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Mechanics Question

Attached, please help. lol. Thanks. Its not I haven't tried, I just didn't get a very convincing answer. My answer was is tan theta < coefficient.
Picture1.jpg26.2KB
Reply 1
Avatar for Chewwy
Chewwy
17
Fmax=µR=µmcos&#952;

in order to stop Fmax > downward force

µmcos&#952; > mgsin&#952;
µ > tan&#952;

given above, resultant = m(µcos&#952;-gsin&#952; )

=>a = µcos&#952;-gsin&#952;

v^2=u^2+2as
u = &#8730;(v^2 - 2as)

v=0,u=v,s=d

v = &#8730;(2d(µcos&#952;-gsin&#952; ))

which looks a bit messy to be right...
Reply 2
Avatar for 7589200
7589200
OP
18
Yea, I got something similar for second part.

I got u = Root(2*g*d*(mew cos theta - sin theta))
Reply 3
Avatar for Chewwy
Chewwy
17
well yeah, obviously.
Reply 4
Avatar for 7589200
7589200
OP
18
obviously what?
chewwy
Fmax=µR=µmcos&#952;

in order to stop Fmax > downward force

µmcos&#952; > mgsin&#952;
µ > tan&#952;

given above, resultant = m(µcos&#952;-gsin&#952; )

=>a = µcos&#952;-gsin&#952;

v^2=u^2+2as
u = &#8730;(v^2 - 2as)

v=0,u=v,s=d

v = &#8730;(2d(µcos&#952;-gsin&#952; ))

which looks a bit messy to be right...

Okay a few errors.
a = gsin&#952; - µgcos&#952;
Reply 6
Avatar for Chewwy
Chewwy
17
Widowmaker
Okay a few errors.
a = gsin&#952; - µgcos&#952;
you're right. the equation wouldn't even be HOMOGENOUS otherwise.
Reply 7
Avatar for Euclid
Euclid
Vazzyb
Yea, I got something similar for second part.

I got u = Root(2*g*d*(mew cos theta - sin theta))


yep, since the normal force to the plane is a component of the packet's weight » mgcos&#952;.
Reply 8
Avatar for 7589200
7589200
OP
18
hmm, who is right, windowmaker or us, which way is it? cos theta - sin theta or sin theta - cos theta?
Reply 9
Avatar for Chewwy
Chewwy
17
Vazzyb
hmm, who is right, windowmaker or us, which way is it? cos theta - sin theta or sin theta - cos theta?
well they have the same magnitude dont they...
Reply 10
Avatar for 7589200
7589200
OP
18
yea but one way around you are left with a negative root....
Vazzyb
hmm, who is right, windowmaker or us, which way is it? cos theta - sin theta or sin theta - cos theta? I think its the latter right?

lol, it's widowmaker. even though I do love windows. :biggrin:
Reply 12
Avatar for 7589200
7589200
OP
18
lol i know :smile: mistype lol.
Reply 13
Avatar for samd
samd
10
Widowmaker is right of course. I will write a full solution out shortly. Closer inspection reveals this not to be necessary, the solution is (almost) perfectly done above.

Final answer is: v = &#8730;[-2(gsin&#952; - µgcos&#952:wink:d]

Obviously this is not a neg square root as the acceleration, (gsin&#952; - µgcos&#952:wink:, is negative.

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