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15 The] helicopter Vicw in Fig: PS.15shows two people pull- ing on stubborn mulc:Thc person on the right] pulls with 1 [orceF of magnitude 120 N and direction of 0,...

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15 The] helicopter Vicw in Fig: PS.15shows two people pull- ing on stubborn mulc:Thc person on the right] pulls with 1 [orceF of magnitude 120 N and direction of 0, 60.0" The person on thc left pulls wilh force F of magni- tude 80.0 Nland direction of 0 75.0" Find (a) thc single force that is cquivalent to the two forces shown and (bJltthe [orce tlat 4 third person would havd TQ exerg 0M Ithe Iile LU make the Tesukant force equa]| tolzero Tha | Forces are mmeaisuuredlin unicsof newcons

15 The] helicopter Vicw in Fig: PS.15shows two people pull- ing on stubborn mulc:Thc person on the right] pulls with 1 [orceF of magnitude 120 N and direction of 0, 60.0" The person on thc left pulls wilh force F of magni- tude 80.0 Nland direction of 0 75.0" Find (a) thc single force that is cquivalent to the two forces shown and (bJltthe [orce tlat 4 third person would havd TQ exerg 0M Ithe Iile LU make the Tesukant force equa]| tolzero Tha | Forces are mmeaisuuredlin unicsof newcons Ksynp bolized NA Fe 02 Figure P3.15



Answers

The helicopter view in Fig. P 3.29 shows two people pulling on a stubborn mule. The person on the right pulls with a force $\overrightarrow{\mathbf{F}}_{1}$ of magnitude 120 $\mathrm{N}$ and direction of $\theta_{1}=60.0^{\circ} .$ The person on the left pulls with a force $\overrightarrow{\mathbf{F}}_{2}$ of magnitude 80.0 $\mathrm{N}$ and direction of $\theta_{2}=75.0^{\circ} .$ Find (a) the single force that is equivalent to the two forces shown and (b) the force that a third person would have to exert on the mule to make the resultant force equal to zero. The forces are measured in units of newtons (symbolized N).

So in this example, we're told that two men are pulling a bull and this is just a rough sketch. I do recommend looking at the picture and the book, but they're pulling on this ball. One man is pulling out a force of 80 newtons, uh 70 degrees north of west of the bull. Right? So we denote that as displacement factor A. Then the second man is pulling with a force of 120 newtons 60 degrees north of east of the bull. Right? So that'll be displacement factor B in part A. I just want to know what the single force equivalent of these two forces would be. Right. So we are trying to find the result in here and in order to do that, we need to find the X and Y components of both displacement factors. So the X and Y component of vector A is going to be the magnitude, which is 80 newtons times co sign of its angle, which is 70 degrees. And then for the Y component it will be 80 times sine of its angle, which is 70 degrees. We'll do the same thing for the components of vector B, which will be 120 times co sign of 60 degrees and 120 of our time sign of 60 degrees. When we plug all these into our calculator, we get the X. Component of A to B 27 0.36 newton's the Y component to be 75.18 newtons. And then for B it is 60 newtons for the X component at 103.92 newtons for the Y component. Yeah. Now, if we look at the vectors, we see that for vector B. It's going in both the positive X and Y direction. But for vector A. It's going in a positive direction, but the negative X direction. So this component here is actually negative 27.36 newtons. All right now, in order to find our resultant, we want to are the X. Component. For our resultant. We're going to take the sum of all of our X components of our displacement factors. So when we do that we get the X component of a result to be 32 point 64 newton's and then the Y component will take the sum of all the white components, far displacement vectors. And we'll get our sub Y to be 100 and 79.1 newton's in order to find the magnitude which is what they're looking for. We take the square root of our X component plus R Y component. Are we take the square root of our sub X squared plus y squared. We plug that nor calculator. We get a magnitude of 100 and 82.5 newtons. And then for part B they ask um what the force that a third person would have to exert on the mule in order to make the resultant equal zero. So what we want to do is we want to find the angle of our resultant first and that is just going to be the our angle will be data which is equal to the inverse tangent of our supply over our subjects. When we do that, we get our displacement angle of 90 r 79.67 degrees. All right. So in order for this third person to make our result in equal zero, they will have to apply the same force as our resultant. Right? So we'll say have to apply the same force but in the equal and an equal and opposite direction. Right? So it would be going downwards. Or you have to be pulling down at an angle of 79.67 degrees. Yeah, with a force of 100 and 82.5 newtons. And if he was able to do that, then he would make the resultant force come out to a total of zero.

