- #1
freespirit
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My problem is to determine the magnitude and direction of the resultant force FR=F1+F2 and it's direction, measured counterclockwise from the positive x direction.
f1=250 lb @ 60 degrees from x
f2= 375 lb @ -45 degrees from x
Ok I got the magnitude by doing this:
(360-2(255))/2=-75 degrees
fr=sqroot of (250^2+375^2-2(250)(375)cos(75)
fr=393.188~ 393
then I got the angle by this:
375/sin x = 393.188/sin 75
x=67.1088
how do i get the resultant angle, what do I need to add to the 67 degrees?
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f1=250 lb @ 60 degrees from x
f2= 375 lb @ -45 degrees from x
Ok I got the magnitude by doing this:
(360-2(255))/2=-75 degrees
fr=sqroot of (250^2+375^2-2(250)(375)cos(75)
fr=393.188~ 393
then I got the angle by this:
375/sin x = 393.188/sin 75
x=67.1088
how do i get the resultant angle, what do I need to add to the 67 degrees?
============ Forum Options ============= Teachers / Educators Forum -- Teacher & Educator Discussion General Physics & Astronomy Forum -- General Physics Discussion -- General Astronomy Discussion High-School Students Forum -- General High-School Discussion -- Homework Discussion Undergrad / College Students Forum -- General Undergrad / College Discussion -- Homework Discussion Grad Students / Postdocs Forum -- General Grad School & Postdoc Discussion Professional / Industrial Physics Forum -- Industrial Physics & Business Discussion -- University Faculty Discussion Special Topics Forum -- Physics, Philosophy & Religion -- Physics, Politics & Ethics ============= Other Options =============== ======================================== --- Log In --- Register --- Search
Forum Home | Back to PhysLink.com | Contact the Webmaster | Terms & Conditions of Use
Copyright © 2003 - PhysLink.com
Powered by < CF FORUM > v.2.0
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Advertisement:
All rights reserved. © Copyright '1995-'2004 PhysLink.com