The Atwood machine consists of two masses suspended from a...

The Atwood machine consists of two masses suspended from a fixed pulley, as shown in $v$ Fig. $4.42 .$ It is named after the British scientist George Atwood $(1746-1807),$ who used it to study motion and to measure the value of $g$. If $m_{1}=0.55 \mathrm{~kg}$ and $m_{2}=0.80 \mathrm{~kg}$, (a) what is the acceleration of the system, and (b) what is the magnitude of the tension in the string?

The Atwood machine consists of two masses suspended from a fixed pulley, as shown in $v$ Fig. $4.42 .$ It is named after the British scientist George Atwood $(1746-1807),$ who used it to study motion and to measure the value of $g$. If $m_{1}=0.55 \mathrm{~kg}$ and $m_{2}=0.80 \mathrm{~kg}$,
(a) what is the acceleration of the system, and
(b) what is the magnitude of the tension in the string?

Key Concepts

Tension in the Rope

Tension is the pulling force transmitted through a string, rope, or cable when forces are applied at its ends. In the context of the Atwood Machine, analyzing tension is essential because it affects both the acceleration of the masses and the dynamics of the entire system, and is calculated by balancing the forces acting on the masses.

Free-Body Diagram

A free-body diagram is a critical analytical tool used to visually represent all the forces acting on each object in a system. For the Atwood Machine, drawing separate diagrams for each mass helps isolate and quantify the gravitational force and the tension, thereby simplifying the process of setting up equations based on Newton's Second Law.

Newton’s Second Law

Newton's Second Law of Motion, expressed as F=ma, is fundamental to understanding the dynamics of the Atwood Machine. By applying this law to each mass in the system, one can derive equations that relate the net forces acting on the masses to their acceleration, thus determining both the acceleration of the system and the tension in the connecting string.

Atwood Machine

An Atwood Machine is a simple mechanical system consisting of two masses connected by a string that passes over a pulley. It serves as a classic experiment in classical mechanics to study uniformly accelerated motion and to illustrate the principles of force and acceleration in a controlled system.

Transcript

00:01 In this problem, we're going to be talking about forces in motion, more specifically, tension forces, and acceleration in an atwood machine.

00:08 So an atwood machine is like a pulley system with two masses on it, and it can usually be drawn and solved to use, like looking like this.

00:19 It's a poor drawing, but i think the easiest way to look at it is in more of a horizontal way, just if you kind of stretch it out and pull these parts.

00:31 Up to the sides and make it horizontal.

00:34 So i'm going to draw it like that, how, with everything that we are given.

00:39 So here's the first mass.

00:42 We'll do m1.

00:45 That's the string connecting going around the pulley connecting to m2.

00:55 Okay.

00:57 So if you think about it, there's going to be this force here, which is going to be the force of gravity of m2.

01:04 So i'm going to write that as f2.

01:07 And then this side is going to be the force of gravity f1.

01:13 And these are the forces of gravity directions because normally they would be, the boxes would be here.

01:21 And this is the force of gravity, but when you spread it out, it's going to be pointing to the side, if that makes sense.

01:29 So then we're going to have the tension force in the middle of the strain.

01:35 And then the direction of the acceleration is this way.

01:43 So i'll label that as a.

01:45 Okay, so this is the drawing i'm going off of.

01:51 So we are first going to solve for f1 and f2, because that is the force of gravity for the masses.

01:58 So for f1, we have 0 .55 kilograms times 9 .8.

02:08 Meters per second, so we get 5 .39 newtons.

02:16 Then for f2, we have 0 .8 kilograms times 9 .8 meters per second squared, and we get 7 .84 now that we have those two outside forces, we can find the net force of the system...

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