Net External Force - Magnitude & Direction

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Homework Help Overview

The discussion revolves around determining the net external force acting on a crate, which is subjected to multiple forces in different directions. The subject area includes concepts of vector addition and force analysis in physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the magnitude of the net external force and expresses uncertainty regarding the direction. Participants question the quadrant in which the resultant force vector lies and discuss the implications of using the inverse tangent function for angle calculation.

Discussion Status

Participants are actively engaging with the problem, exploring the correct quadrant for the force vector and discussing the mathematical approach to finding the angle. There is a recognition of the need to adjust calculations based on the signs of the components.

Contextual Notes

There is a mention of potential confusion regarding the use of the inverse tangent function and the interpretation of the angle based on the quadrant, which may affect the final answer. The original poster's calculations and assumptions are being scrutinized.

ughschool
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1. A crate is pulled to the right with a force of 82 N, to the left with a force of 115N, upward qith a force of 565N and downward with a force of 236N.



2. what is the magnitude and direction of the net external force



3. i solved for magnitude(330N) but I am not sure how to solve for direction cus i did tan^-1(329/33) and got 84° and i should be getting 96°...?
 
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"i solved for magnitude(330N) but I am not sure how to solve for direction cus i did tan^-1(329/33) and got 84° and i should be getting 96°...?"
----
On which side of the y-axis does the force vector reside?
In the first quadrant, or in the second?
 
its on the negative y aXIS i think
 
oh wait no sorry its in the second quadrant!
 
ughschool said:
oh wait no sorry its in the second quadrant!

That's right!
So:
Since your ordinary inverse tangent function gives values between -90 degrees and +90 degrees, these lie in the first and fourth quadrant.

Furthermore:
You chose to suppress the minus sign in the x-value for the tangent value, and therefore your calculator gave you the answer 84 degrees.
You would then have to add the deviation from 90 degrees in order to get the correct value 96 degrees.

If you had kept the minus sign, you would have gotten -84 degrees as your answer; adding 180 degrees to this (remember tanget is periodic with 180 degrees as its period!), would also have given the correct answer.
 
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