very fiddly.

- anonymous

@IrishBoy123, thats a good point about the relative velocity... what an inconspicuous mess of a problem

- IrishBoy123

@LifeEngineer
"inconspicuous mess of a problem"
how beautifully put!!!

- anonymous

Hmmm. I can somehow understand it, but I would have to write it out myself to fully understand it

- Vincent-Lyon.Fr

The question asking for the force F required so that M3 does not move is WAY MUCH simpler to answer, and was already dealt with on OS.
I will try to solve it but I need time. The equations are very messy.

- anonymous

yeah. mkay

- anonymous

@IrishBoy123 I noticed that the 1/2 from the kinetic energy equation disappeared. I cant see how that happened

- IrishBoy123

\( \frac{d}{dt} [\frac{1}{2} M \dot X ^2] = 2 (\frac{1}{2}) M \dot X \ \ddot X = M \dot X \ \ddot X \).

- Michele_Laino

In my reasoning, I supposed that the mass of pulley is neglectable in comparison with m_1, m_2, and m_3. I think that it is a reasonable approximation. @IrishBoy123

- IrishBoy123

@Michele_Laino yes, pulley's mass and inertia is to be ignored, agree.

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