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(11) The cords accelerating the buckets in Problem 33 $\mathrm{b}$ ,Fig. 37 , each has a weight of 2.0 $\mathrm{N}$ . Determine the tension ineach cord at the three points of attachment.

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1. the tension in the cord at first point is $[35 \mathrm{N}]$2. the tension in the cord in between the two buckets is $[38 \mathrm{N}]$3. the tension in the bottom of the second cord is $\overline{73 \mathrm{N}}$

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

Rutgers, The State University of New Jersey

Hope College

University of Sheffield

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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each point of attachment. So we can first, um, draw the first free body diagram. This would be the force tensions of one at the bottom and then going straight down would be the mass A B G. And so we can say that he's some of forces in the wind direction would be equal to the mass times acceleration of the Y direction which would be equal to the forced tension. Someone bot bottom minus. Um so be dreams. So we can say that the force tensions of one at the bottom would be equal to the mass times the mass sabi times G plus a sad boy. So at this point, we can say that we can actually solve and say force attention. Someone in the bottom would be equal to 3.2 kilograms massive be multiplied by 9.8 plus 1.25 meters per second squared. And this is giving us ah, 35 0.36 new tints. Let's take the rope. So we have this route going down would be the force tensions of one bottom. So this would be the same force as here and then going down would be the mass of the rope times G and then going up would, of course, be force tensions of one at the top. Uh, again, we hear we can say that, um, some of forces in the wind direction would equal the mass times acceleration of the Y direction, and so this would be equal to the force. Tensions of one top minus force tensions of one bottom minus m sub r G. And so we can say that the force tension someone at the top would be equal to the force. Tension. Someone at the bottom, plus the weight of the rope times gravity divided by gravity, where the rate of the rope divided by gravity, is equaling theme ass of the rope. And so gravity Counsel's out. And this is ah, plus the weight of the rope times the acceleration of the wire direction divided by G once again and we can solve. So the force tension Someone top would be equal to 35.36 plus the weight of the rope to Newton's 35.36 Newton's my apologies plus 2.0 noons and then plus the weight of the rope. Two points their own Newtons, uh, multiplied by the acceleration of 1.25 meters per second squared. And this is going to be divided by 9.8 meters per second squared. And we finally have that force of tensions of one at the top 37.6 Newtons. So this would be your second answer. This would be your first Ellen. Finally we have forced tensions of two and then going straight down would be forced tension, someone at the top and then going straight down again would be the massive block, massive, massive, massive, the block times G. So this would be the sum of forces in the right direction. This would be equal to the mass times acceleration of the Y direction. Aah! This would be equal to the force tensions of two minus the force tensions of one at the top minus the mass of the block times G. And so we can say that the force tension said to would be equal to the force tensions of one at the top, plus the mass of the block times ace of Y plus G. And so we can solve at this point, so forced tensions of two would be equal to 37.615 Newtons Ellen, plus 3.2 kilograms. My apologies, uh, multiplied by the summer 1.25 plus 9.8 again meters per second squared. And this is giving us 72.98 Newtons. So this would be our final answer. That is the end of the solution. Thank you for watching.

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