Fido pulls with a force of 525 Newtons at 160 degrees while Spot pulls with a force of 175 Newtons at 65 degrees and Rover pulls with a force of 825 Newtons at 315 degrees. What is the resultant force and degree?
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what did you tried? – cand Oct 08 '19 at 00:44
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+1 for vector war. – darij grinberg Oct 08 '19 at 00:46
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My teacher said that you can't multiply cos(160)*525. You have to put it in a coordinate graph, up and down is y and left and right is x. 525 Newtons is the hypotenuse. But I don't know what degree to use instead of 160. – Dylan Oct 08 '19 at 00:49
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You certainly can multiply $525 \cos(160^\circ)$ and there is even a way to use that multiplication as part of a correct method for this problem. But since you have not shown what you attempted to do with that multiplication, we don't know whether it was correct. There could be a reason your teacher said you did it wrong, but we cannot guess unless you show a lot more detail of what you did. Since you will have to write a lot of formulas to show your work, go here to see how to make them readable: https://math.stackexchange.com/help/notation – David K Oct 08 '19 at 01:09
1 Answers
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The forces along the $x$- and $y$-direction are, respectively,
$$F_x=175\cos 65+525\cos 160 + 825\cos 315=163.98$$
$$F_y=175\sin 65+525\sin 160 + 825\sin 315=-245.20$$
The resulting force is
$$F= \sqrt{F_x^2+F_y^2} = 294.98 \>\text{N}$$
The resulting angle is in the 4th quadrant,
$$\theta = 360 +\arctan\frac {F_y}{F_x}=303.8 \>\text{degrees}$$
Quanto
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I divided it into 4 quadrants, with each being 90 degrees. Thus, in total, 90*4=360 degrees. The 160 degrees one was in the second quadrant, and I subtracted 90 degrees from it to get 70 degrees. My FINAL answer is 8 degrees, in the first quadrant. – Dylan Oct 08 '19 at 01:18
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why do you need to substract 90 from 160? You could just 160 degrees directly. – Quanto Oct 08 '19 at 01:27