Calculating Torque and Angle of Force on a Particle

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    Angle Calculation
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Homework Help Overview

The discussion revolves around calculating the torque and angle of a force acting on a particle, with specific focus on the definitions and applications of torque and vector relationships in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore different methods to calculate the angle between vectors using both the dot product and torque formulas. There is discussion about the correctness of the angle derived from these methods and potential quadrant issues affecting the results.

Discussion Status

Participants are actively sharing insights and alternative methods for calculating the angle, with some expressing uncertainty about the results. There is no explicit consensus, but suggestions for checking calculations and considering different approaches have been provided.

Contextual Notes

Participants note that the problem involves online homework with variable changes for each user, and there is a mention of potential discrepancies in the expected answer format (degrees vs. radians).

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Homework Statement



Force F= (-6.93 N) + (7.85 N) acts on a particle with position vector r = (4.66 m) + (1.36 m). What are (a) the magnitude of the torque on the particle about the origin and (b) the angle between the directions of and ?

Homework Equations





The Attempt at a Solution



So I find (a) to be 46.0058 N*m which is correct and for (b) I use t/f*r

So 46.0058/sqrt(6.93^2+7.85^2)*sqrt(4.66^2+1.36^2) = .9050610046

sin-1(of that) = 64.8315006 = angle. But this is incorrect?
 
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Remember this definition of the dot product?

[tex]\vec{A}\cdot\vec{B}=|A||B|\cos\theta[/tex]

Try using this to solve for the angle and see if it gives the same value. (I think the cross product formula should work, but for some reason I'm getting a different value for the angle in each one...)
 
Does that work with a vector? I've been using t = rFsin(Theta)
 
Only thing I could think of is maybe its in the wrong quadrant but that doesn't seem to be the problem.
 
AnkhUNC said:
Does that work with a vector? I've been using t = rFsin(Theta)

Yes. It's the formula for the dot product between two vectors. Theta is the angle between the vectors.


Also, you have to know that:

[tex]\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y[/tex]

in order to use the dot product formula to find the angle between the vectors.

Let A=F and B=r and see what angle the dot product formula gives you.

(Sorry if you haven't learned what I'm talking about here. I also got the same angle you did with the formula your using. So I'm trying to check that answer with another equation.)
 
Really appreciate the help, not sure what its looking for :(
 
AnkhUNC said:
Really appreciate the help, not sure what its looking for :(

Well. I think the method your using should be correct. Here is what I suggest.

If your answer for the torque is correct, and the angle you get from the torque formula is wrong, I suggest the following alternative method:

Find the angle of the F vector from the x axis. (You should be able to do this using the components.)

Do the same for the r vector.

Now you should be able to subtract the two angles to find the angle between them.

If you get the same answer as you got before. Chances are that your probably right, especially if your answer to part a) is correct. I suggest asking your instructor about this problem. It is possible that the answer in the back of the book is wrong.

Also, this may seem simple after all of this, but we should make sure just in case: Are you sure the answer in the back of the book is in degrees and not radians?
 
Its not in degrees, its actually online homework that changes the variables for each person. I know (a) is right because it told me so and it also said the units for (b) were degrees :(

I tried it the way you suggested and again still wrong. I'll ask the instructor tomorrow but its really not a big deal I guess.
 

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