I was given the following notes in class:
4.32 A skier of mass $65.0\text{ kg}$ is pulled up a snow-covered slope at constant speed by a tow rope that is parallel to the ground. The ground slopes upward at a constant angle of $26.0^\circ$ above the horizontal, and you can ignore friction. (a) Draw a clearly labeled free-body diagram for the skier. (b) Calculate the tension in the tow rope.
This is a problem where there is motion. There is still no acceleration though. Let's look at a quick diagram (with a block as our skier).
There are two ways we could do a problem like this…
$$\begin{align}
T\cos\theta – n\sin\theta &= ma_{x'} = 0 \tag{1a} \\
n\cos\theta + T\sin\theta – mg &= ma_{y'} = 0 \tag{2a}
\end{align}$$$$\begin{align}
T – mg\sin\theta &= ma_x = 0 \tag{1b} \\
n – mg\cos\theta &= ma_y = 0\tag{2b}
\end{align}$$
and I'm having a little trouble understanding how he came up with those equations, although I think I have an idea.
Do equations (1a) and (2a) represent like…the $x$ and $y$ components of all the forces in the diagram? Is that an accurate way to think about it? For (1a), I understand the $T\cos(\theta)$ is the horizontal component of the tension, and $n\sin(\theta)$ is the horizontal component of the normal force, and I guess you subtract them from each other because they go in different directions, but why is the $mg$ force not considered at all in this one? My guess is that there is no horizontal component to gravity since it can only pull down, but I'm not 100% sure if that's why.
Finally, for equation (1b) I see that the normal force isn't factored into the equation at all, and for (2b), the force of $T$ isn't considered at all either, and I don't really think I understand why.
Can anyone help me understand this a little better?
Best Answer
I think you are pretty close to understanding it.
The "A" set of equations represent the components in the primed coordinate system shown with axes $x'$, $y'$. Here $y'$ is what's usually considered "vertical" (normal to the tangent plane to Earth where you happen to be , if you will).
You are exactly right, $mg$ is not considered in Equation 1a because the $x'$ axis is perpendicular to the direction that $mg$ acts (parallel to $y'$, which has the equation 2a).
Now the "B" set of equations represent the components in the unprimed coordinate system. Unprimed $x$ is tangential to the face of the block touching the inclined plane, unprimed $y$ is normal to that face. To get from the primed to the unprimed coordinate system,you rotate by exactly $\theta$, the angle of the inclined plane.
So from the diagram, tension $T$ comes into play only in the $x$ component, while normal force $n$ comes into play only in the $y$ component.