Finding an unknown applied force with friction, Fnet and more.

  • Thread starter QuickSkope
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  • #1
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Homework Statement


a 20kg Toaster is on a level surface (Cooefficient of friction is 0.1). A rope applies an unknown force to the toaster at 30° from the horizontal upwards to the right. This force causes the toaster to accelerate at 0.277 m/s^2. What is the magnitude of the applied force.



Homework Equations


Fnet = m * a
Fn = Fg - Fay
Fax - Ff = Fnet
Ff = μFn (Not sure if thats the symbol for mew)


The Attempt at a Solution


Found Fnet
Fnet = 0.277 * 20
Fnet = 5.54 N


Friction Along the Floor

Fn = Fg - Fay
Fay = FaSin30°
Fn = (9.8 * 20) - (FaSin30°)
Fn = 196 - FaSin30°

Finding Fa
Fax - Ff = Fnet
Ff = μFn
Fax = FaCos30°


Fax - (0.1*(196 - FaSin30°)) = 5.54 N
Fax - 19.6 - (0.1 * - FaSin30°) = 5.54 N
FaCos30° - 19.6 + (0.1 * FaSin30°) = 5.54 N
FaCos30° + (0.1 * FaSin30°) = 25.11 N
Cos30° + (0.1 * Sin30°) = 25.11 N / Fa
0.866 + (0.1 * 0.5) = 25.11 N / Fa
0.916 = 25.11 / Fa
0.916Fa = 25.11
Fa = 27.4 N

---------------------------------------------------------

EDIT: Fax = FaCos30°, making a common variable.
EDIT2: Factored, checking if its right ATM.
EDIT3: YAY :D, Working backwards it works :). Excentric, if anyone can double check my work, would be so awesome.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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1,901
Fax - (0.1*(196 - FaSin30°)) = 5.54 N
Fax - 19.6 - (0.1 * - FaSin30°) = 5.54 N
FaCos30° - 19.6 + (0.1 * FaSin30°) = 5.54 N
FaCos30° + (0.1 * FaSin30°) = 25.11 N
Cos30° + (0.1 * Sin30°) = 25.11 N / Fa
0.866 + (0.1 * 0.5) = 25.11 N / Fa
0.916 = 25.11 / Fa
0.916Fa = 25.11
Fa = 27.4 N

---------------------------------------------------------

EDIT: Fax = FaCos30°, making a common variable.
EDIT2: Factored, checking if its right ATM.
EDIT3: YAY :D, Working backwards it works :). Excentric, if anyone can double check my work, would be so awesome.
It is correct at the end :smile:

ehild
 
  • #3
60
0
It is correct at the end :smile:

ehild
Sweet, thank you :)
 

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