Find Force Vector for Given Position & Torque

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Homework Help Overview

The discussion revolves around finding the force vector F given a position vector r and a torque vector. The original poster presents the problem of determining F such that the torque equals the cross product of r and F.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the determinant form of the cross product to express the relationship between the vectors. There are attempts to derive equations from the determinant, leading to questions about the correctness of the expansion and the cancellation of unknowns.

Discussion Status

Some participants have provided guidance on correctly expanding the determinant and interpreting the resulting equations. The original poster expresses a moment of clarity regarding the problem, indicating progress in understanding.

Contextual Notes

There appears to be confusion regarding the correct application of the determinant and the resulting equations, which may stem from the complexity of the cross product operation.

DOMINGO79
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I am having a huge problem trying to answer the following question:
Given the position vector r = 3i + 4j + 5k and torque = 16i - 20j - 5k, find the force vector F which will give the correct result, so that torque = r x F
Can anyone please help me solve this!:confused:
 
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DOMINGO79 said:
I am having a huge problem trying to answer the following question:
Given the position vector r = 3i + 4j + 5k and torque = 16i - 20j - 5k, find the force vector F which will give the correct result, so that torque = r x F
Can anyone please help me solve this!:confused:
Write out the cross product as a determinant with the force vector's components a,b,c. Then you can solve:

\det{\begin{bmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k} \\ 3&4&5 \\ a&b&c\end{bmatrix}}=16\mathbf{i}-20\mathbf{j}-5\mathbf{k}
 
16i-20j-5k = i j k = i(4c-5b)-j(3c-5a)+k(3b-4a)
3 4 5
a b c

16i = i(4c-5b) -3(16 = 4c-5b) -48 = -12c + 15b
-20j = -j(3c-5a) → 4(20 = 3c-5a) → 80 = 12c – 20a
-5k = k(3b-4a) 32 = 15b -20a

32 = 15b-20a 32 = 15b-20a
5(-5 = 3b-4a) → -25 = -15b+20a, so what am I doing wrom?

why are the unknowns totally canceling out?
 
They don't cancel out. You're not expanding the determinant correctly.

For example, your first equation should read: 16 = 4c - 5b and nothing else.
 
Wow...

Hey, thanks a lot, I think I just opened my eyes:bugeye: and understand how to complete the question...

If I have anymore Q's, I will let you know, thanks...
 

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