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

1. Jun 3, 2009

### aps0324

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

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)

2. Jun 3, 2009

### PhanthomJay

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.

3. Jun 4, 2009

### aps0324

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

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?

4. Jun 4, 2009

### LowlyPion

Re: Is this assumption correct? (force vectors and moment)

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

5. Jun 4, 2009