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anonymous
 one year ago
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
 one year ago
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)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How should I approach this? I have no idea where to start

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you 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
 one year ago
Best ResponseYou've already chosen the best response.0since there is no friction, the total momentum along the xaxis has to be conserved: dw:1434562048822:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here we have to compute the acceleration of the object #1

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434562306655:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0in 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434562777709:dw where \tau is the tension of the string connecting objects #1 and #3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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?

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0This can't be right. If M1=M3 (which is possible), the acceleration would be infinite (which is physically impossible).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm. how do I solve for a1 then? :/

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0Do you have numerical values for M1, M2 and M3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to state the answer in terms of M1 M2 M3 and g

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw: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
 one year ago
Best ResponseYou've already chosen the best response.0what do you mean by constraining m3 in the vertical plane?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok. So this is what I did: dw:1434691193424:dw \[T = m _{1}a _{1}\ \\WT = m _{3}a _{3}\] \[m _{3}g  m _{1}a _{1} = m _{3}a _{3}\\m _{1}a _{1} = m _{3}g  m _{3}a _{3}\\a _{1} = m _{3}g  m _{3}a _{3}\] am I do this right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0almost. you have to divide the right hand side of your last equation by m1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh. oops, forgot about that. So then it would be: \[a _{1} = \frac{ m _{3}g  m _{3}a _{3} }{ m _{1}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How can I then get rid of a3 so the answer is only in terms of masses and gravity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 = (Tg)/m3 Unfortunately, we just introduced another unknown (T) These new relations bring us to: T/m1 = (Tg)/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
 one year ago
Best ResponseYou've already chosen the best response.0Since 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
 one year ago
Best ResponseYou've already chosen the best response.0(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
 one year ago
Best ResponseYou've already chosen the best response.0not sure if you're still here, but that should be everything you need :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes im still am. im writing them down cuz i cant really see it in text form lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if so, thank you! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, definitely write it out and make sure you understand the process! thats the most important part!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have a question. How did a3y become equal to T/m3 I thought its either a3y = a1x = T/m1 or a3y = (Tg)/m3

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \left\{ \begin{gathered} a = \frac{{{m_3}}}{{{m_1} + {m_3}}}g \hfill \\ \hfill \\ \tau = \frac{{{m_1}{m_3}}}{{{m_1} + {m_3}}}g \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0Now 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
 one year ago
Best ResponseYou've already chosen the best response.13 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
 one year ago
Best ResponseYou've already chosen the best response.0so from that, I solved for A and got: \[A = \frac{ m _{1}m _{3}g }{ m _{2} (m _{1}+m _{3})}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm. two different answers....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i see what you mean now @IrishBoy123

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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
 one year ago
Best ResponseYou've already chosen the best response.1well 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
 one year ago
Best ResponseYou've already chosen the best response.0oops.. 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
 one year ago
Best ResponseYou've already chosen the best response.1@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 Lagrangianlite 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
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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
 one year ago
Best ResponseYou've already chosen the best response.1that 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the answer from yahoo answer is right?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1no 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
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123, thats a good point about the relative velocity... what an inconspicuous mess of a problem

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1@LifeEngineer "inconspicuous mess of a problem" how beautifully put!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm. I can somehow understand it, but I would have to write it out myself to fully understand it

VincentLyon.Fr
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 I noticed that the 1/2 from the kinetic energy equation disappeared. I cant see how that happened

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\( \frac{d}{dt} [\frac{1}{2} M \dot X ^2] = 2 (\frac{1}{2}) M \dot X \ \ddot X = M \dot X \ \ddot X \).

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0In 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
 one year ago
Best ResponseYou've already chosen the best response.1@Michele_Laino yes, pulley's mass and inertia is to be ignored, agree.
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