# Determine the angle of incline

by idlackage
Tags: angle, determine, incline
 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
HW Helper
PF Gold
P: 3,440
 Quote by idlackage 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.
P: 11
 Quote by kuruman 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?

 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.
 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°
 HW Helper PF Gold 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.
 P: 11 Awesome, thank you so much!

 Related Discussions Introductory Physics Homework 1 Introductory Physics Homework 6 Introductory Physics Homework 2 Introductory Physics Homework 11 Introductory Physics Homework 8