# Is there a way to solve this complex statics problem algebraically?

This problem can be solve in 2 ways either I solve it with vectors which would be relatively painful and more time consuming and the other faster way is algebraically but I faced a problem when trying to find the points of intersection with the axes:

After I calculated $$R_x=403.3$$ N , $$R_y=-131.81 N$$ and Moment about G(origin): $$M_G=-460cos(15)*0.47+100*0.59+120cos(70)*0.47-120sin(70)*0.19+100+135=88.89 N$$

Then I said that the sum of moments of forces about G = The moment of the resultant force about G

$$R_x*y+R_y*x=88.98$$ $$\therefore -131.81x+403.3y=88.98$$

Now when I plug $$y=0 :x=0.675m=675mm$$

And when $$x=0$$ : $$y=-0.2207m=-220.7mm$$

Apparently there is no answer with the signs that I got ,What did I do wrong here.

The idea was correct, and all your calculations were correct. You only neglected to consider the sign of the generating moment, i.e. in the 2D case the moments of a force are given by the following equation (notice the minus) :

$$M= -F_x\cdot y + F_y\cdot x$$

So for horizontal forces

• when a positive horizontal force is applied on a positive y distance then the resulting moment is negative
Positive y Negative y
Positive $$F_x$$
- M

+M
Negative $$F_x$$
+M

-M

Similarly, vertical forces:

Positive x Negative x
Positive $$F_y$$
+ M

-M
Negative $$F_y$$
-M

+ M

So when you were calculating the x coordinate, only the y component of the force generates moment (i.e. \$R_y), so what you should have calculated was:

$$M_G = +R_y\cdot x \Rightarrow$$ $$x = +\frac{M_G}{ R_y}=+\frac{88.98}{ −131.81}= -220.65 mm$$

and similarly for the y calculation

$$M_G = -R_x\cdot y \Rightarrow$$ $$y = +\frac{M_G}{ R_x}= - \frac{88.98}{ 403}= -675 mm$$

Let's sum forces about point "G":

$$\sum F_x = 460cos15 - 120cos70 = 444.3 - 41 = 403.3$$, pointing lt to rt.

$$\sum F_y = 460sin15 + 120sin70 - 100 = 119.1 + 112.8 - 100 = 131.9$$, pointing down.

Let clockwise rotation be positive:

$$\sum M_G = 444.3(0.47) - 119.1(0.05) - 41(0.47) + 112.8(0.19) - 100(0.59) = 146$$ - rotate clockwise.

Because there is a linear equation with two variables, with neither being tied to a reference datum, thus, all points along the line bounded by $$(0,0.36)$$ & $$(1.11, 0)$$ can satisfy the equilibrium requirement.

My note from yesterday stays -

Note: My solution does not invalidate your calculations. As I don't think there is a unique answer to the original question that asks a "single equivalent force...".