Resultant Magnitude of Two Forces


by ut4ever8
Tags: forces, magnitude, resultant
ut4ever8
ut4ever8 is offline
#1
Sep23-08, 07:42 AM
P: 7
1. The problem statement, all variables and given/known data
Two forces, 464 N at 10 degrees and 397 N at 35 degrees are applied to a car in an effort to accelerate it. What is the magnitude of the resultant force. Answer in units of N.
The picture is attached.



2. Relevant equations
Pythags Theorem


3. The attempt at a solution
Well I thought this should be an easy problem but Ive run into a wall here. I started out by taking the first force of 464 N and Finding its X and Y components. X= 464*cos10 and Y= 464*sin10 which gave me X1=465 and Y1=80.57. Then I did the same for F2 by using X=397*cos35 and Y=397*sin35 which gave me X2=325.20 and Y2=-227.71. Then I added up the sum of the forces in the X direction and got 776.15 and the Y direction= -147.14. Now that I have the sum of the forces in both directions I used Pythagoras theorem:
776.152+-147.142=F2 and I came out with a final answer of 762.08 which is incorrect. Can someone tell me where I went wrong. Am I way off? I thought I was on the right track here. My only other thinking was to place one of the forces on the positive x axis and go from there, but if I bring the 35 degree angle up to that axis, does that change the 10 degree angle to 45 or do I still figure that force out with an angle of 10? Thanks for any help, I really appreciate it.
Attached Thumbnails
Phy 211 pic.JPG  
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Kurdt
Kurdt is offline
#2
Sep24-08, 07:52 PM
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I think the mistake comes in working out the x-component for the 464N force. The component is greater than the actual force.
ut4ever8
ut4ever8 is offline
#3
Sep24-08, 10:02 PM
P: 7
That was actually a typo, the X1 is 350 something. I got it right, I just goofed in the Pythagorean Theorem and made a magnitude negative. Thanks for taking a look at that though. I certainly appreciate that.


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