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# M1 Calculating the coefficient of friction Help Needed? watch

1. The question is " A particle is sliding with acceleration 3ms-2 down a 40 degree slope. Calculate the coefficient of friction."

I don't really know how to answer this question without been given the mass of the particle. The only thing i could do is give the coefficient of friction in algebra form and not as a number, is this the correct way of approaching this?
2. Must be without a mass.
3. (Original post by Robert Green)
The question is " A particle is sliding with acceleration 3ms-2 down a 40 degree slope. Calculate the coefficient of friction."

I don't really know how to answer this question without been given the mass of the particle. The only thing i could do is give the coefficient of friction in algebra form and not as a number, is this the correct way of approaching this?
You do not need the mass

You will get R in terms of m
You will get friction in terms of m

And then you will divide them and the m will cancel

This has to be the case otherwise mu would depend on m and it doesn't, it os the same for all m
4. Maybe not then ^ this is probably why I got that M1 score
5. just find it in terms of m, they should cancel.
6. (Original post by TenOfThem)
You do not need the mass

You will get R in terms of m
You will get friction in terms of m

And then you will divide them and the m will cancel

This has to be the case otherwise mu would depend on m and it doesn't, it os the same for all m

I drew this quickly to model whats happening.

So i got R = mgCos(40)

And F =mgSin(40) - 3m

Is this correct so far? And could you help in telling me what i'd do next?
7. (Original post by Robert Green)

Is this correct so far? And could you help in telling me what i'd do next?
yes

so F = mu R

therefore mu = F/R

So do the division
8. (Original post by TenOfThem)
yes

so F = mu R

therefore mu = F/R

So do the division
I got

mgSin(40) - 3m / mgCos(40)

Therefore would that cancel to

Sin(40) - 3m / Cos(40) = mu

?
9. (Original post by Robert Green)
I got

mgSin(40) - 3m / mgCos(40)

Therefore would that cancel to

Sin(40) - 3m / Cos(40) = mu

?
You have not cancelled the m in the 3m

Where has g gone?

You just have to cancel the m
10. (Original post by TenOfThem)
You have not cancelled the m in the 3m

Where has g gone?

You just have to cancel the m
So i end up with...

mgSin(40) - 3 / gCos(40) ?
11. Don't you end up with gTan(40)-3?
12. (Original post by Robert Green)
So i end up with...

mgSin(40) - 3 / gCos(40) ?
You still are not cancelling the m

mgSin40 - 3m = m(gSin40 - 3)
13. (Original post by Fool In The Rain)
Don't you end up with gTan(40)-3?
no you don't
14. f= μr
15. (Original post by TenOfThem)
no you don't
Do you end up with gSin40-3/gCos40? Since m(gsin40-3)/m(gcos40) will cancel the m's out. Apologies in advance for my terrible M1 skills .
16. ..Just do F=ma downwards......
mgsin40-uR=3m
now R=mgcos40
mgsin40-umgcos40=3m
cancel all m's and rearrange for u.
17. (Original post by TenOfThem)
You still are not cancelling the m

mgSin40 - 3m = m(gSin40 - 3)
Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
18. (Original post by Robert Green)
Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
yes
19. (Original post by Fool In The Rain)
Do you end up with gSin40-3/gCos40? Since m(gsin40-3)/m(gcos40) will cancel the m's out.
yes
20. (Original post by Robert Green)
Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
Yeah, I got that answer as well. Using gsin(40)-3/gcos(40). Let's hope it's right!

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