
#1
Jun309, 03:01 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 (This i know how to do) iii) The line of action of R cuts the Yaxis at (0,d). Find d iv) Find the equation of this line of action How do I solve iii) and iv) 



#2
Jun309, 07:45 PM

Sci Advisor
HW Helper
PF Gold
P: 5,959

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.




#3
Jun409, 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 Yaxis 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? 



#4
Jun409, 03:07 PM

HW Helper
P: 5,346

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


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