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

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To find the line of action of the resultant force R = (3,6) and its intersection with the Y-axis at (0,d), it's crucial to calculate the total moment about the origin, which is determined to be -5 Nm. The assumption that R is applied at (0,d) can be valid, provided the correct signs are maintained for the coordinates. To find d, one must ensure that the moments calculated align with the resultant force's application point. Consistency in the direction of d on the Y-axis is essential for accurate results.
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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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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 (X = \Sigma{(F_y(x))}/R_y, etc.), the slope of which resultant is R_y/R_x.
 
Is this assumption correct? (force vectors and moment)

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

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?
 


aps0324 said:
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

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
 
I merged two duplicate threads.
 


aps0324 said:
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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