AS AQA Mathematics Mechanics 1 MPM1 16 June 2014

OP

Can someone explain to me the F=uR equation in terms of when it's < or =< or = or > or >=

???

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Reply 81

TLHroolz

3

Original post by CD223

Can someone explain to me the F=uR equation in terms of when it's < or =< or = or > or >=

???

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It's F=uR in motion or just about to move off (limiting equilibrium) or when it is equilibrium or not moving it is F=< uR

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Reply 82

OP

Original post by TLHroolz

It's F=uR in motion or just about to move off (limiting equilibrium) or when it is equilibrium or not moving it is F=< uR

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Cheers! So it's only F=uR when moving. In all other cases it's =<
?

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Reply 83

TLHroolz

3

Original post by CD223

Cheers! So it's only F=uR when moving. In all other cases it's =<
?

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Yeah, and it makes sense. The frictional force could be anything just up to the point of it moving when it isn't moving, so it will be equal to and anywhere up to the equal to... make sense?

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Reply 84

OP

Original post by TLHroolz

Yeah, and it makes sense. The frictional force could be anything just up to the point of it moving when it isn't moving, so it will be equal to and anywhere up to the equal to... make sense?

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Yep! Thanks!

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Reply 85

Jimmy20002012

4

I seem to be struggling which way the arrows go, how are you meant to know like in this question:

ImageUploadedByStudent Room1402136563.525473.jpg

ImageUploadedByStudent Room1402136563.525473.jpg

How are you meant to know for the horizontal the arrow is right or left, and for vertical if it is up or down in order to solve for components??

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Reply 86

WesterosWildling

3

Today I'm just going over theory. I cant just do questions like core maths cos i seem to always get Ds.

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Reply 87

OP

Original post by Jimmy20002012

I seem to be struggling which way the arrows go, how are you meant to know like in this question:

ImageUploadedByStudent Room1402136563.525473.jpg

How are you meant to know for the horizontal the arrow is right or left, and for vertical if it is up or down in order to solve for components??

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Think of the arrows as being on an axis. If it's pointing in the direction you've shown as an example, it's going to be in the top right quadrant of the axis, in which case the vertical points up and the horizontal points to the right :smile:

Reply 88

OP

Original post by WesterosWildling

Today I'm just going over theory. I cant just do questions like core maths cos i seem to always get Ds.

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That's sensible. The theory is the hardest bit for some people so getting your head round it will help!

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Reply 89

Jimmy20002012

4

Original post by CD223

Think of the arrows as being on an axis. If it's pointing in the direction you've shown as an example, it's going to be in the top right quadrant of the axis, in which case the vertical points up and the horizontal points to the right :smile:

I don't seem to get what you mean by it being in the top quadrant of the axis?

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Reply 90

OP

Original post by Jimmy20002012

I don't seem to get what you mean by it being in the top quadrant of the axis?

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Ignoring the numbers, this is what I mean. The arrow you showed if you picture it from the origin it points in the north easy direction (the top right quadrant). This means if you look below, the x axis points to the right and the y axis points upwards in this quadrant.

ImageUploadedByStudent Room1402137608.378655.jpg

ImageUploadedByStudent Room1402137608.378655.jpg

Reply 91

Jimmy20002012

4

Original post by CD223

Ignoring the numbers, this is what I mean. The arrow you showed if you picture it from the origin it points in the north easy direction (the top right quadrant). This means if you look below, the x axis points to the right and the y axis points upwards in this quadrant.

ImageUploadedByStudent Room1402137608.378655.jpg

Okay I sort of get it, what about a inclined plane, Mg going downwards, which way would sin and cos being going now?

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Reply 92

OP

Original post by Jimmy20002012

Okay I sort of get it, what about a inclined plane, Mg going downwards, which way would sin and cos being going now?

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1) mg acts downwards
2) mg cos theta acts perpendicular to the slope
3) mg sin theta acts parallel to the slope

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Reply 93

RoughWritings

9

I am taking this exam too. I am ****ting bricks. I sort of everything now. I am yet to master Connected Particles.

Reply 94

OP

Original post by RoughWritings

I am taking this exam too. I am ****ting bricks. I sort of everything now. I am yet to master Connected Particles.

I've got physics on Monday so I've been devoting time to that - hope that doesn't backfire!

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Reply 95

toonervoustotalk

20

I am soo ready for this exam. Did a june 09 paper today and got 58/75. :biggrin:

Reply 96

OP

Original post by toonervoustotalk

I am soo ready for this exam. Did a june 09 paper today and got 58/75. :biggrin:

Well done

what did you drop marks on :tongue:

Reply 97

toonervoustotalk

20

Original post by CD223

Well done

what did you drop marks on :tongue:

most of the marks i lost were on silly mistakes and i lost a few on resolving vectors. it was a 5 marker that i couldn't do.

Reply 98

OP