I'm a high school student and I'm studying Newton's second Law. While my teacher is calculating a net force of an object he always treat all vectors as positive numbers. I think what he meant is the magnitude of the vector which is always positive, but the problem is that he didn't use the magnitude symbol, for example like $||Fg||$. He said the acceleration due to gravity , g, is positive, but shouldn't it be negative since it is pointing down. This also cause a problem: every time when I calculate the net force I can't add all forces together, instead I need to choose to use minus sign when the vector is pointing down. So in general which way is correct?
[Physics] Sign of the vector (especially $g$) while calculating
conventionscoordinate systemsvectors
Related Solutions
I think you just forgot that the $\int_A^B F\,dl$ is not a scalar expression. Rather it should be written in a form $\int_A^B \vec{F}\cdot d\vec{l}$. Then it comes to the sign of the scalar product: $$\vec{F}\cdot d\vec{l}=F\,dl\,\cos\theta$$ where the angle $\theta$ is taken between the vector $\vec{F}$ and the direction of the tangent to the integration path from $A$ to $B$. Then, in your first example,
$W_{A \to B}=\int_{r_A}^{r_B} F(r) dr = \int_{r_A}^{r_B} \left(-\frac{GMm}{r^2} \right)dr$
the path could go with any slope, but the gravity is always directed downwards, along the $r$ axis. That means, we can always take $(\pi-\theta)$ as the angle between the vector $d\vec{l}$ and the $r$ axis, that is $$dl\,\cos(\pi-\theta)=dr$$ but $\cos(\pi-\theta)=-\cos\theta$ and thus we have $$\vec{F}\cdot d\vec{l}=-F\,dr=-\frac{GMm}{r^2}dr$$
For your second example:
...we also should change the sign, because the gravitational force is always a force of attraction.
what the authors actually mean is that: the Coulomb's and Newton's forces have exactly the same expressions, but the sign conventions for them are different. The Newton's force is defined that if all the quantities ($M$, $m$ and $r$) are positive, then the vector of the force is directed towards the other body. But for the Coulomb's force, if all the quantities ($q_1$, $q_2$ and $r$) are positive, then the vector of the force is directed away from the other charge. That becomes manifest if we take the vector expressions for these forces: $$\vec{F}_N=-\frac{GMm\,\vec{r}}{r^3}\qquad\vec{F}_C=\frac{q_1q_2\,\vec{r}}{4\pi\epsilon_0\,r^3}$$ Now the different signs are clearly seen.
"...from point $A$ to point $B$..." - ...as I understand it - the work that I must do is always $U_B-U_A$. However the work that the force that is being created by the field do_es_ is always $U_A-U_B$, am I correct?
Yes this is correct.
The mnemonic rule is very simple: $U$ is like the height of the slope. When you go up, $U_B>U_A$, and it is you who does the work. But when you go down, $U_A>U_B$, and it is the field force who does the work.
Here is another approach that is essentially the same as the one suggested by John, but I find it to be less error-prone when doing calculations by hand.
You are given
$$ \mathbf{A} = 5\mathbf{i} - 6.5\mathbf{j}\\ \mathbf{B} = -3.5\mathbf{i} + 7\mathbf{j} $$
You know that $ \mathbf{C} \perp \mathbf{A}$. Therefore,
$$ \mathbf{C} = s \cdot (6.5 \mathbf{i} + 5\mathbf{j}) $$
with $s$ a scale factor. This is because in 2D, for any vector $\alpha = a\mathbf{i} + b\mathbf{j}$, the vector $\beta = -b\mathbf{i} + a\mathbf{j}$ satisfies $\alpha \cdot \beta = 0$ and thus $\alpha \perp \beta$.
Given that $\mathbf{B} \cdot \mathbf{C} = 15$, it should be fairly easy to solve for the scale factor $s$, thus solving for $\mathbf{C}$.
Best Answer
The teacher's point is that all vector magnitudes are positive, and you only add signs because of directions.
So yes, you are right. There will be minus signs as well.
But you can't say that for example $g$ should be negative. It is a positive value on itself, and it only becomes negative, if the direction is opposite to the axis. If you in some specific case choose your coordinate system to point downwards, then the $g$ is positive.
If you had been taught that $g$ is always negative, you might have been confused in that case.