What Applied Force is Needed to Push a Box Up a Ramp at 50 Degrees?

[bratleroni]

Homework Statement

The table (below) shows the force needed to push a 10-kg box up a ramp as the angles increase. From the trend in the table, what is the applied force needed if the angle of the ramp increases to 50 degrees?

Applied Force (N) ; Ramp Angle (Degrees)
19N ; 10 degrees
30N ; 20 degrees
40N ; 30 degrees
49N ; 40 degrees

Homework Equations

applied force = mg sin(degree of angle)

The Attempt at a Solution

I honestly don't know about this, and I'm not really looking for the answer to the problem, just the "how-to". I tried, with the equation above, to see if it worked for the 19N ; 10 degrees (in the table). My work looked like:

applied force = (10kg)(9.81)sin(10) = 17.0348...

This obviously isn't the 19N that I was supposed to get. I can't find any decent explanations for this problem, and would appreciate any help.

 

[PhanthomJay]

Well, there could be friction acting, but something is wrong with the problem data. For example, when theta = 30 degrees, then you need at least 50 N to get it to move, but the table indicates 40 N.

 

[bratleroni]

Well, there could be friction acting, but something is wrong with the problem data.

Alright well, that was all the info I was given for the problem, so in that case I'm assuming my teacher transposed the numbers or something? Other than that, was I at least correct in what formula to use in that type of problem, and in the execution? Its just that I really don't know how to do this, and would at least like to know if I was going about it the right way, even if the data provided is wrong.

 

[Kabbotta]

I agree, there seems to be something wrong with the problem statement.

If you are sure there is no friction then you have exactly the right setup and you know the data is off, but even if you try to account for friction the numbers given produce a coefficient of kinetic friction that is not constant.

With friction,

F_{push} = f_{k} + F_{g}Sin(\theta)

So that extra friction force can be set up to give you the 19 N. at 10 degrees, but then the rest of the data needs a different coefficient of friction to produce the stated forces. If the data were correct, the coefficient should stay constant and allow you to predict the force needed at 50 degrees.

That's all I got.