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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).

force diagram of 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?

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  • $\begingroup$ In general you should develop the equations with $\ddot{x} \ne 0$ and $\ddot{y} = 0$. Then in the end use the constant motion $\ddot{x}=0$ in order to find the tension $T$. $\endgroup$ – ja72 Sep 15 '14 at 19:06
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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.

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For equation 1a we don't take component of $ mg $ because it's component in this direction is zero (your guess is right) $$ mg\cos (90) = 0$$

The same thing happens for normal in 1a and $ T $ in 2b, since direction "x" is perpendicular to direction of normal and direction "y" is perpendicular to direction of $ T $ their components in respective directions become 0.

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