WEBVTT physics/ap-physics-c-mechanics/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I am Dan Fullerton and in this lesson we are going to talk about Atwood machines.
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Our objectives include drawing and labeling a free body diagram, showing all forces acting on an object.
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Understanding the tension is constant and a light string passing over a mass-less pulley.
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Analyzing systems of 2 objects connected by a light string over a mass-less pulley.
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Atwood machines, what is an Atwood machine?
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Basically what it is, it is a couple of masses on by light string over a pulley there on a ramp or straight up or on the table.
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The whole goal is to help students practice and learn about Dynamics Newton’s laws and how to apply these different concepts.
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Here is an example.
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We have 2 objects mass m1 and m2 connected by a light string over a mass less pulley over radius r.
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Pretty obvious to see if you look at the M1 if that is a bigger mass than m2 is going to be accelerating down as M2 comes up.
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We are going to analyze some of these.
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Now, properties of that Atwood machines.
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Ideal pulleys are frictionless and mass less, they had no inertia to the system.
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We will get into real pulleys here pretty soon.
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Tension is constant the light string passing over an ideal pulling.
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The tension here is = the tension there.
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How do we set up these problems?
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First, adopt a sign convention for positive and negative motion.
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In this case I see that we are going to be going down on the M 1 side so
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I would probably draw an arrow like this and I am going to call that my positive y direction.
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That means if M1 is moving down in that direction I'm going to call that positive.
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If M2 is moving up on the side I am going to call that positive.
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Let us get into these.
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How do we solve them?
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We are going to draw free body diagram for each mass.
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Write Newton’s law equations for each mass and solve for the unknowns.
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We will also talk about an alternate way to solve these after we have the basics down.
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First, let us take a look will call this the positive y direction or call this T1, the tension 1 on the M1 side.
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This T2 on the M2 side.
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I am going to draw free body diagram for M1 first.
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There is a dot from my object.
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We have T1 pulling it up.
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Down is our positive y direction because we are here and we called down the positive y on this side.
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And we have M1 G.
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Doing the same thing for M2 to draw free body diagram.
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We have our dot for M2.
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We have M2g pulling down.
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We have T2 up.
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But now on this side up is the positive y direction.
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We can write our Newton’s second law of equations.
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For M1, let us write that down here that force in the y direction is = M1 G - T1 which is all equal to M1 a.
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We can do the same thing for out second mass.
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Net force in the y direction is = T2 - M2 G which is = M2 a.
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We can try to put these together to eliminate some variables to see if we can solve for the acceleration of the system.
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The way I am going to do that there are bunch of different procedure to do so.
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In this problem, let us start by taking this equation here.
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You are going to write it up here M1 G - T1 = M1 a.
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Underneath that I'm going to write the corresponding equation for M2.
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T2 - M2 G = M2 a.
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I'm just going to add those together I am going to add the left hand sides in the right hand side.
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They are equalities as some of the left hand side has to equal some other right hand side.
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On the left hand side, I'm going to come up with M1 G - T1 + T2 - M2 G is all equal to.
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On the right hand side, we have a × M1 + M2.
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If I pull the a out of those is I add them together.
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Now, because tension 1 = tension 2, tension on each side and we have an ideal pulley - T1 + T20 that is going to be 0.
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I now have M1 G - M2 G = a M1 + M2.
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Let us factor in G out the left hand side.
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GM1 - M2 = a M1 + m2 and I want the acceleration of the system A therefore = G × M1 - M2 / M1 + m2.
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I know the acceleration of the system by using free body diagrams applying Newton’s second law
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and also the fact that we know that the tensions are equal in an ideal pulley.
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There is an alternate way we could analyze this too.
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If we look at the whole thing is a system calling this the positive y direction let us define our system here as we got an extended system.
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I'm just going to draw it inside the dashes there.
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We are just going to look at all of the forces that crossed the boundary of our system that are not internal forces.
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I'm going to redraw the system up here just to make it even more apparent what is going on.
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If I were to take those and spread them out this way we are going to have is something that looks kind of like this.
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We will have M1 on the left it is connected by a string to M2.
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There is our system we are calling this way the positive y direction and if I look at the forces
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that are crossing that boundary of our system on the M1 side we have M1 G.
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On the m2 side we have m2g.
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If I were to look at this in applying Newton’s second law to something like this or call the net force in the y direction
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I'm going to have M1 G and a positive y direction - M2 G.
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The net has to equal the mass of our system M1 + M2 × acceleration a or just rearranging for a, A = G × M1 - M2 / M1 + m2.
