Is a force to the right positive?

Question

I am trying to do this question and I have everything but one.

Solving for $A_x$, $x$ force at $A$.

$$\begin{gather} F_x = 6\cdot20 + A_x \\ A_x= -120\text{ kN} \end{gather}$$

Is this correct?

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Sign normally doesn't matter, if you're consistent, but it's causing an error in my working out the moment.

Evaluating it, I get -1852.5 which is incorrect. The answer is 1132.5.

If I change the sign of $6\cdot20$ to $-6\cdot20$, I get the ‘correct’ answer, -1132.5.

$$\begin{gather} F_x = 20\cdot6 + A_x = 0 \\ A_x = -120 \\ M_D = -30\cdot7\cdot3.5+2\cdot E_y = 0 \\ E_y = 367.5 \\ F_y = D_y + E_y - 30\cdot7 = 0 \\ D_y = -157.5 \\ F_y = -D_y - 60 + A_y = 0 \\ A_y = 60-157.5 = -97.5 \\ \sum M_A = M_A + 6\cdot20\cdot3-60\cdot4-D_y\cdot11=0 \end{gather}$$

Answer 1

A moment is defined as the cross-product of distance and force vectors:

$$\begin{align} M &= r\times F \\ &= \det\left(\left|\begin{matrix} i & j & k \\ r_x & r_y & r_z \\ F_x & F_y & F_z \\ \end{matrix}\right|\right) \\ &= (r_yF_z-r_zF_y)\hat{i} -(r_xF_z-r_zF_x)\hat{j} + (r_xF_y-r_yF_x)\hat{k} \end{align}$$

In a 2D frame, $r_z = F_z = 0$, which simplifies this to $M = (r_xF_y-r_yF_x)\hat{k}$. So, a positive vertical force at a positive horizontal distance (to the right of the studied node) results in a positive moment. Meanwhile, a positive horizontal force at a positive vertical distance (above the studied node) results in a negative moment.

A simple way of thinking about it is that force components are positive to the right or upwards and distances are positive if a positive force would generate a positive (counter-clockwise) rotation.

In your last equation, you made the moment due to the distributed load along $\overline{AB}$ positive, when it should be negative. After all, a positive horizontal force at a positive vertical distance (above $A$) results in a negative moment.

It should, however, be:

$$\begin{align} \sum M_A =& M_A \\ & - (3)\cdot(6\cdot20) \\ & + (4)\cdot(-60) \\ & + (14.5)\cdot(-7\cdot30) \\ & + (13)\cdot (F_{E,y})= 0 \\ \therefore M_A =& -1132.5\text {kN} \end{align}$$

I personally hate writing equations like this, with signs in the middle of the multiplications, but it makes all the variables explicit. So, in this calculation:

• all the distances are positive (everything is either above or to the right of $A$)
• the horizontal distributed load is positive
• the vertical loads are negative
• $F_{E,y}$ is whatever it will be (positive, in our case)
• When calculating the moment due to vertical forces, the sign is positive
• When calculating the moment due to horizontal forces, the sign is negative

Answer 2

You are correct that sign doesn't matter as long as you are consistent.
In that final stage you are calculating the torque about point A, you need to be consistent as to the direction of that torque, is it clockwise or counterclockwise?

When looking at point A if you use positive for horizontal forces to the right then you need to use positive for vertical forces that are down. Mixing it and using positive for left and down or for right and up will give the wrong answer because you're being inconsistent as to the direction of the torque.

This does mean that for torque calculations you may have to change the sign for a given force depending upon it's position relative to the point you are analyzing.