Net External Force in a Direction

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The discussion focuses on calculating the net external force acting on a crate in the x-direction, given opposing forces. The user initially attempted various calculations but was confused about the correct method. It was clarified that the same subtraction method used for the y-direction should apply to the x-direction as well. The correct approach involves subtracting the force pulling to the right (92.3 N) from the force pulling to the left (116.8 N). The net external force in the x-direction can be determined by this straightforward subtraction.
Smiley.Teddy
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Homework Statement


A crate is pulled to the right with a force of 92.3 N, to the left with a force of 116.8 N, upward with a force of 508.8 N, and downward with a force of 248. 6 N
What is the net external force in the x direction? Answer in units of N




Homework Equations


When I solved for the Y-Direction I simply subtracted the upward force from the downward force and got the answer correct. I don't know what to do for the x-direction.



The Attempt at a Solution


X direction: I attempted these:
-92.3N-116.8N = -209.1
116.8N-92.3N = 24.5
116.8N^2+92.3N^2 and then square root that answer to get = 148. 867...

Please help me understand what it is that I'm doing wrong. Thank you. =]
 
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Welcome to PF!

Hi Smiley.Teddy ! Welcome to PF! :smile:
Smiley.Teddy said:
A crate is pulled to the right with a force of 92.3 N, to the left with a force of 116.8 N, upward with a force of 508.8 N, and downward with a force of 248. 6 N
What is the net external force in the x direction? Answer in units of N

When I solved for the Y-Direction I simply subtracted the upward force from the downward force and got the answer correct. I don't know what to do for the x-direction.

(why did you think they might be different? :confused:)

For the x-direction, just do exactly what you did for the y-direction. :smile:
 
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