Forces and Motion - Sliding a Box Up an Inclined Plane w/ Friction

[Scimitar]

Homework Statement

You are helping a friend move, and you need to load a 2.65x10^2 kg box of books. You slide the box up a ramp, which has an incline of 30 degrees and a coefficient of static friction of 0.45. You apply the force on the box at an angle of 39 degrees with respect to the ramp.

Calculate the minimum force needed to slide the box up the ramp.

Homework Equations

Net force = ma

The Attempt at a Solution

I'm not too sure how to draw the FBD for this situation since the static friction may be working in the same direction as the Fapp since the box would slide downwards if no Fapp was affecting it. But I don't really know.

 

[tiny-tim]

Hi Scimitar! Welcome to PF!

… the static friction may be working in the same direction as the Fapp since the box would slide downwards if no Fapp was affecting it. But I don't really know.

Ah, I see what you mean …

before anyone pushes, the static friction is pointing up, since it's stopping the box from sliding down.

You have to remember that static friction is not equal to µN, it is ≤ µN.

In fact, it's a "left-over" force that you have to calculate without using µ at all … it's the force that balances all the other forces.

So when you start to pull the box, the static friction gets less, and eventually it becomes 0.

Then it gets bigger, but down the ramp, until it equals µN.

After that point, it's dynamic friction, and so it is exactly µN (different µ of course ).

When you draw the FBD, you can forget about the history, of the static friction changing direction, and just apply the rule that the friction opposes the motion, which by definition (ie from what the question says) is that the box is about to move upwards​

 

[akaay123]

I was wondering how you solve the problem. And also what the answer is. (what equations to use and how to arrange them)

 

[haruspex]

I was wondering how you solve the problem. And also what the answer is. (what equations to use and how to arrange them)

Then you need to take the role of the original poster, i.e. post an attempt.

 

[ava ray]

This is what I tried, but I did not get the right answer.

fnetx=max=facos39-Usfn=0
fnety=may=(fn+fasin39)-fgcos30=0

fn+((Usfnsin39)/cos39)-fgcos30=0
fn((1+Ussin39)/cos39)-fgcos30=0
fn=1362.1N

ffs=fax

Usfn=fax
fax=612.945N

fa=fax/cos39
fa=788.7N

but the correct answer is 2.1x10^3N
can someone tell me where I went wrong?
I'm thinking maybe my assumption that Ffs=Fax is incorrect, but I don't know what to do if it is incorrect.

 

[haruspex]

fnetx=max=facos39-Usfn=0

What about gravity?

 

[ava ray]

Does gravity play a factor in the x- direction?
I have aligned my x-axis with the 30 degree incline to simplify.

 

[haruspex]

Does gravity play a factor in the x- direction?
I have aligned my x-axis with the 30 degree incline to simplify.

If you had taken x as horizontal then gravity would not have had an x component, but as you say you have aligned x parallel to the ramp, so gravity does have an x component.

With more experience, it would strike you as wrong that you have an equation with g cos(θ) but no equation with g sin(θ).

 

[ava ray]

These are my revised fnet desriptions:

facos39-Usfn-fgsin30=0
fasin39+fn-fgcos30=0

how should i go about solving for fn?

 

[haruspex]

These are my revised fnet desriptions:

facos39-Usfn-fgsin30=0
fasin39+fn-fgcos30=0

how should i go about solving for fn?

You don't need to solve for Fn; you want to eliminate Fn so as to solve for Fa.
Standard procedure: get one equation into the form Fn=... and use that to replace Fn in the other equation.

 

[ava ray]

It finally worked. Thanks!