Calculating acceleration on an inclined plane (with friction)

  • Thread starter Bobazoide
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  • #1
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Homework Statement


A 38kg block is on a slope that is 50 degrees above the horizon. The box is attached to a 9kg weight that rests on a flat surface (connected to the top of the slope). The coefficient of friction for both surfaces is 0.12. Calculate the acceleration of the system.


Homework Equations


Fnet = Fapp - Ff
Fgx = (sin[angle])(gravity)(mass)
F = ma

The Attempt at a Solution



Fnet = Fapp - Ff
Fnet = (38 * 9.81 * (sin50)) - (9*9.81*.12) - ((Cos50)*9.81*38*0.12)
Fnet = 246.1

F = ma

a = f/m
a = 246.21/(38+9)
a = 5.24m/s^2

My professor got 2.9m/s^2 and I can't see where (if) I'm going wrong. Help would be absolutely fantastic. Thanks!
 

Answers and Replies

  • #2
haruspex
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I can't find a fault in your reasoning or calculation either.
 
  • #3
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He was using the formula "Fnet = Fapp - Ff (of the 38kg object) - T2" where T2 is tension in the rope that is connecting the two weights, if that helps. I can't really explain much about about it though, since it was the first time I had ever seen it used in my life and he didn't really explain it either.

I have a test about this kind of thing tomorrow and it's freaking me out right now. >_> Thanks so much!
 
  • #4
haruspex
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He was using the formula "Fnet = Fapp - Ff (of the 38kg object) - T2" where T2 is tension in the rope that is connecting the two weights,
That's a perfectly reasonable approach, but it should produce the same answer:
M = 38, m = 9, kinetic friction coefficient = μ, slope = α, accn = a.
Ma = Mg sin(α) - T - Mgμ cos(α)
ma = T - mgμ
Summing:
Ma + ma = Mg sin(α) - mgμ - Mgμ cos(α)
a = g(M sin(α) - mμ - Mμ cos(α))/(M+m)
= 9.81*(38*sin(50*pi()/180) - 9*.12 - 38*.12*cos(50*pi()/180))/(38+9) = 5.24
 
  • #5
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Well I'll show that to him before the test, then. Thank you so much; I really appreciate it!
 

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