# Pushed up ramp

1. Oct 1, 2009

### shar_p

1. The problem statement, all variables and given/known data
A box, initially at rest, is pushed up a ramp by a constant force acting 20 degrees above the ramp’s surface (pushing into the ramp from behind the box). The magnitude of the force is 50.0 N. The coefficient of friction is 0.30. The box has a mass of 5.0 kg and the ramp has an angle of 25 degrees. (a)Find the force of friction. (b)Find the net force. (c)Find the acceleration. (d)Find the final speed of the box after 6.0 s. Find how far up the ramp the box will travel in 6.0 s.

2. Relevant equations
2 different angles th1 and th2
Ff = ?, Fnet = ?
Calc Fn first then from Ff = (Mu)Fn, calc Ff, that would be (a)
Sum of Fx would be (b)
Using Sum of Fx = m. a we can get acc
using d = 1/2 a t ^ 2 (because vi = 0) we get d at 6 sec
using vf^2 = vi^2 + 2ad we can get vf
But I am not getting the correct value of Ff

3. The attempt at a solution
Sum of Fx = Fpx - Ff - Fgx = ma
Sum of Fy = Fn - Fpy - Fgy = 0
Fn = 50sin20 + 5 cos20 = 17.10
Ff = (0.3) Fn = 6.54 but the correct ans is 18 N and so everything else is going wrong.. pleaase help.

2. Oct 4, 2009

### rl.bhat

Fn = 50sin20 + 5*9.8*cos25.
Now find Ff

3. Oct 4, 2009

### Staff: Mentor

You have the wrong sign for Fpy; it acts in the same direction as the normal force.

Thus this value for Fn is wrong.

4. Oct 4, 2009

### shar_p

Thanks for the reply. rl.bhat thanks for catching my silly mistake. Glad this problem was not on the test. Though I don't agree with Doc Al that the sign -Fpy is wrong in
Sum of Fy = Fn - Fpy - Fgy = 0
because Fpy is downward because the Fp is pushed up the ramp and because of that Fn = Fpy+Fgy and with 50sin20 + 5*9.8*cos 25 I get the correct answer. Thanks

5. Oct 4, 2009

### Staff: Mentor

I see what you mean now, but your problem statement said:
I guess you meant that the force was pointing in a direction 20 degrees below the ramp surface. My bad.