How do I find the line of action of this resultant force?


by aps0324
Tags: action, force, line, resultant
aps0324
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#1
Jun3-09, 03:01 PM
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Three forces are applied to a body. They are :

F1 = (4,5) applied at (1,2)

F2 = (2,-1) applied at (3,-1)

F3 = (-3, 2) applied at (-2,1)


i) Find the resultant force

Answer : R = (3,6)

ii) Find the total moment about the origin

(This i know how to do)

iii) The line of action of R cuts the Y-axis at (0,d). Find d

iv) Find the equation of this line of action



How do I solve iii) and iv)
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PhanthomJay
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#2
Jun3-09, 07:45 PM
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Did you try calculating the moments about the origin from the the x and y components of the forces separately? Then you can find a set of coordinates through which the resultant passes ([tex] X = \Sigma{(F_y(x))}/R_y[/tex], etc.), the slope of which resultant is R_y/R_x.
aps0324
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#3
Jun4-09, 02:36 PM
P: 9
Three forces are applied to a body. They are :

F1 = (4,5) applied at (1,2)

F2 = (2,-1) applied at (3,-1)

F3 = (-3, 2) applied at (-2,1)


i) Find the resultant force

Answer : R = (3,6)

ii) Find the total moment about the origin

Answer: I calculated this and it is -5 Nm

iii) The line of action of R cuts the Y-axis at (0,d). Find d

[B]Question: Can I assume that R = (3,6) is applied at (0,d) and its moment about the origin is - 5Nm? In order to find what d is?


If that assumption is wrong then how do i find d?

LowlyPion
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#4
Jun4-09, 03:07 PM
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How do I find the line of action of this resultant force?


Quote Quote by aps0324 View Post
Three forces are applied to a body. They are :

F1 = (4,5) applied at (1,2)

F2 = (2,-1) applied at (3,-1)

F3 = (-3, 2) applied at (-2,1)


i) Find the resultant force

Answer : R = (3,6)

ii) Find the total moment about the origin

Answer: I calculated this and it is -5 Nm

iii) The line of action of R cuts the Y-axis at (0,d). Find d

[B]Question: Can I assume that R = (3,6) is applied at (0,d) and its moment about the origin is - 5Nm? In order to find what d is?

If that assumption is wrong then how do i find d?
Can you show your work on determining the moment in part ii)?

Isn't it the cross product of the F X r ?

Doesn't that yield

(4*2 + 5*1) + (2*(-1) + (-1)*3) + ((-3)*1 + 2*(-2)) = ... but ≠ -5
berkeman
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#5
Jun4-09, 05:16 PM
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I merged two duplicate threads.
PhanthomJay
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#6
Jun5-09, 03:54 PM
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Quote Quote by aps0324 View Post
Can I assume that R = (3,6) is applied at (0,d) and its moment about the origin is - 5Nm? In order to find what d is?

Yeah, as long as you are consistent with your plus and minus signs, that is to say, is d on the positive or negative y axis?


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