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Determine the angle of incline

by idlackage
Tags: angle, determine, incline
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idlackage
#1
Dec8-10, 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/s2.

The answer is supposed to be 4.10.

2. Relevant equations

Fnet = ma

3. The attempt at a solution

Mass 1 (FBD):

Perpendicular (y):

Fny = +Fn
Ty = 0
ay = 0
Fgy = -m1gcosθ
= (428)(-9.80)cosθ
= -4194.4cosθ
Fn + Fgy = 0
Fn = +4194.4cosθ

Parallel (x):

Fnx = 0
= -m1gxsinθ
= -4194.4sinθ
Fnet = m1a
Fn + Fgx + T = m1a

+4194.4cosθ (-4194.4sinθ) + T
= (428)(0.147)
= 62.9

Mass 2 (FBD):

Fg2 = (20.0)(9.80)
= 196
Fg2 + T = m2a

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.
Attached Thumbnails
fbd3_m1.jpg   fbd3_m2.jpg   diagram3.jpg  
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kuruman
#2
Dec9-10, 09:29 AM
HW Helper
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kuruman's Avatar
P: 3,440
Quote Quote by idlackage View Post
Parallel (x):

Fnx = 0
= -m1gxsinθ
= -4194.4sinθ
Fnet = m1a
Fn + Fgx + T = m1a
Stop!!!. This equation is wrong. The normal force is along y (as you correctly point out) and does not belong in the x-equation. Furthermore, the tension is up the incline and the component of the weight is down the incline, so they must have opposite signs.
idlackage
#3
Dec9-10, 09:46 AM
P: 11
Quote Quote by kuruman View Post
Stop!!!. This equation is wrong. The normal force is along y (as you correctly point out) and does not belong in the x-equation. Furthermore, the tension is up the incline and the component of the weight is down the incline, so they must have opposite signs.
Thank you! I omitted Fn and changed the equation to this:

-Fgx + T = -m1a
-4194.4sinθ + T
   = (428)(-0.147)
   = -62.9
So that the final calculations were:

    T = -62.9 + 4194.4sinθ
(-) T = -490
---------------------------
    0 = 427.1 + 4194.4sinθ
-427.1 = 4194.4sinθ
-427.1/sinθ = 4194.4
sinθ = -427.1/4194.4
   θ = sin-1(-427.1/4194.4)
   θ = 5.84
Is this just a big percentage error, or am I still doing something wrong?

kuruman
#4
Dec9-10, 09:50 AM
HW Helper
PF Gold
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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.
idlackage
#5
Dec9-10, 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?:

Fg2 - T = -m2a

196 - T = (20.0)(-0.147)
        = -2.94
-T = -2.94 – 196
T = 198.94

Final calculations:

    T = -62.9 + 4194.4sinθ
(-) T = 198.94
---------------------------
    0 = -260.94 + 4194.4sinθ
260.94 = 4194.4sinθ
260.94/sinθ = 4194.4
sinθ = 260.94/4194.4
    θ = sin-1(260.94/4194.4)
    θ = 3.57
kuruman
#6
Dec9-10, 10:21 AM
HW Helper
PF Gold
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P: 3,440
That looks about right considering round offs. I got 3.55o.

*** On edit ***
Note that according to your drawing, arcsin(0.162/2.28) = 4.09o. If that's not close enough, you may have to explain the discrepancy.
idlackage
#7
Dec9-10, 10:25 AM
P: 11
Awesome, thank you so much!


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