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Three forces on a box in different directions what is the acceleration

  • Thread starter Lucaburra
  • Start date
  • #1
3
0

Homework Statement



Force A 250 degrees @ 200 N
Force B 150 degrees @ 300 N
Force C 50 degrees @ 400 N
mass = 40kg
mew = .14
what is acceleration

Homework Equations



Fa^2 = Fy net^2 x Fx net^2
Fn = Fg
Ff = Fn x mew
Fnet = Fa - Ff
F= ma


The Attempt at a Solution



Fax=Cos 20 (200) = -187.93
Fay = Sin 20 (200) = -68.40

Fbx = Sin 30 (300) = 150
Fby = Cos 30 (300) = -259.81

Fcx = Cos 50 (400) = 257.11
Fcy = Sin 50 (400) = 306.42

Fx net = 219.18 N
Fynet = -21.79 N

Fa^2 = Fxnet^2 + Fynet^2
Fa^2 = 219.18^2 + -21.79^2
Fa^2 = 48039.87 + -474.80
Fa^2 = 47565.07
Fa = 218.09 N

Fn = 9.81 m/s^2 x 40Kg = 392.40N

Ff = Fn (mew) = 392.40N (.14) = 54.94 N

Fnet = Fa - Ff = 218.09 - 54.94 = 163.15 N
a = F/m = 163.15/40 = 4.08 m/s

I'm just not sure Ive done this correct
 

Answers and Replies

  • #2
6,054
390
The components of Fa and Fb are not computed correctly. BTW, you could make your life easier by choosing x to coincide with one of the forces, say Fc, then Fcx = 400 N and Fcy = 0, and the angles of the other forces are relative to Fc, i.e., 200 and 100.
 
  • #3
3
0
Sorry voko I dont follow. I drew my x,y graph and entered the force vectors. I then used the angles relative to the x axis, (as the text says) hence F ax = Cos 20 and F bx = Sin 30. I'm not sure about your meaning to "choose x to coincide with one of the forces." I'm in grade 11 and this is what we were taught to do.
 
  • #4
6,054
390
No problem, you can ignore this part. But the components of Fa and Fb are still not computed properly. For example, Fax = Fa cos 250 = -68.4 N.
 

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