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. Muscle attachment. When muscles attach to bones, they usually do so by a series of tendons, as shown in Figure 1.27 In the figure, five tendons attach to the bone. The uppermost tendon pulls at $20.0^{\circ}$ from the axis of the bone, and each tendon is directed $10.0^{\circ}$ from the one next to it. (a) If each tendon exerts a 2.75 $\mathrm{N}$ pull on the bone, use vector components to find the magnitude and direction of the resultant force on this bone due to all five tendons. Let the axis of the bone be the $+x$ axis. (b) Draw a graphical sum to check your results from part (a).

13.3661 $\mathrm{N}$$\theta=40.2^{\circ}$

Physics 101 Mechanics

Chapter 1

Models, Measurements, and Vectors

Physics Basics

Audry B.

August 28, 2020

Cornell University

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

04:16

In mathematics, a proof is a sequence of statements given to explain how a conclusion is derived from premises known or assumed to be true. The proof attempts to demonstrate that the conclusion is a logical consequence of the premises, and is one of the most important goals of mathematics.

09:56

In mathematics, algebra is one of the broad parts of mathematics, together with number theory, geometry and analysis. In its most general form, algebra is the study of mathematical symbols and the rules for manipulating these symbols; it is a unifying thread of almost all of mathematics.

03:20

When muscles attach to bon…

03:28

Refer to the following:

03:58

Muscle $A$ and muscle $B$ …

04:02

$\bullet$ Tendons. Tendons…

01:01

When a large bone such as …

03:55

Bones and muscles. A patie…

02:03

Muscles are attached to bo…

01:11

The large quadriceps muscl…

04:09

For Exercises 87 and 88, r…

07:05

this's Chapter one problem. 56. So in this problem, we're given five vectors. We're told the magnitude of each vector when which is all the same among all the vectors we've got. That's 2.75 mutants, and we're told the directions. The vectors, which are different reach one Call us are the magnitude of the vector in every case. And so we're asked to add up the vectors, even victor components, to find the magnitude and direction of the resultant force when we add those altogether. So in Part B, were asked to do a graphical some I'm actually going to start with that just to get a sense of how that will look. So starting from the right, most vector in the in the figure given in the text Let's have this length we 2.75 Newton's and let's say it's a needle of 20 degrees in the next one is bent little 10 degrees to the right, the next one ten 10 degrees to the right of that 10 degrees to the right of that, these air supposed to really small angles, and you could draw better if you have a straight edge and 10 degrees to the right of that. So this last one should be. I've drawn maybe a little too curd, but should be 60 degrees from vertical 20 plus 10 plates, 10 plates, 10 plates. 10. So if there were a vertical, this should be 60 degrees and this one 20 degrees and the others are at all the multiples of 10 degrees in the in between and the resulting vector. See, these were all drawn out tip to tail. Now's the result in vector goes from one end of the line straight to the other. All right, so what we want to do is add up components. So in the extraction, the ex component of the result in vector and actually, if you want, are to be the man with you. The result. Invest very better. Call this someone something else. It's like call it f its just the attitude of one force. So the ex component of the resultant is the sum of the ex components of all of these other factors. And remember, for the diagram given X is going straight up. So all these angles here, the 20 degrees that are marked the 60 degrees it's marked and the 30 40 50 in between. The's are measured clockwise from the counter clockwise from the X axis on the counter. Clockwise from the X axis is what we need when we're using expressions to find the components where our exes are. Co signed data data is that angle measured counter car cars from the X axis or are signed data for the wife opponents? So, in this case, are ex Component is going to be of the first factor is 2.75 Newtons Times Co. Side of 20 degrees plus 2.75 Newton's Times Co. Sign of 30 degrees plus 2.7 time five Newton's Times Co. Signed for injuries and so on and so on. So actually, write this a little more simply by factoring out the 2.75 Newtons and doing co sign 20 degrees plus Co. Signed 30 degrees. And so on a live 60 degrees, I notice that you can factor out the 2.75 Newton's klutz. It's really just a number. Multiply it by the sum of co signs. You cannot factor out. The co sign itself, cause co sign is a function That's the kind of thing that people might be tempted to dio bits. But do not you have to compute each of these co signs individually and add them all up, plus co sign of 60 degrees. Great. So that's some of co signs will dip, lied by the magnitude of just one of the vectors is going to get us our ex component, and that is 10.2 Newtons. And so that's going to be put it in red on the diet around that's gonna beat this. This component this distance here for the Y component will do the same thing except replace the ex with Why replace all the co signs with sign, etcetera, etcetera, and she'll end up with a slightly different value and even the angles involved. On average, these individual force factors are a little more aligned with the X axis on the Y axis. So expect the Y component to be a little smaller. And in fact it is. We had met with 8.57 mutants. It's a seven here. Great. And so once you've got these components now need to convert them back into a magnitude and direction. So what for that, we use our formula for man getting magnitude from components. The magnitude is the square root of the sum of the squares of the components. So our ex squared plus are y squared. We substitute in 10 point to Newton's for RX 8.57 Newtons for our why, and you end up with 13.3 Nunes, which looks reasonable. If you look at the vector, the diagram above and remember that one of those black vectors was 2.75 Newton's and this blue back there should be 13.3, and then our angle here and in the diagram that's going to be this angle drawn and blue fi is the inverse tangent of our Y Over rx, and so we're expecting something between zero and 90 degrees here. Plug in are y over rx and you get exactly 40 degrees. And it's no coincidence, given how these five individual actors were evenly spaced, that this angle comes out, too. The average of the five different angles involved and that is, since we've already done Part B really got the graphical some, and you could draw it. You could draw it carefully with a ruler and protractor picking some scales in length scale to be equivalent to one Newton on your paper at home. You could do it really carefully to check that the magnitude and angle come out right away. We have drawn it. We just have, ah, sense that it looks looks reasonable. So if he is, we already done for B and that stand of the problem.

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