Need Help on Physics Problem!!

Test Preparation AP Test Preparation
1

<p>Started a review that my honors physics teacher gave me for the summer which he said will help for AP Physics. It's basically all the tests we took during the year but with different numbers. When I first did this problem in the begininning of the year I got it wrong on the test. But after redoing it for the review I am not sure if I am right or not, can someone compare my answer to theirs for this problem?</p>

<p>Two 70 lbs forces are holding up a 50 lbs object. What angle do they make to each other?
(I know it's simple but I won't be able to talk to my physics teacher for a long time and this problem is irking me)</p>

<p>My answer was that the angle they make to each other(the angle between the two "ropes holding the 50 lbs object up each with 70 lbs of force) is 159 degrees.</p>

2

<p>Out of curiosity, are you taking Physics B or Physics C next year?</p>

<p>Any ways, here is how I approached your problem.</p>

<p>First drew a diagram of the situation. (Diagram=god in physics as you probably know from honors physics)</p>

<p>So the object they hold up is 50lbs*10m/s^2=500N (If you haven’t started using g=10, trust me you will next year. Its very convenient lol).</p>

<p>The forces are 70lbs<em>10m/s^2=700N a piece. The upward force they apply on the object is given by F=T</em>cos(x). The horizontal component of the force is useless as the two ropes cancel each other out. Hence I’m using cos here.</p>

<p>We know F must equal the weight of the object (500N)</p>

<p>So 500N=T*cos(x)</p>

<p>We know each rope exerts 700N, and 2*700N=1400N so:</p>

<p>500N=1400N*cos(x)
500N/1400N=cos(x)
cos^-1(500/1400)=x
69.08 degrees = x</p>

<p>But x here is the angle made by one rope and the perpendicular. So to find the angle they make to each other or the angle between the two ropes, double it and you get 138.16 degrees. Let me know if I am way off…</p>

3

<p>Physics C, haha don’t worry I know this problem is like basic stuff but I assure you, I do extremely well in electricity/magnetistm and mechanics</p>

4

<p>We should somehow team up then lol. I’m selfing Physics C next year and while I’m good at mechanics and problems like this, you would not believe how bad I am in electricity/magnetism.</p>

5

<p>and wait thats kinda wrong, you wouldnt multiply 50 lbs times gravity, pounds itself is the english system measurement for force, and you certaintly can’t multiply the SI units m/s^2 for gravity times the english unit lbs</p>

6

<p>I get 138 degrees. Resolve it in components:</p>

<p>sumFy = 0</p>

<ul>
<li>(70lb)(cosa) + (70lb)(cosa) - 50 lb = 0</li>
</ul>

<p>where a is the angle that the 70 lb vector makes with the vertical. Solve for a and multiply by 2 to get the angle between the forces. I’ll upload a diagram if you don’t follow.</p>

<p>

</p>

<p>You’re dividing by 2.2 then multiplying by 10 to everything, but the end result will be the same even in Newtons.</p>

7

<p>

Whoops. But it won’t matter. You don’t even need to multiply by gravity and at the end it cancels out.</p>

<p>

</li>
</ul>

<p>Wait An0maly, I solved your second equation and get a=69.07 degrees??? So the final answer would end up being 138 like I got. I don’t see where 88 comes from?</p>

8

<p>well again anomaly if you are trying to find the Forces along the y axis you’d use sine not cosine, but basically I did same thing, 70sinx + 70sinx - 50 = 0 then i solved for x</p>

9

<p>

</p>

<p>It was a miscalculation. I edited it to avoid confusion :)</p>

<p>

</p>

<p>I treated my alpha as the angle from the vertical, which is why I used cosine.</p>

10

<p>and then the angle between the forces is 180-2x not just 2x</p>

11

<p>Can’t use sin here. You have to use cos. Sin gives you the horizontal component of the force (the two ropes cancel out the horizontal force). You want the vertical force. The reference angle x we are using is between each rope and the vertical.</p>

12

<p>

</p>

<p>He is using the angle from the horizontal as the reference angle, so sine is appropriate. He just does 180 - 2x instead of straight 2x, which is why I prefer the angle from the vertical.</p>

13

<p>

</p>

<p>k. That explains it lol…</p>

<p>

</p>

<p>Should be just 2x. Draw a diagram for yourself…</p>

14

<p>

</p>

<p>Then I misunderstood sorry. I thought he was saying the vertical as a reference angle and I was like hmm…</p>

15

<p>sine is the verical component of an angle and cosine is the horizontal component dude</p>

16

<p>

</p>

<p>But its all relative…h/o let me try to make a diagram…</p>

17

<p>In senario #1, you would use cos. But in senario#2, you would use sin like you argue. Me and An0aly are drawing #1, while you are drawing #2.</p>

<p><a href=“http://i48.■■■■■■■.com/dzimfs.jpg[/url]”>http://i48.■■■■■■■.com/dzimfs.jpg&lt;/a&gt;&lt;/p&gt;

18

<p>Your method:</p>

<p><a href=“http://img248.imageshack.us/img248/8138/physicsproblem002.jpg[/url]”>http://img248.imageshack.us/img248/8138/physicsproblem002.jpg&lt;/a&gt;&lt;/p&gt;

<p>Our method:</p>

<p><a href=“http://img682.imageshack.us/img682/907/physicsproblem001.jpg[/url]”>http://img682.imageshack.us/img682/907/physicsproblem001.jpg&lt;/a&gt;&lt;/p&gt;

<p>That should clear it up :)</p>

19

<p>An0maly’s diagrams are much clearer than my scribbles in Paint… But as you see its relative based on your reference angle.</p>