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M1 help please- coefficient of friction and motion



This is in the M1 Jan 2007 paper. When I tried part a), I used F=ma because the box is in motion. After looking at the mark scheme, they've used F=μR. I can see why you might not use F=ma now- is it because there is no acceleration as it is moving at a constant speed?

Part b) was fine

Answer for a) is 110N I ended up with 137N:frown:
Reply 1
Could a Moderator move this thread into the maths subforum for me? Sorry, didn't know how else to request this.

Need help urgently
Reply 2
Thank you for your help, I've solved it by myself now :tongue:

My original method worked- i.e. you can use F=ma. I substituted in 1 ms^2 for the acceleration when it should have been 0 due to the constant speed. No one will read this, but this thread can be closed if Moderators wish to do so.
Reply 3
yeah, since its moving at a constant speed, acceleration = 0 so
PSin20 = 30g
Reply 4
I just had it, cant remember how, was just playing with calculator
Reply 5
pwdrin
I just had it, cant remember how, was just playing with calculator


Try typing in 1,1 and then 0. :p:

Don't worry about it because I figured it out. Unless you need help...
Reply 6
Ye could you explain, I really want to know lol.
Reply 7
pwdrin
Ye could you explain, I really want to know lol.

Right, I did the following:

Resolve vertically
R+Psin20=30g
R=30g-Psin20

Resolve horizontally using F=ma (Taking direction of motion as the positive)
Pcos20-0.4(30g-Psin20)=30x0
Pcos20-117.6+0.4Psin20=0
Pcos20+0.4Psin20=117.6
2.3492P+Psin20=294
2.69125P=294
P=110 (N)

Sorry if it doesn't make sense. I've copied down what I wrote. Hope it helps :smile:
Reply 8
Yep it does, I just attempted to make it harder than what it really was, by dividing by the cos making it tan etc
Thanks :biggrin:

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