AQA Physics A - I suck at mechanics - need help

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spleenharvester
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Always had a very poor understanding of mechanics and trigonometry. In PHYA2 I got a C grade, as while I got most of the marks on waves and materials, my mechanics section dragged me waaaay down.

I don't understand this question in the slightest.

"A car of mass 1200kg is parked on a hill inclined at 20 degrees to the horizontal. The maximum frictional force between the tyres and the road is 5000N. Will the car remain in equilibrium?"

Please excuse the crude diagram. This is what I have so far:

Image

So I understand I need to resolve it into perpendicular vectors, as above.

1) How are we supposed to decide which vector is the hypotenuse and which is the adjacent? I get this stage wrong time and time and time again. According to the textbook, "?" is given by Wsin20, and the red arrow (whatever vector that is) is given by Wcos20.

But I can't make any sense of that. I understand it's a C-angle so the angle of the red arrow away from mg = 20 degrees also. But how do we decide whether mg is the hypotenuse or if the red arrow is the hypotenuse?

2) What the hell is that "maximum frictional force" stuff about? Maximum before what?

Thankyou all in advance, it is very much appreciated.
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spleenharvester
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Okay I think I figured 1) out. So mg is a resultant vector right? And it's being split into its horizontal and verticular components perpendicular to the slope. So using parallelogram rule, mg is the resultant/longest line, and therefore is the hypotenuse?
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OliverG
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Right.

Consider it in terms of forces parallel to the plane (along the slope in layman's terms)

The force acting down the slope (as a component of the mass acting vertically down) MUST be less than the force than the force acting vertically down. So the force vertically down is ALWAYS going to be the hypotenuse, as that is the largest side of the triangle. Your 20 degree angle acts between mg and the contact force, R.

You, however, want the force acting DOWN the plane of the slope, so mgSIN20, not up and thus cos20. (That would be for finding the co-efficient of friction, u, but you don't need that for Physics, only A Level Mechanics)

So, the question is basically asking, is the force down the slope greater than the MAXIMUM force (friction) up the slope. Note how I say maximum, as friction will not always give the greatest force possible. If it did, then it would become greater than the force acting down the slope, and the car would start moving up the slope, (and we should not go to that world, for it is a silly place). Imagine that you are pulling a box along a floor with a force on 1N. It doesn't move. Then 2N. Still doesn't move. Then 3N. Still doesn't move. But after 4N, it will move. That is because friction increases from 1, to 2, to 3, to 4N, then can increase no further, and thus there is a resultant force = Mass x Acceleration.

You should find the answer to be yes, the object does remain in equilibrium. Try working it out for yourself.


Sorry if that is a little bit long winded, but these kind of things require you to look at the situation and figure out what forces are acting, and it's best to fully describe it.
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spleenharvester
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Fantastic explanation mate, thankyou
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