# Force, friction problem. Can't start

1. Sep 14, 2011

### 1MileCrash

1. The problem statement, all variables and given/known data

[PLAIN]http://img718.imageshack.us/img718/5747/wtfproblem.png [Broken]

2. Relevant equations

3. The attempt at a solution

I can't start because I have no idea what the heck ".470mg" means. I thought .470 times accel due to gravity, but that doesn't make sense, because then what is m? Or is it just .470mg as in .470 milligrams? :P

F = ma
F = .470(m)(9.8)

That makes no sense because it just wops the equation. How could the force equal .470 times the mass, THEN times the acceleration? That's just wrong.

So, what the hell are they talking about??? What is the *actual* force in, ya'know, newtons/a unit of force/something sensible?

Last edited by a moderator: May 5, 2017
2. Sep 14, 2011

### Delphi51

I think .470mg means .47 times the mass times 9.81.
That makes sense, since the mass isn't given and the m has to cancel out in the calculation.

3. Sep 14, 2011

### 1MileCrash

So you're basically saying that accel is sin(theta).47g?

EDIT: Actually I think it would be cosine?

4. Sep 14, 2011

### 1MileCrash

Changed my mind, I have no idea how to do this. I don't know how to incorporate both a kinetic and static friction. What should I do first?

5. Sep 14, 2011

### Delphi51

Begin by finding both horizontal and vertical components of .47mg.
The vertical part will affect the normal force, so be sure to include it when calculating the friction force. Use .47mg*cos(θ) - Ff as the net force horizontally when finding the acceleration.

6. Sep 15, 2011

### 1MileCrash

Okay, I can do that. So static friction plays no part??

7. Sep 15, 2011

### PeterO

If the horizontal component of the applied force is smaller than static friction, the block will have zero acceleration.
If the Static friction is exceeded, then the dynamic friction is all that opposes the accelerating force, so we use the effects of that when calculating the acceleration.

8. Sep 15, 2011

### SteamKing

Staff Emeritus
The problem clearly states, "a block of mass m". The Force F shown in the diagram has a magnitude of 0.47 times the mass m of the box times the acceleration due to gravity g, or in algebraic terms |F| = 0.47mg.

9. Sep 15, 2011

### 1MileCrash

I worked the problem like six times today and simply cannot get it. I will post a picture of my work & notes, I latexed the whole thing but the forum won't keep me logged in while typing all of it.. not doing that again.

10. Sep 15, 2011

### 1MileCrash

Nevermind, I got it this time. I need to compare it to my previous work to see what I did differently, but I think I didn't distribute u correctly.