So in this problem we're going to use vector edition but apply it to forces not just displacements this time. And so we have were given to separate forces with two different magnitudes f. Supp one and F two. And we want to find the force that we could that is that the object is essentially feeling from the two separate forces. And this one force we typically call the net force and it is the sum of the individual forces. Various individual forces can be acting on a single object, but to the object is that it's as if there is one force which is the sum of all of the individual forces. And this is the one that we need to find. This is the one force that the object feels. Uh And so to do this we need to add the components of each of defectors. So we have the x component here, f sub one X and the Y component here. F sub one. Y. Mhm. And F someone will then be the addition of the component vectors and the magnitude of F sub one X will have to be F sub one times the co sign 60 degrees. Using this right triangle here. And that will have to point in the positive exit direction since the X component points in the positive X direction Plus F sub one times to sign of 60 degrees. And that points in the positive light had direction just looking at the direction and the the The fact that the sign of 60 is related to the y component. Yeah, F Sub two Very similar. We have this angle here is 75°. We have this x component and this white component except ey F sub two. X F Sub two will then be the addition of the component vectors which is F sub two Times The co sign of 75 degrees X. hat. And that points the negative X that direction. So we put a negative sign plus F sub two times design of 75° in the white hat direction. And now we just have to add these two vectors using their components. So we have that this is F1 times to co sign of 60° X hat plus F one Times The Sine of 60°. Why had -F sub two times the co sign of 75° except Plus F sub two times to sign of 75°. Why have And now we add the uh magnitudes of each of the vectors that at that point in the same direction. So we have to in the X direction into in the white hat direction, so we add their magnitudes separately. Uh such that we have mhm a total vector pointing in the X hat and a white hat direction. The two magnitudes to add up in the X direction. So, F sub one which has a magnitude of 120 Times The coastline of 60 -F sub two, which has a magnitude of 80 Times The co sign of 75 will give us a net force of 39.3 newtons in the X hat direction, plus add the components in the Y direction, which are underlined in red. So f sub one has a magnitude of 120 times the sine of 60 And then plus F said Tuesday as a magazine of 80 Times The Sine of 75. That gives a net force of 181.2 newtons in the white hat direction. And so uh we need to we really want to figure out what we know the components. Now, we can work backwards from what we did over here to figure out the magnitude and direction of this net force. So the magnitude of any vector will be the the square root of the, of the sum of the squares of the components. So the square root of 39.3 squared Plus 181.2 Squared. So if we plug that into calculator, the square root of 39.3 squared plus 1 81.2 squared, Well, give us about 185.4 newton's. And what direction is that in? Well, we have so say we this force points will point in the upper right because it has a positive X. Component and a positive light component. So this will be ethnic and say we can and this is a B F Y. This will be fx say we want to label or want to figure out this angle theta. And so we can just take the tangent of data. The tangent of theta will be the opposite magnitude, which is F. Sub Y. Over the adjacent, which is F sub X. And we can take the arc tangent on both sides. To find that data must be the are tangent of S. M. Y over F sub X. And we can figure this out. So we know that the magnitude of the net force in the X direction is 181.2 and the magnitude in the Y direction was 39.3. So given these values And plug in the respective values into this equation, you find that this angle has to be about 77 0.8 degrees. And since this points to the upper right and we found this angle, there will be 77.8° north of east, where the second component in this sort of phrases the direction that you would be going. And the first is the direction that you have to really account for by adding this angle. So north of east, what? Uh And so this is the magnitude and the direction of this net force. So this will be the really single force, a single net force that the object would experience based on the two individual forces. Because forces add and forces are represented by vectors. And the final part we need to figure out is the force that another person would have to exert Uh to make the net force equal to zero. Well, let's try to sketch the net force on our Diagram here. So we know the magnitude, it's about 185 newtons so represented like that. And this angle is about 77.8°. And this will be F net. And again, this is the addition of the two individual vectors. Well, if we have F net and we want to find this third force, you want to add it such that we get zero on the other side, while uh we can rearrange this to find that ah the third force would have to be If my program would catch up, the third force would have to be the negative of the net force, which would be a vector that looks something like this, since the negative of vector is just the same vector pointing in the opposite direction. And so it would have the same magnitude and this angle would be the same as the 77.8. So this would be F sub three and this third force Would have the same mag suit of 185.4 Newton's. Yeah. And the direction would be 77.8° this time for considering this angle, we would be going west, we have to move south to get to the sector. So this would be uh south of west. So we were able to find the net force based on two individual forces, the magnitude and direction, as well as the magnitude of the direction of a third force, such that it makes the net force The total force on the object equal to zero.

We have two people pulling on a mule. We have ups of one equaling 120 Newtons. We have except to Equalling 80 Newtons. If this angle here of 75 degrees this angle here data and this angle here of 60 degrees now to find data, this would of course be 180 degrees minus the sum of 75 plus 60 degrees. And this is giving us 45 degrees. We can then say that in order to find their magnitude of the resultant force factor essentially this The result is force from these two individual forces. We can say that this would be our equaling the square root of EFS of one squared. So 120 Newtons squared plus F sub two squared plus two multiplied by 80 Newtons, 120 Newtons and finally multiplied by co sign of data or 45 degrees. So this is equaling 185.4 Newtons. We can then find the direction of the resultant force. We could say that Tangent of Alfa equals F sub to sign of Leda, divided by its of one plus s up to co sign of Fada. And this is of course, just do the trigonometry. And this would be Equalling 80 Newtons, multiplied by sine of 45 degrees, divided by 120 Newtons, plus 80 Newtons multiplied by co sign of 45 degrees. And so also would be equaling art 10 of 0.32 0377 and this is giving us 17.76 degrees. So the direction of the single force from the positive X axis we could safe I would be Equalling Alfa plus 60 degrees equally, 77.76 degrees. This would be counter clockwise from be positive X access. So this would be our answer for heart? A. My apologies for Part B. It's similar in the sense that we simply have to find a third force to count to cancel out the some of the 1st 2 forces. The magnitude of the Third Persons Force will be equal to the magnitude of the force that we found in part a. So 185 point for Newton's. And this is because, of course, we need to cancel out completely. Now the direction is what changes in the sense that we have to apply it in the opposite direction. So we can say that then, fi off the third person would be equaling 180 degrees, minus 77.76 degrees. This is giving us 100 to 0.24 degrees. This would be clockwise from the positive X access. So this would be our answers for part B. That is the end of the solution. Thank you for watching.

We would have f It was if one goes I Instead, I want us if to go scientific to I thus if one signed that I want us if to sign that I do j And this would be 39.3. I had thus 1 81 Jay had. So the manager off f will be square at this two numbers and take a spread out. It will be 185 new done and the angle would be tanning. Worse. Off this over that equals 77.8 duty and F three. It was minus F equals minus 39.3. I had minus 1 81 jihad.


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