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(II) Three blocks on a frictionless horizontal surface are incontact with each other as shown in Fig. $42 .$ A force $\vec{\mathbf{F}}$ isapplied to block $\mathrm{A}$ (mass $m_{\mathrm{A}} )$ . (a) Draw a free-body diagram for each block. Determine $(b)$ the acceleration ofthe system (in terms of $m_{A}, m_{B},$ and $m_{C} ),(c)$ the net forceon each block, and $(d)$ the force of contact that each block exerts on its neighbor. (e) If $m_{A}=m_{B}=m_{C}=10.0 \mathrm{kg}$ and$F=96.0 \mathrm{N},$ give numerical answers to $(b),(c),$ and $(d)$ .Explain how your answers make sense intuitively.

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a.) In the free-body diagrams below, $\overrightarrow{\mathbf{F}}_{\mathrm{Aa}}=$ force on block $\mathrm{A}$ exerted by block $\mathrm{B}, \overrightarrow{\mathbf{F}}_{\mathrm{BA}}=$ force on block $\mathrm{B}$ exerted by block $\mathrm{A}, \overline{\mathrm{F}}_{\mathrm{BC}}=$ force on block $\mathrm{B}$ exerted by block $\mathrm{C},$ and $\overrightarrow{\mathrm{F}}_{\mathrm{cl}}=$ force on block C exerted by block B. The magnitudes of $\overrightarrow{\mathbf{F}}_{\mathrm{BA}}$ and $\overrightarrow{\mathbf{F}}_{\text {AB }}$ are equal, and the magnitudes of $\overrightarrow{\mathbf{F}}_{\mathrm{BC}}$ and $\overrightarrow{\mathbf{F}}_{\mathrm{ca}}$ are equal, by Newton's third law.b.) $a=\frac{F}{m_{A}+m_{\mathrm{B}}+m_{\mathrm{C}}}$c.) $F_{\mathrm{A} \mathrm{net}}=F \frac{m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$, $F_{\mathrm{B} \text { net }}=F \frac{m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$, $F_{C \mathrm{net}}=F \frac{m_{\mathrm{c}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$d.) $F \frac{m_{c}}{m_{A}+m_{B}+m_{c}}$, $F \frac{m_{c}}{m_{A}+m_{\mathrm{n}}+m_{\mathrm{c}}}$, $F_{\mathrm{AB}}=F \frac{m_{\mathrm{B}}+m_{\mathrm{C}}}{m_{\mathrm{A}}+m_{\mathrm{B}}+m_{\mathrm{C}}}$, $F \frac{m_{2}+m_{3}}{m_{1}+m_{2}+m_{3}}$e.) $3.20 \mathrm{m} / \mathrm{s}^{2}$, $32.0 \mathrm{N}$, $64.0 \mathrm{N}$, $32.0 \mathrm{N}$

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

Cornell University

University of Washington

Hope College

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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three blocks apart. A is simply asking us to draw the free body diagram, so this it would be in with this would be the answer for per day. We just have to late rather draw the mass and label of the forces that are acting on that mass. Now. For Part B, however, we need to find the acceleration of the total system. So we can say that the sum of forces would be equal to the force minus the force of Block AM Block B plus the force on block a block B and block A minus the force of Block C on Block B plus the force of Block B on Block C. And then this would be equal to F because all of these would cancel out due to Newton's third Law. And so we can say that the sum of forces would then be equal to the sum of the masses. Time's A which is giving us that the acceleration would simply be the force divided by the sum of the masses. So this would be your answer for a part B now, for a part. See, they want us to find the net force on Block C Whether the net force on each block So we can say that the force the net force on block A would be equal to the force times massive a divided by the total mass um massive A plus Master B plus mass subsea on B It would follow the same pattern So this would be the force times amsa be This would be divided by M of a plus B plus m sub sea and then for C, This would, of course, be forced times the mass of C divided by mass again. So this would be those three answers for apart. See, Now Part D is asking us to find, um ah, essentially the expression Oh, are the magnitude for all of the ah intermediate forces. So essentially efs of a B s, A b a B c. So we can say that for d, we can say that force of block B on sea would be equal to force C net and this would be equal to again the forced times m subsea divided by that total mass. And at this point, we can say that this is going to be equal to force the BC due to Newton's third Law of Motion. Now we have to apply it to block A. So we can say that force minus force of a B ah would be equal to, of course, force a net. We've we know that this is going to be equal. Uh, rather weekend actually say that. Force A B equally, of course, foursome be a with equal force minus. And then that force a net so we'd be forced times I'm sub a divided by the total mass. And after this, we can ah substitute. And so we can say that the force of a B equal in the force force of B A. This is giving us force times the sum of the mass of Block B and Block C divided by again that total mass. So at this point for a party, they actually want us to find the force in the net force on each mass and the acceleration. So the acceleration is quite easy. We found the expression in the first page and so we can just plug in. And so this would be 96 Newtons force and then the sum of all the masses would be 30 kilograms. Let's I need six points. Are Nunes uh, divided by three 0.0 kilograms. And so this is giving us 3.20 meters per second squared now. The net force on each mass force net would be the same for all of them and so this would be m A. So this would be 10 kilograms multiplied by 3.20 meters per second squared, giving us 32 Nunes. So that would be the net force on each mess. And to find the magnitude of a B equaling f sub b A, this would be equal to the acceleration times, massive B plus massive sea, and this would be equal to 20 kilograms multiplied by 3.20 meters per second squared. And this is giving us ah 64 new needs now at this point for four sub BC. This is equaling for some C B. And at this point it's pretty apparent that this illegal again 32 Nunes so this would be our final answer for a Part E and our final answer for the second final answer for a party that is the end of the solution. Thank you for washing

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