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Same answer just taking it and looking at it from a systems approach.
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You can use a system like that as well as when you go to solve these problems.
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It is fun to do it both ways and check and make sure you get same answer.
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Let us take an example of the basic Atwood machine.
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Find the acceleration of the 20 kg mass given the masses are connected by light string over an ideal mass less pulley.
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It looks to me like the 20 kg mass is the 1 that is going to win here.
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Let us call this our positive y direction and we will do this when the old fashioned way.
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Let us first take a look we will call this M1 and M2.
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We will draw our free body diagram for M1.
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I got my object I have T1 and T2.
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We have T1 or T because they are the same on both sides and M1 G down that would be a positive y direction.
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For our second mass, M2 down will be our positive y direction and we have tension upwards and M2 G down.
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Now, drawing or writing Newton’s second law equations for M1 we can see they were going to have T - M1 G = M1 a.
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For our second mass, we are going to have M2 G - T = M2a.
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We will combine these together T - T that is going to be 0.
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We will end up with on the left hand side M2 G - M1 G = M1 a + M2 a or G × M2 - M1 = A × M1 + M2.
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Or a =G M2 - M 1 / M1 + M2.
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And substituting our values is going to be =10 m / s² M2 20 kg – M1 15 kg a/ 20 + 15 or 35 kg.
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Gives us an acceleration of about 1.43 m / s².
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Let us take a look at another one.
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Masses are hung on a light string attached to an ideal mass less pulley has shown a diagram here below left.
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The total mass hanging from the left string is = that on the right to their equilibrium here, 1 kg 1 kg.
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At time T = 0 the 0.2 kg mass is moved to the right side.
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This one gets bigger that one gets a little bit smaller.
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How far does each mass move in 1 second?
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I think we need a little bit more room to solve this one.
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Let us move to the next slide we got a little bit more room and we will do our analysis.
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Let us call this a M 1, we will call this T1 and that will be a positive y direction.
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Here we will have M2 and T2.
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Looking at our first mass M 1, we have T1 up and M1 G down.
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For M2 we have T2 up and M2 G down.
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If I put these together with Newton’s second law equations make sure we define our positive and negative directions that is positive Y and that is positive y.
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We have T1 - M1 G = M1a and we have M2 G - T2 = m2a.
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T1 and – T2 can get to 0 because the tension is equal on both sides their T 1 = T2.
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We come up with M2 G - M1 G = M1 + M2
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or A = G M 2 - M1 / M2 + M1 which is again 10 × the difference in the masses which is going to be a 0.4 kg ÷
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the sum of the masses 2 kg or 2 m / s².
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Hopefully, you are starting an idea how these repeat in that general pattern.
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That has just how far does it move?
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Now, that is a kinematics problem are going to look in the y direction and the vertical direction
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We call down the positive y direction the initial equal 0 we do not know the final we do not know Δ y.
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But we know the acceleration in the y is 2 m / s² I want to know how far it goes in 1second.
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Using kinematics Δ y = the initial T + 1/2 to 80²
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or that is going to be since the initial is 0 that term goes away.
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Δ y = ½ × are acceleration 2 m / s² × our time 1/2⁺or Δ y = 1 meter.
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We will we can not adjust our Atwood machines a little bit to make a touch more interesting.
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Here we have 2 masses and M1 and M2 connected by a light string / masses pulley.
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Assuming a frictional surface find the acceleration of M2.
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We are going to define that is our positive y direction tensions are the same T and T and let draw our free body diagrams.
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For M1, we have normal forced up, we have M1 G down and we have T to the right which for M1 is going to be what we call the + 1 direction.
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For M2, we have tension up, we have M2 G down, we are calling down the positive y direction over here.
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Finding a 2nd Newton’s law equation in the direction of motion for N1, the net force.
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What we are calling the line that is really horizontal for M1 is going to be = just T which must be = M1a.
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Doing the same thing for mass 2 net force = M2 G - T = M2a.
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Let us see if we can put this together we have T = M1a and we are going to combine that with M2G - T = M2a and what do we get?
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M2 G = M1 + M2 × a or a = G × the quantity M2 / M1 + M2.
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Just slight variation on the same theme.
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We can even do this with their ramp.
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2 masses M1 and M2 are connected by light string over masses pulley.
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A sum of frictional surfaces finding acceleration of M2.
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Let us start with our free body diagrams again will do M1 cassette looks like it is a more complicated
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We are pretty good doing free body diagrams for objects on ramps.
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Let us do our x over here there is our y.
