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2.1 Pre-LabIn this week s upcoming lab, you will be looking at the acceleration of two coupled objects; a cart and hanging mass See Fig: 2.1.PASCO Cart(hanging mas...

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2.1 Pre-LabIn this week s upcoming lab, you will be looking at the acceleration of two coupled objects; a cart and hanging mass See Fig: 2.1.PASCO Cart(hanging massFigure 2.1 A sketch of the experimental apparatus, showing the basic components_[2.1] Question: Draw by hand, and attach digitally to your Pre-Lab, one free-body dia- gram for each of the two objects: the cart and the hanging mass. Be sure to include all possible forces acting on the objects b_ For each free-body diagram, write out N

2.1 Pre-Lab In this week s upcoming lab, you will be looking at the acceleration of two coupled objects; a cart and hanging mass See Fig: 2.1. PASCO Cart (hanging mass Figure 2.1 A sketch of the experimental apparatus, showing the basic components_ [2.1] Question: Draw by hand, and attach digitally to your Pre-Lab, one free-body dia- gram for each of the two objects: the cart and the hanging mass. Be sure to include all possible forces acting on the objects b_ For each free-body diagram, write out Newton's Second Law (Fnet ma) for that particular diagram. Be careful that your coordinate systems are consistent between the two diagrams Your equations should depend on mass, mass of the cart, mc' acceleration, a, and gravita- hanging mh, tional constant, g.



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In Atwood's machine, two masses $m_{1}$ and $m_{2}$ hang vertically, connected over a frictionless pulley, as in Figure P4.62. Assume $m_{2}>m_{1}$. (a) Draw force diagrams for both masses. (b) Find the magnitude of the masses' acceleration, in terms of $m_{1}, m_{2},$ and $g$. (c) Evaluate the acceleration numerically if $m_{1}=0.150 \mathrm{~kg}$ and $m_{2}=0.200 \mathrm{~kg}$.

So here we control all the system. It's is this a frictionless massless pulley? Aah! This would be, ah, maths. A going up would be the force normal of a going down would be, of course, um sub a g and going to the right butts. This would be massive. Be going to the right here would be forced. Tension going up here would be forced. Tension going down would be m sabi tree and this is the full free body diagram for the system. Now we can say that the force tension is going to be equal to rather this would actually my apologies. This would be your answer for part A. So just party just asked us to draw the free body diagram. This would be the full free body diagram for part B. However, now we need to solve for the acceleration. So here. Ah, for block A. There is no motion in the vertical direction. Therefore forced normal of a wood equal the mass of a g again. This plane right here. This table is perfectly horizontal. Therefore, the normal force of block a on block a rather would be equal to the weight of block A We can apply Newton's second long to block a in the ex direction. This would equal force tension and this would equal the mass sub a times acceleration of a in the X direction. We can say that the sum of forces for block B in the UAE direction would be equal to M C B G minus forced Sebti Aah! This would equal and sub B times a. So be in the wind direction. Now the two blocks air connected. Therefore, the acceleration of block A in the ex direction with the equal to the acceleration of block B in the wind direction. And we're going to call this simply a So we're gonna lose the sub scripts for a and then we can combine the two force equations. We can essentially combine these two equations to solve for a so we can say that forced tension would then be equal to m sub a time I'm so be she with minus force sub t would be equal to m sub b a and then we're going to say that m c b g minus m c a a cool m sub b times a uh this would become massive eight times today plus massive B times A would equal mass Sabi G And we can say that the acceleration would then be equal to the massive be times G divided by the sum of the masses. And this would be your answer for the acceleration. At this point, we want to find force tension. We know forced tension would be equal to again m sub a times, eh? Therefore, this would be equal to g times m sub a m sabi divided by again the sum of the masses m sub A plus I'm sabi So this would be your force tension here That is the end of the solution. Thank you for watching.

