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(III) The double Atwood machine shown in Fig. 48 has fric-tionless, massless pulleys and cords. Determine $(a)$ the acceleration of masses $m_{A}, m_{B}$and $m_{C},$ and $(b)$ thetensions $F_{T A}$ and $F_{T C}$ inthe cords.

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a.) $\frac{3 m_{\mathrm{A}} m_{\mathrm{C}}-m_{\mathrm{B}} m_{\mathrm{C}}-4 m_{\mathrm{A}} m_{\mathrm{B}}}{4 m_{\mathrm{A}} m_{\mathrm{B}}+m_{\mathrm{A}} m_{\mathrm{C}}+m_{\mathrm{B}} m_{\mathrm{C}}} g, \frac{m_{\mathrm{A}} m_{\mathrm{C}}-3 m_{\mathrm{B}} m_{\mathrm{C}}+4 m_{\mathrm{A}} m_{\mathrm{B}}}{4 m_{\mathrm{A}} m_{\mathrm{B}}+m_{\mathrm{A}} m_{\mathrm{C}}+m_{\mathrm{B}} m_{\mathrm{C}}} g, \frac{m_{\mathrm{A}} m_{\mathrm{C}}+m_{\mathrm{B}} m_{\mathrm{C}}-4 m_{\mathrm{A}} m_{\mathrm{B}}}{4 m_{\mathrm{A}} m_{\mathrm{B}}+m_{\mathrm{A}} m_{\mathrm{C}}+m_{\mathrm{B}} m_{\mathrm{C}}} g$b.) $\frac{4 m_{A} m_{B} m_{C}}{4 m_{A} m_{B}+m_{A} m_{C}+m_{B} m_{C}} g, \frac{8 m_{\mathrm{A}} m_{\mathrm{B}} m_{\mathrm{C}}}{4 m_{\mathrm{A}} m_{\mathrm{B}}+m_{\mathrm{A}} m_{\mathrm{C}}+m_{\mathrm{B}} m_{\mathrm{C}}} g$

12:14

Averell Hause

Physics 101 Mechanics

Chapter 4

Dynamics: Newton's Laws of Motion

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

Moment, Impulse, and Collisions

Amirul H.

August 18, 2020

what is ar?

University of Washington

Hope College

University of Sheffield

University of Winnipeg

Lectures

03:28

Newton's Laws of Motion are three physical laws that, laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. These three laws have been expressed in several ways, over nearly three centuries, and can be summarised as follows: In his 1687 "Philosophiæ Naturalis Principia Mathematica" ("Mathematical Principles of Natural Philosophy"), Isaac Newton set out three laws of motion. The first law defines the force F, the second law defines the mass m, and the third law defines the acceleration a. The first law states that if the net force acting upon a body is zero, its velocity will not change; the second law states that the acceleration of a body is proportional to the net force acting upon it, and the third law states that for every action there is an equal and opposite reaction.

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

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Okay we need to find the acceleration of each of the masses and both of the tensions. So this is the free body diagram you have gravity pulling down on math a tension pulling up the acceleration will pull the mass upwards. Same for mass B. Gravity pulling it down the tension pulling it up the acceleration will pull it upwards. Here is the tension for see gravity is going to pull it down the acceleration. Okay so we have these three equations of emotion and we can see that The attention at C. is equal to two times the tension at a. Okay so we need to get some kind of comparison between the acceleration. So we can do that with length. So let's let this length B. Y. One. This length here let's call it Y. A. This length why two. And then the length To see we're gonna call Y three. So we have those links So we know that why 1 -3. A. Plus Y tu minus white A. So why one minus Y. A. So we can just get this little distance and just get this little distance. So we're looking for that blue distance. This is equal to the length at A. And the length of this second rope is Y. A. Plus Y. Three is equal to Let's just call it length be. So to find the accelerations, let's take the second derivative. So to take this when we take the second derivative we get oh Okay 1 2nd. Let's change our naming convention so that it is a little less confusing. All right So this is going to be Y. A. This is going to be why one. This is going to be white B. It's going to be Y. C. So why A minus Y. One plus Y B minus white one. And this is Why one Plus by Sea. Okay so now we get the acceleration at -1. Plus the acceleration of b minus a. Two is when you take the first derivative this length will be equal to zero. So the second derivative will also be able to zero. This is why one. So we get the acceleration at A plus the acceleration at B is equal to two times the acceleration one. Let's use this. The acceleration At one plus the acceleration at sea Is equal to zero. So combine both of them and we get that for our last equation that we need negative two. A. C is equal to a plus A B. Okay to find acceleration three Is quite a long process. So let's number the equations and call this one 23 four and five. So You will substitute equations one and 2 into equation five. When you do that you will get one over Mass A plus one over nasty is equal to the force attention at a minus two G. Oh, excuse me, times the force of tension Is equal to -2 A. C. Yeah. So now sub equation three into equation for and after quite a bit of rearranging You get that a three is equal to negative last three C. Is equal to negative mass. See, Mass A let's mass B -4. Mass a must be over to pass a mess. The all of this is multiplied by two. Massey must be over messy times Mass A plus Mass B was for mass A. Mass B times G. Which reduces to so we get A C. Is equal to negative and see in A. Was in B -4 & A. And be over and see M A plus MB. That's four M. A and B times G. So this is the acceleration of C. You can plug this into equation three to find the attention at B. I'm sorry, the tension at see, so if you plug this in, you get that, the attention At C is equal to eight M. A. And B. And C. G over and see. Plus sometimes in A S M P Plus four M. A can be that's the tension at sea and then the tension at A can be found by looking at the equation for So this is four. I see SB nasty G over. I see. Mass A plus must be plus four times f. A. Mass B What's okay? It looks like it's glitch ng. So let me do a new page to find the acceleration at A. And the acceleration at B. Okay, so we have those values Subsequuting that tension into equation one is how we will find the acceleration of A. And when we do that We get four MBB m c minus in a minus m c m A plus n B. All of this will be times G over and see in a plus N B. Was four and A and be within to find A two or A B. Plug it into equation too. And we get four. That's a messi minus Mass B minus Massey, S A plus mess. Be over Massey as a plus math B was for Massey, nasty.

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