Force, friction problem. Can't start

In summary, the homework statement says that there is a problem with the equations, and the Attempt at a Solution says that the block has zero acceleration if the Static Friction is exceeded.
  • #1
1MileCrash
1,342
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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?
 
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  • #2
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
So you're basically saying that accel is sin(theta).47g?

EDIT: Actually I think it would be cosine?
 
  • #4
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
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
Okay, I can do that. So static friction plays no part??
 
  • #7
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.
 
  • #8
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
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.
 

FAQ: Force, friction problem. Can't start

1. What is force?

Force is a physical quantity that can change the state of motion or shape of an object. It is a vector quantity, meaning it has both magnitude and direction.

2. What is friction?

Friction is a force that opposes motion between two surfaces in contact. It is caused by the irregularities in the surfaces and the interlocking of the molecules on those surfaces.

3. Why do objects experience friction?

Objects experience friction because of their contact with other surfaces. The roughness of the surfaces and the strength of the intermolecular forces determine the amount of friction between two objects.

4. How does friction affect motion?

Friction acts in the opposite direction of the applied force, thus slowing down or stopping the motion of an object. It also causes objects to heat up and wear down over time.

5. What can be done to reduce friction?

Friction can be reduced by using lubricants, polishing the surfaces to make them smoother, or by using materials that have low coefficients of friction. Additionally, reducing the weight of an object can also reduce the amount of friction it experiences.

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