Hello. And this problem, we have an Atwood machine which is basically a bully and accord With two masses hanging on each side. So here we have a mass of seven kg. And on the other side we have another mass of nine kilograms. The question here is to determine the exploration of this group here and also to find attention in the court. So let's first Draw the forces on each object here. So here we have the weight of this nine kg object. F weight is equal to the mass of the object, multiplied by the gravitational exploration. This looks like jesus who let's remove this part here. So that's the first force. The second force here is the force of weight also of the other block, And it's equal to seven, multiplied by 9.81 here, ignoring the friction of this bully here makes the tension equal on each side. So we have a force of tension here Acting on this seven kg block. And another Force of tension which is equal to this one. Also acting on this nine kg Flock from here, Let's get the equation of motion of each one of them. Or simply applying Newton's second law to each block in individually. So let's start by the nine kg block. We know that the summer forces on that block will be equal to the mass times the acceleration of that block. So here we assume that the direction of motion will be Favored to the nine kg. So this one is going down. So let's assume that down is positive and upward is negative here. So in that case we have the force of weight of that nine kg object minus the force of tension And that's equal to the mass of this block, which is nine kg multiplied by the acceleration. Let's start blogging in numbers. The force of weight here is nine multiplied by nine 0.81 minus the force of tension, And that's equal to nine eight. And let's call this equation number one. Now let's do the same. But for The other object here, the seven kg object, this object will be moving upward and this is just a physicist intuition. So because the intuition says here that That nine kg will be the heavier side and it will move down side and this one will be moving upward. So here we have the force of tension minus The force of weight of that seven kg Block is equal to the mass times the acceleration of this block. Here we need to note that both both blocks being connected with cord, they will have the same acceleration regardless of the opposite direction. This one will be moving downward with the same acceleration and this one will be moving upward with the same magnitude of acceleration. So here let's blufgan numbers, we have the force of tension which is also the same for both sides And here we have the weight of the seven kg mess. No, and again here we have seven multiplied by the exploration. And this will be equation number two. Now let's add equation want to equation too. So we get rid of the force of tension here. So one plus two, we get nine times 9.81 -7 times 9.81. So that's two Times 9.81. And we have negative the force of tension and hear positive the force of tension. So they cancel each other. And finally we have on the right hand side nine Times the exploration plus seven times the same exploration. So that's 16. The exploration. From here, we can directly calculate the exploration and we find out that the acceleration is equal to two times 9.81, divided by 16, Which is equal to 1.2, 3 meters per second square. So that's the final answer for the first part. And that's the acceleration. The second requirement here is to calculate the force of tension. And that's pretty easy because we can substitute with the value of the acceleration we got here in any of the equations number one or number two. And we get the force of tension. So let's substitute an equation number one. And we will have that the force of tension is equal to nine times 9.81 -9 times the acceleration, Which is equal to approximately 77 Newton's. So that's the fincen in the court. And that's the exploration

Haider. So in this problem we need to determine the angle that the strings make with the vertical. So in here we have a dice. And the first thing that we need to do is to draw all the forces that are acting on the dies. And then appliance need to second law we can obtain equations for the Y. And the X. Company. So doing that we find that we have attention that we call team that is due to this string. And we also have the weight of the dice that we call. W. We also note that the system has an acceleration because the dice is inside of a car and it axillary res from from velocity in six seconds. That velocity is that we call final velocity Is equal to 28 m/s in a time of six seconds. So um we first are going to find the equations of for the dynamics of this. So some of the forces in the X company we have that this is detention sign of tita because the angle is with respect to the vertical and this because the car is moving in the horizontal way it is moving in the ads access. So dad we will have an acceleration in this company. So this is mm which is the mask of the dice times the acceleration for the white company of the forces. We will have two forces which are T casino tita in the positive part of wine and minus their weight in this case the dice is not moving in In this company. So it is equal to zero. Now from this we obtained that the tension times the cosine of theta is equal to the weight. We know that the weight is equal to the mass times the acceleration due to gravity. With these two equations, we're going to call this one in these two we are going to determine the angle tita and what we can do is to eliminate detention in the other equation so that we can divide want over to Yeah, Equation 1/2. So that we could see that these will give us deal sign data over T of consign of tita and for the right hand side we will have and A times A. M. Of G. So we have that. We can eliminate this too. And we also eliminate the tension so that we know that sign of tita. Over a dozen of teacher is the tangent of tita. And then we will have a over T. So to obtain the angle tita, we take the inverse function of the tangent function for both of the size of these equations. So we won't have Tangent of -1. All of the acceleration over the acceleration due to gravity. We know the acceleration due to gravity because this is the value of 9.8 m/s. But we don't know the acceleration of the dies or the car in which the diet is in it. Sm So we need to determine the acceleration but we know from the beginning that we are given a velocity and the time. So you think any memories? Yes. In key markets we can solve this by using the following equation. We know that the final velocity is equal to the initial velocity plus the acceleration times at time T. And we know that the initial velocity is zero and the final velocity is the one that we are given. So solving for a, we know that this is the final velocity over the time. We know that the final velocity is 28 meters per second and the time is six seconds. So from here we obtain about you off. Yeah. Mhm. For 0.6, see seven meters per second is square. So now we know the body of the acceleration. So we just simply need to substitute that value into the equation for tita. And then we can obtain the value for the angle tita. So it does regions substitute those values. The acceleration is 4.66 m/s square and the acceleration due to gravity is 99.8 m/s square. So good in this, plugging this into the calculator, we obtain about you. Yeah, 25 point 48 degrees. And this is the angle the district make with the particle

Free by diagrams party in the first block over here on the surface. So you have normal force going upwards. You have mass times grab me the weight going down and then you have the force attention being pulled on. Then for second block, you have force of tension being pulled upwards and then you have mass two times gravity weight being pulled down. As for the free body doctors since two blocks are connected, the maxims of their accelerations will be the same. So what a one. The extraction equal a to in the y direction the T equals the overall acceleration. So if we combine forces equations in second law, we get this is on for part de now. So force attention equals and one times a that's two times gravity. My eyes force attention equals mass to times acceleration. So rearrange subs to you get massive too Times gravity, my eyes, massive block One times acceleration equals mass to times acceleration, the neck business and to times a being and one times a plus and two times a equals and to times she so you re range and solve for a celebration equals gravity times. That's what to over and one plus into forced tension came equals mass and one times a equals g times m one and to over and one plus. And to this is for your part a free, wide diagrams over here. And then we saw four acceleration terms. Love Newton's second law, all right?


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