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Balanced and unbalanced forces

by pt20army
Tags: balanced, forces, unbalanced
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pt20army
#1
Nov26-12, 11:36 AM
P: 3
The following forces act on a hockey puck (round, rubber disk) sitting on a frictionless surface: F_1 = 15.5 N at 15 degrees; F_2 = 27.9 N at 125 degrees; and F_3 = 31.7 N at 235 degrees. All the forces are in the plane of the ice. Determine the net force on the puck.


Fnetx= F1x-F2x-F3x=max, Fnety= F1y+F2y-F3y=may



F1x=15.5Ncos15=14.97
F1y=15.5Nsin15=4.01
F2x=27.9Ncos125= -16
F2y=27.9Nsin125=22.85
F3x=31.7Ncos235=-18.18
F3y=31.7Nsin235= -25.96

Fnetx=14.97+16+18.18=49.15
Fnety=4.01+22.85+25.96=52.82

Im getting this problem wrong and I am not sure where I am messing up.
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Doc Al
#2
Nov26-12, 11:43 AM
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Don't ignore the signs of the components when you add them up.
jedishrfu
#3
Nov26-12, 11:55 AM
P: 2,781
Quote Quote by pt20army View Post

Fnetx= F1x-F2x-F3x=max, Fnety= F1y+F2y-F3y=may
You should simply sum the forces with the values you've computed for the x and y components:

Fnetx = F1x + F2x + F3x and similarly for Fnety


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