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(III) Suppose two boxes on a frictionless table areconnected by a heavy cord of mass 1.0 $\mathrm{kg}$ . Calculatethe acceleration of each box and the tension at each end of the cord, using the free-body diagrams shown in Fig. $49 .$Assume $F_{\mathrm{P}}=35.0 \mathrm{N},$ and ignore sagging of the cord. Compare your results to Example 12 of "Dynamics:Newton's Laws of Motion" and Fig. $22 .$

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$1.52 \mathrm{m} / \mathrm{s}^{2}$, $18.3 \mathrm{N}$, $19.8 \mathrm{N}$

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

University of Washington

Simon Fraser University

University of Sheffield

McMaster University

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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(III) Suppose two boxes on…

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Suppose two boxes on a fri…

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Problem_7 Box A and box B …

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The figure below shows two…

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In Fig. 3-22, the two boxe…

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(II) Figure 45 shows a blo…

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As shown in Fig. $3-11(a)$…

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In Fig. 5-64, a force $\v…

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(II) Figure $4-49$ shows a…

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Figure $5-56$ shows a box …

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Two boxes connected by a l…

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In Fig. $5-64,$ a force $\…

so we're gonna initially treat the two boxes and the rope as a single system. So if we treat everything all masses of a single system, we can say that the Onley accelerating force would be the force of force A p. So here, if we use Newton's second, while we can say that the acceleration would be equal to force a Pete divided by the total mass of the entire system, this would equal 35 Newtons divided by 23.0 kilograms, giving us an acceleration of 1.52 meters per second squared. So this would be one answer That would be the acceleration of the system of each part of the system. Now we have to consider massive be alone. So after we consider massive be alone, we can say that the force of we can say force of tension on mass on the mass be would be equal to the massive be times the acceleration of the system. So this would be 12.0 kilograms multiplied by 1.52 meters per second squared. This is giving us a force tension on be 18.3 Nunes and then for a we can say that we're going to consider now the rope alone. Therefore, the net force on the rope would be equal to of nets on sea or the cord would be equal to force tension on a minus the force tension on B and we can say that this would equal the massive cord times the acceleration of the entire system. Therefore, forest tension of a would be equal to m subsea, eh? Plus force tension on B ah, we can solve and say that the force tension on a would be equal to 18.26 Newton's. That's not round to the very end, plus the mass of the rope essentially one kilogram times the acceleration of the system as a whole 1.52 meters per second squared. This is gonna give us 19.8 noons. So this would be your force tension on a That is the end of the solution. Thank you for watching

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