What is the Angle of Incline on a Frictionless Air Track?

[idlackage]

Homework Statement

Determine the angle of incline on https://www.physicsforums.com/attachment.php?attachmentid=30517&stc=1&d=1291856979" frictionless air track.

Acceleration = 0.147 m/s².

The answer is supposed to be 4.10°.

Homework Equations

F_net = ma

The Attempt at a Solution

Mass 1 (https://www.physicsforums.com/attachment.php?attachmentid=30515&stc=1&d=1291856979"):

Perpendicular (y):

F_ny = +F_n
T_y = 0
a_y = 0
F_gy = -m₁gcosθ
= (428)(-9.80)cosθ
= -4194.4cosθ
F_n + F_gy = 0
F_n = +4194.4cosθ

Parallel (x):

F_nx = 0
= -m₁gxsinθ
= -4194.4sinθ
F_net = m₁a
F_n + F_gx + T = m₁a

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

Mass 2 (https://www.physicsforums.com/attachment.php?attachmentid=30516&stc=1&d=1291856979"):

F_g2 = (20.0)(9.80)
= 196
F_g2 + 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.

 

[kuruman]

Parallel (x):

F_nx = 0
= -m₁gxsinθ
= -4194.4sinθ
F_net = m₁a
F_n + F_gx + 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]

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 F_n 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]

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]

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]

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.

 

[idlackage]

Awesome, thank you so much!