Force, friction problem. Can't start

  • Thread starter Thread starter 1MileCrash
  • Start date Start date
  • Tags Tags
    Force Friction
Click For Summary

Homework Help Overview

The discussion revolves around a force and friction problem involving a block of mass m, where participants are trying to interpret the expression ".470mg" and its implications for calculating forces and accelerations in the context of static and kinetic friction.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the meaning of ".470mg" and whether it represents a force or a component of force. There is discussion about how to incorporate both static and kinetic friction into the problem-solving process. Some participants suggest breaking down the forces into horizontal and vertical components to analyze the problem further.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered guidance on how to approach the problem by considering the components of the forces involved, while others express confusion about the role of static friction and how to proceed with the calculations.

Contextual Notes

There is a lack of explicit values for mass and angles, which may affect the calculations. Participants are also navigating the constraints of homework rules and the challenge of accurately representing their work in the forum format.

1MileCrash
Messages
1,338
Reaction score
41

Homework Statement



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

Homework Equations





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:
Physics news on Phys.org
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.
 
So you're basically saying that accel is sin(theta).47g?

EDIT: Actually I think it would be cosine?
 
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?
 
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.
 
Okay, I can do that. So static friction plays no part??
 
1MileCrash said:
Okay, I can do that. So static friction plays no part??

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.
 
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.
 
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
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.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K