Equation for Fourth Force and its Line of Action?

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    Equivalent Force
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Homework Help Overview

The discussion revolves around a problem in statics involving forces and equilibrium. Participants are tasked with finding a fourth force that balances three given forces, along with determining the vector equation of its line of action.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of the fourth force and the subsequent vector equation for its line of action. There are attempts to take moments about a specific point to derive the equation, and some participants express uncertainty regarding the correctness of their results compared to a reference book.

Discussion Status

The discussion has revealed some productive exchanges, with participants clarifying their misunderstandings about the problem's requirements. There is acknowledgment of a typo in one participant's calculations, which has led to confusion regarding the results. Multiple interpretations of the problem setup are being explored.

Contextual Notes

Participants are working under the assumption that the fourth force is meant to create equilibrium in the system, rather than replacing the existing forces. There is also mention of discrepancies between personal calculations and the reference book's answers, which are being scrutinized.

gnits
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Homework Statement
Find the equivalent force
Relevant Equations
Balance of moments
Could anyone please help with the following?

Three forces of magnitudes 10, 3 * sqrt(5) and 25 Newtons act along lines whose vector equations are respectively:

r1 = i - 2j + k(4i + 3j) , r2 = -2i + 4j + k(2i - j) and r3 = 4i + k(7i - 24j)

A fourth force F4 is introduced which reduces the system to equilibrium.

Find F4 and the vector equation of its line of action:

I get the same answer as the book for the first part, F4 = -21i + 21j but not for the second part.

I take moments about -2i + 4j to find the y-intercept of the line of action and I am led to the equation r=-(8/21)i+k(-i+j) but the book answer is r=(22/21)i + k(-i+j).

Can anyone help confirm if it is the book or I (as I suspect) that is worng?

Thanks,
Mitch.
 
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gnits said:
I take moments about -2i + 4j to find the y-intercept of the line of action and I am led to the equation r=-(8/21)i+k(-i+j) but the book answer is r=(22/21)i + k(-i+j).
I get 92/21.
Please post your working.
 
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We know from the first part that:

F1 = 8i + 6j
F2 = 6i-3j
F3 = 7i-24j

Resolving horizontally and vertically gives us F4 = -21i+21j

Taking anti-clockwise moments about (-2,4) gives:

8*6 + 6*3 + 7*4 - 21 * 6 = 21 * (y' - 4) + 21 * 2

This leads to y' = -8/21 and so to an equation of y = -x - 8/21 and so when y = 0 then x = -8/21 hence my answer of r=-(8/21)i+k(-i+j)

Thanks,
Mitch.
 
gnits said:
8*6 + 6*3 + 7*4 - 21 * 6 = 21 * (y' - 4) + 21 * 2
THere's a typo, but probably just in making the post. I think you meant
8*6 + 6*3 + 7*4 - 24 * 6 = 21 * (y' - 4) + 21 * 2
But shouldn’t it be
8*6 + 6*3 + 7*4 - 24 * 6 + 21 * (y' - 4) + 21 * 2 = 0
?
That would match my answer.
 
Thanks so much (yes it was a typo in posting). I misread the question as saying that the forces were replaced by F4 which was to be equilvalent to the initial forces, but of course that's not what it is asking, it is saying that F4 is added into the system to create equilibrium.

Thanks again,
Mitch.
 
gnits said:
Thanks so much (yes it was a typo in posting). I misread the question as saying that the forces were replaced by F4 which was to be equilvalent to the initial forces, but of course that's not what it is asking, it is saying that F4 is added into the system to create equilibrium.

Thanks again,
Mitch.
And I misread the book answer as 22/21 when it is 92/21
 
gnits said:
And I misread the book answer as 22/21 when it is 92/21
That's a relief. Thanks for letting me know.
 

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