An object is pulled up a slope at constant speed. I'm trying to find the Tension on the rope.

$$m = 65kg \Rightarrow W=-637$$ $$\mbox{constant speed} \Rightarrow a = 0$$ $$\mbox{angle of slope} = 45^\circ$$ $$T=?$$

I use Newton's first Law to determine that there is no acceleration because of constant speed: $$\sum F=0$$

Online I found that I should use this: $$W \sin\theta=T$$

I don't understand how they came up with that equation. I tried using mine like this but the answer is way too much to be correct. My equation: $\sum F_y=-W+\sin\theta*T_y=0$

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    $\begingroup$ Hint: forces are vectors. $\endgroup$ – dmckee --- ex-moderator kitten May 25 '14 at 3:16
  • $\begingroup$ @dmckee I know, but that doesn't help. If you look at the equation I created, I use the Force as a vector on the y component. That is why my equation has $sin\theta * T_y$ $\endgroup$ – CharlieK May 25 '14 at 3:17
  • $\begingroup$ You forget to consider normal reaction in your equation. $\endgroup$ – velut luna May 25 '14 at 3:34
  • $\begingroup$ Second hint: There is no way that the tension and the weight alone can add up to vector zero (because they are not collinear). There must be at least one more force in the problem. $\endgroup$ – dmckee --- ex-moderator kitten May 25 '14 at 3:34
  • $\begingroup$ Orient co-ordinate axis so that X-axis lies on the surface. $\endgroup$ – Schrödinger's Cat May 25 '14 at 3:52

This is because you're thinking of the wrong problem. What you're thinking is this:

enter image description here

That's not what the problem states. Actually, if you assume 0 N in the vertical direction, this would cause it to accelerate horizontally, not what we want. The problem is about a ramp:

enter image description here

You have to break the weight into components that are "x/y" if we treat the surface of the ramp as "y=0". Then, as you can see, $F_{net \, x}=T-W\sin \theta=0$, which gives you the solution. Also note for future that the "vertical" component of the weight cancels the normal force.

By the way, tip for the future: know $mg \sin \theta$ well. I remember that it's the "horizontal" by thinking: "s" in "sin" for "sliding."

  • $\begingroup$ Yes that is exactly how I had drawn my diagram, good catch. Now it this makes sense. +1 For the diagrams. $\endgroup$ – CharlieK May 25 '14 at 3:59

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