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Dakkers
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Homework Statement
A box is trying to be pulled by a rope up a hill that is 30 degrees above the horizontal. The coefficient of friction ([tex]\mu[/tex]) is 0.25. The acceleration is 1.5m/[tex]s^{2}[/tex] down the hill. The mass of the box is 2.0 kg. Find the force of tension in the rope.
Homework Equations
Fg = mg
Fn = Fg
Ff = [tex]\mu[/tex]Fn
Fnet = ma
The Attempt at a Solution
I started by calculating each of the forces that I could.
Fg = mg = (2.0)(9.8) = 19.6 N
Fn = Fg = 19.6 N
Ff = [tex]\mu[/tex]Fn = (0.25)(19.6) = 4.9 N
I drew a diagram and broke everything into components.
[URL]http://img231.imageshack.us/i/physicsforces.png/[/URL]
I forgot to add that if we do components of each force we have to do the components of acceleration, which becomes either 1.5cos30
or 1.5sin30 [down].
I plugged the components into the Fnet = ma equation but it didn't work, like so:
Fnet = ma (up/down)
Ftsin30 + 19.6sin60 - 19.6 - 4.9sin30 = (2.0)(-1.5sin30)
Ftsin30 - 5.0759... = -1.5
Ftsin30 = 3.5759
Ft = 7.1518
buuut that's not right because, at the least, Ff > Ft since the box is sliding down the hill.
Fnet = ma (left/right)
Ftcos30 - 4.9cos30 - 19.6cos60 - 0 = (2.0)(1.5cos30)
Ftcos 30 + 5.5564... = 2.5980...
Ftcos 30 = -2.95...
Ft = -3.41
once again, wrong.
pleeeease help!
I plugged the components into the Fnet = ma equation but it didn't work, like so:
Fnet = ma (up/down)
Ftsin30 + 19.6sin60 - 19.6 - 4.9sin30 = (2.0)(-1.5sin30)
Ftsin30 - 5.0759... = -1.5
Ftsin30 = 3.5759
Ft = 7.1518
buuut that's not right because, at the least, Ff > Ft since the box is sliding down the hill.
Fnet = ma (left/right)
Ftcos30 - 4.9cos30 - 19.6cos60 - 0 = (2.0)(1.5cos30)
Ftcos 30 + 5.5564... = 2.5980...
Ftcos 30 = -2.95...
Ft = -3.41
once again, wrong.
pleeeease help!
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