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Determine the angle of incline 
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#1
Dec810, 07:17 PM

P: 11

1. The problem statement, all variables and given/known data
Determine the angle of incline on this frictionless air track. Acceleration = 0.147 m/s^{2}. The answer is supposed to be 4.10°. 2. Relevant equations F_{net} = ma 3. The attempt at a solution Mass 1 (FBD): Perpendicular (y): F_{ny} = +F_{n} T_{y} = 0 a_{y} = 0 F_{gy} = m_{1}gcosθ = (428)(9.80)cosθ = 4194.4cosθ F_{n} + F_{gy} = 0 F_{n} = +4194.4cosθ Parallel (x): F_{nx} = 0 = m_{1}gxsinθ = 4194.4sinθ F_{net} = m_{1}a F_{n} + F_{gx} + T = m_{1}a +4194.4cosθ – (4194.4sinθ) + T = (428)(0.147) = 62.9 Mass 2 (FBD): F_{g2} = (20.0)(9.80) = 196 F_{g2} + T = m_{2}a 196 + T = (20.0)(0.147) = 2.94 T = 2.94 – 196 T = 490 Final calculations: +4194.4cosθ – (4194.4sinθ) + T = 62.9 () T = 490  +4194.4cosθ – (4194.4sinθ) = 552 Not only do I not know how to solve that final equation, I'm also sure the answer will be wrong. I really don't get how to properly go about this question. 


#2
Dec910, 09:29 AM

HW Helper
PF Gold
P: 3,440




#3
Dec910, 09:46 AM

P: 11




#4
Dec910, 09:50 AM

HW Helper
PF Gold
P: 3,440

Determine the angle of incline
The equation from the second FBD is also incorrect. The weight and the tension are in opposite directions. When you fix it, make sure that the direction of the acceleration is consistent between FBD1 and FBD2.



#5
Dec910, 10:08 AM

P: 11

I just realized I had no idea how I got 2.94 – 196 = 490 or other glaring errors. Is this correct?:



#6
Dec910, 10:21 AM

HW Helper
PF Gold
P: 3,440

That looks about right considering round offs. I got 3.55^{o}.
*** On edit *** Note that according to your drawing, arcsin(0.162/2.28) = 4.09^{o}. If that's not close enough, you may have to explain the discrepancy. 


#7
Dec910, 10:25 AM

P: 11

Awesome, thank you so much!



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