Why is weight the hypotenuse (A Level Year 1 Physics)

In the answers, they do the force down the slope as (M+2m)gsin35. So it's the weight times the sine of the angle. If you do SOH, that means the weight is the hypotenuse. But if you draw the triangle, the weight is clearly the opposite... right? This has happened to me in so many similar questions and I don't get it! Am I being really stupid? Could someone please explain this to me because my teachers haven't been very helpful lol. Thanks.

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Reply 1

Callicious

Try to think of it in terms of vectors, which are just magnitudes that are assigned a direction. Here's a terribly drawn picture to show; I've attached it.

If you think of the weight as a vector, and then decompose it onto a coordinate system of convenience, i.e. one where there are two axes with one parallel and one orthogonal to the slope, you can find the component parallel to the slope.

Remember that you can pick any coordinate system you want, as long as the magnitude of the weight vector remains constant in your decomposition.

fedrheherh.PNG53.6KB

Reply 2

Starra

Edit: I am thinking in vectors, that's why I'm drawing triangles and stuff.
But why isn't the weight the vertical? If it's only acting in the vertical plane, how can we split it up into horizontal and vertical? Why isn't this the triangle?

(edited 5 years ago)

Reply 3

Callicious

Sorry, I wasn't particularly clear in my diagram and assumed it was taken, but, the vertically downward arrow is the weight, the two side/angled arrows are the decomposed vectors.

Also, for your vector diagram, assuming that it shows the weight decomposed, the weight is the final vector, the hypotenuse for example, you can't just make an arbitrary decomposition, there is a set way to go about it in this manner. Naturally the weight could be represented by vectors in the analogue above, but it wouldn't be right to say that the component of weight parallel to slope exceeds the magnitude of the weight itself, that'd be crazy;

The force you decompose should always be kept to the hypotenuse of whatever you're decomposing, at least for these kinds of questions. If it were a component of another force, then you could feel free, but that'd depend on the component, the angles involved, etc.

Original post by Starra

(edited 5 years ago)

Reply 4

Starra

I'm not decomposing the weight in my diagram. I'm decomposing F, the resultant force down the slope, which is affected by the weight (its vertical component) and tension. Idk which part of F would be horizontal though.

Original post by Callicious

Reply 5

Callicious

Original post by Starra

The force F down the slope also has a magnitude, and if it comes from the weight, it stands to reason it can't have a magnitude greater than it, which is what your diagram appears to show;

F is a component of W, not the other way around, as the component of W parallel to the slope.

Reply 6

Starra

Mhm, so F is the hypotenuse, because F is the biggest. W is a component of F in this situation, right?

Original post by Callicious

Reply 7

Callicious

Original post by Starra

Mhm, so F is the hypotenuse, because F is the biggest. W is a component of F in this situation, right?

It's a dubious way to say it, and I'm not sure if that's mathematically accurate since I've actually just ran across a problem with decomposition where a force given was actually part of another force (it turned out, but wasn't said initially) and so might bite you in the future,

However as far as common sense goes, that should work.

It just takes practice to be honest, and you'll get a common sense about how to decompose forces.