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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. enter image description here

Youssef Mohamed
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2 Answers2

1

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 $F_x$ enter image description here
- M		 enter image description here
+M
Negative $F_x$ enter image description here
+M		 enter image description here
-M

Similarly, vertical forces:

Positive x Negative x
Positive $F_y$ enter image description here
+ M		 enter image description here
-M
Negative $F_y$ enter image description here
-M		 enter image description here
+ 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

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

NMech
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-1

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.

enter image description here

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...".

Please check my calculation and advise for mistakes.

r13
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