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So x y and their object M1 we have attention of pulling it up the ramp we have normal force and we have M1 G.
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M1 G does not lineup with the an x again you know the drill let us turn it into a pseudo free body diagram.
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It will be a little bit more useful to us for problem solving.
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There is our x, there is our y, and let us define that direction as + y.
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We are not certain it is actually going to go that way.
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It could be the other direction but let us define it that way and if we come up with the negative acceleration we know we chose.
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Tension up the ramp, normal force perpendicular to it and M1 G we got to breakup the components
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and that is our angle θ that means that is also angle θ.
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This would be the adjacent side M1 G cos θ.
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This would be the opposite side M1 G sin θ.
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I am going to write our Newton’s second law of equation here calling up the x axis for calling + y T - M1 G sin 30° = M1 a.
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Or solving for T, t = M1 G sin 30° + M1 a.
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Let us do our free body diagram from as 2 here.
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For M2 down as our positive y direction we have tension up and M2 G down.
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M2 G - T must equal M2 a in the Newton’s second law M2 G - T is = M1 G sin 30° + M1 a.
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All of that must be = M2 a or a little bit of math here.
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M2G - M1 G sin 30° - M1 a = M2 a.
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Factoring G × M2 - M1 sin 30° must equal a × M1 + M2.
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Or finally solving for acceleration that is going to be G × M2 - M1 sin 30° ÷ M1 + M2.
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We can keep getting trickier here.
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Let us finish off with one last question our ranking test.
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Rate from least the greatest the acceleration in net force of these 6 different Atwood machines.
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The first thing I think I'm going to do here with all of that data is I'm just going to solve a generally without plugging in the values.
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What I'm going to do is I'm going to do this one from a systems approach.
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Let us draw our systems.
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We will call this on the left M1 that one on the right is M2 and we will define our system M1 is that way.
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We will call them + y direction.
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We will define our system like we did with our ultimate solution earlier just to change things up as looking like that.
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When I do that the net force in the y direction are going to have m1 G and the positive y direction - M2 G
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and all of that has to be equaled Ma where our total mass is M1 + M2 A.
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Or acceleration is going to be net force in the y ÷ M1 + m2.
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The net force is just M1 G - m2 G or net force will be = G × M1 - M2.
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Now, all of this different data seems to be the easiest way to handle it might be to make a table.
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Let us do that.
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We will have a row for ABCDE and F.
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Our data will have a value from s1, a value from S2.
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Once we have those we can calculate the net force by multiplying G × the difference of those.
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Let us make a column for net force and then we will know the acceleration.
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Let us make a column for acceleration which is the net force ÷ the sum of m1 and m2.
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We can start filling in our data.
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For situation A, M1 is 5, M2 is 1.
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The net force is going to be M1 - M2 4 × G10 or 40.
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The acceleration was going to be the net force ÷ the total mass 40 / 6 which is 6.67.
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I can just keep going to fill it in there is my table in that manner.
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For B, we have M1 is 3, M2 is 1, the net force is going to be the difference of those × 10 or 20 in the acceleration 20 ÷ 4 or 5 m / s².
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For C, mass 1 is 4, mass 2 is 2.
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Again, we will have a net force of 20 that our acceleration is going to be 20 ÷ the total mass 6 or in this case 3.33.
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For D, we have initial mass m 1 is 1, M2 is 4, so the magnitude of the net force 4 -1 is 3 × 10 is going to invest 30 N.
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30 ÷ the total mass 5 will be 6 m / s².
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For part E, 1 and 2 for masses.
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Our net force is just going to 10 and accelerations 10 ÷ total mass of 3 is just 3.
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For F, we have 8 as our first mass, 2 as our second mass, the difference is 6 × 10 =60 N for force.
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Our acceleration is 60 ÷ 10 or 6 m / s².
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We are asked to write these from least to greatest for acceleration and for net force.
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I just use my table to rank them from least to greatest for acceleration.
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It looks like we have C and E together those are the smallest.
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Then we go up to B than it looks like D and F for both 6 those are together.
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And finally, our highest acceleration A.
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Same thing for the net force that looks like our starting point is E our lowest then B and C are both 20 then we come to D 30 then we come to A 40.
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Our highest net force F with 60 N.
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Hopefully, that gets you a good start with Atwood machines.
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Get you feel comfortable with them and lots more practice with Newton’s second law and how you apply that with free body diagrams and pseudo free body diagrams.
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Thank you so much for watching www.educator.com.
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We will see you again soon and make it a great day everyone.