enter image description here

I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.

So if I call the top one F1 and the bottom loading F2 I got:

$$F_1+F_2=0= -4b\dfrac{1}{2}+2.5(b+a)\dfrac{1}{2}$$

solving this i got $a=0.6b$

Moment around the free end of the bar (not $A$, but opposite side) due to the loadings is (counter clockwise positive):

$$M= -F_1\cdot\text{center of triangle} + F_2\cdot\text{center of triangle}$$

Because you can replace the loading with force $F_1$ and force $F_2$ and its line of action is through the center of the triangle area ($\dfrac{1}{3}\text{base}$):

$$M= -8= -4b\dfrac{1}{2}\cdot\dfrac{1}{3}b+2.5(b+a)\dfrac{1}{2}\cdot\dfrac{1}{3}(b+a)$$

So what am I doing wrong? Because M can never be negative with me, plus the answer should be $b= 5.625$ and $a= 1.539$. But this to me makes no sense, because then $F_1+F_2\neq0$. And if I should take the reactive forces into account at $A$ then you can never have still a moment, because then it is not static anymore.

  • $\begingroup$ I am really sorry, but the other question that I uploaded about this, had the wrong image. Sorry for the inconvenience $\endgroup$
    – strateeg32
    Feb 25, 2016 at 15:18
  • $\begingroup$ Did the answer to your previous question not help you with this one as well? $\endgroup$
    – hazzey
    Feb 25, 2016 at 15:24

1 Answer 1


The answer can be found with the same process used in the previous question.

Defining $F_1$ as the downwards load, we have

$$F_1 = -\dfrac{4b}{2} = -2b$$

Defining $F_2$ as the upwards load, we have

$$F_2 = \dfrac{2.5(a+b)}{2} = 1.25(a+b)$$

As you stated, $F_1+F_2 = 0 \therefore b = \dfrac{5}{3}a$.

Now, the moment due to a force couple is $M = F \times D$, where $D$ is the distance between the forces in the couple. Now, the centers of action of $F_1$ and $F_2$ are (from the free end):

$$\begin{align} D_{F_1} &= \dfrac{b}{3} \\ D_{F_2} &= \dfrac{a+b}{3} \end{align}$$ Therefore $D = \dfrac{a+b}{3} - \dfrac{b}{3} = \dfrac{a}{3}$. Thus, $M = 2b \times \dfrac{a}{3} = \dfrac{10}{3}a \times \dfrac{a}{3} = 8 \therefore a = \sqrt{7.2} \therefore b = \dfrac{5}{3}\sqrt{7.2}$.

Checking our work: $$\begin{align} F_1 &= -\dfrac{10}{3}\sqrt{7.2} \\ F_2 &= 1.25(\sqrt{7.2}+\dfrac{5}{3}\sqrt{7.2}) = \dfrac{10}{3}\sqrt{7.2} \\ &\therefore F_1 + F_2 = 0 \text{ OK!}\\ M &= \dfrac{10}{3}\sqrt{7.2}\cdot\dfrac{\sqrt{7.2}}{3} = \dfrac{72}{9} = 8\text{ OK!} \end{align}$$

  • $\begingroup$ But so then the book is wrong in saying that a= 1.539 an b=5.625? $\endgroup$
    – strateeg32
    Feb 25, 2016 at 15:44
  • $\begingroup$ And with your couple you get a counterclockwise moment of 8 kN*m right? While the text states a clockwise moment. $\endgroup$
    – strateeg32
    Feb 25, 2016 at 16:02
  • $\begingroup$ Yes, I believe your book is wrong. As you yourself noticed, with those values for $a$ and $b$, you get $F_1\neq F_2$. And yes, the moment should be counter-clockwise. You have two equal forces and obviously the downwards force will be to the left of the upwards force since the latter has a longer base. This creates a counter-clockwise moment. $\endgroup$
    – Wasabi
    Feb 25, 2016 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.