How Do They Calculate the 9 Degrees in the Force Diagram?

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Homework Help Overview

The discussion revolves around understanding the calculation of an angle in a force diagram, specifically how the 9 degrees is derived in part c of the problem. The subject area includes vector components and trigonometry in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using inverse tangent to find the angle related to the resultant of three forces, questioning the calculation of the angle and the magnitude of a force. There is also uncertainty about the origin of certain values in the calculations.

Discussion Status

The discussion is active with participants exploring different interpretations of the calculations. Some guidance has been offered regarding the use of inverse tangent, but there remains confusion about specific values and their derivations.

Contextual Notes

Participants mention a lack of access to calculators and express uncertainty about certain force values, indicating potential constraints in their problem-solving process.

circuscircus
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Homework Statement


http://img61.imageshack.us/img61/4585/8m1cdhx1.gif

I understood every except, How do they get the 9 degrees in part c?

Homework Equations





The Attempt at a Solution

 
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Look at the components of the resultant of the three forces: using an inverse tangent you should be able to find the angle it makes with the x axis. (Draw a right triangle.)
 
So then is arctan(30/190) 9 degrees?

(I don't have access to a radian calculator right now), also how do they get the 19N?
 
circuscircus said:
So then is arctan(30/190) 9 degrees?
Yes.
(I don't have access to a radian calculator right now), also how do they get the 19N?
You don't need a radian calculator. (If ever you don't have a calculator handy, Google has one.)

The 19 N makes no sense. But the components are close enough, so you should be able to figure the magnitude properly from that.
 
So the real F is sqrt(130^2+30^2) am I correct?
 
circuscircus said:
So the real F is sqrt(130^2+30^2) am I correct?
Where did the 130 come from?
 

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