anonymous
  • anonymous
Can anyone help? http://oi64.tinypic.com/300fh4x.jpg
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1447301503241:dw| \[\huge F_{\text{static friction}}=\mu_{s} F_{\text{normal}}= \mu_s mg\] \[\huge F_{\text{kinetic friction}}= \mu_{k} F_{\text{normal}}= \mu_k mg\] You'll notice that the frictional force only depends on the object's weight and not on the force being applied
anonymous
  • anonymous
F static = .555(37.6704)=.555(9.81*3.840) Fkinetic = .255(37.6704)=.255(9.81*3.840)
anonymous
  • anonymous
I'm confused

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anonymous
  • anonymous
@matt101
anonymous
  • anonymous
`F static = .555(37.6704)=.555(9.81*3.840)` `Fkinetic = .255(37.6704)=.255(9.81*3.840)` F static is the friction on the block before the box moves and F kinetic is the friction on the box while it is moving. It looks good to me.
matt101
  • matt101
Well so far you're on the right track. You've calculated the maximum force of static friction and the maximum force of kinetic friction that can be applied to the block. What you need to do now is determine how the applied force compares to the frictional force. Remember that the applied force needs to first overcome the static friction force for the box to start moving, so we need to consider static friction first and then kinetic friction. In the first case, the applied force is 12.9 N. How does this compare to the force of static friction you calculated?
anonymous
  • anonymous
The static is strong than the applied force so it's not going to move.
anonymous
  • anonymous
*is stronger
anonymous
  • anonymous
Yep! What about kinetic?
anonymous
  • anonymous
Actually, since it can't overcome the force of static friction, you won't have kinetic friction since it won't be moving. The answers to both parts to the first question should be the same I believe.
anonymous
  • anonymous
Right
anonymous
  • anonymous
So N would just be the static force?
anonymous
  • anonymous
N as in normal force or N as in Newtons? In your blank space for your answers, the N stands for Newtons.
anonymous
  • anonymous
But should the answer be the number for the static force?
anonymous
  • anonymous
At t=0 for both scenarios you have static friction. In the first scenario, the static friction is greater than the force being applied, so for t>0, the box will not move, thus you will only have static force. In the second scenario, I think the static force is less than the applied force, therefore it will move and now for t>0, the box will have kinetic friction. And yes, it's asking about what is the frictional force being applied during those times.
anonymous
  • anonymous
So for the T>0 I take the number for Fkinetic which is 9.61
anonymous
  • anonymous
For which scenario?
anonymous
  • anonymous
The first one still
anonymous
  • anonymous
No -- like I said before `At t=0 for both scenarios you have static friction. In the first scenario, the static friction is greater than the force being applied, so for t>0, the box will not move, thus you will only have static frictional force.`
anonymous
  • anonymous
Okay so they both have the static friction as the answer t=20.9 and t>20.9
anonymous
  • anonymous
@ganeshie8
ganeshie8
  • ganeshie8
Thats right!
anonymous
  • anonymous
Okay so what about the second part?
ganeshie8
  • ganeshie8
For second part, the moment you apply the force, the static friction opposes it. At t=0, the block is not moving, it is "just" about to move, so the frictional force is "static".
anonymous
  • anonymous
So it's still 20.9 for t=0?
ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
Okay so what about the t>0
anonymous
  • anonymous
It obviously has to be greater than statics max.
ganeshie8
  • ganeshie8
It is difficult to move a large stationary object. But once it starts moving, it is easy to keep it going.. so we expect the kinetic frictional force to be less than the static
ganeshie8
  • ganeshie8
\(F_{k} = \mu_k mg\) simply plugin the given numbers and see
anonymous
  • anonymous
Fk=(.255)(3.840*9.81)
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
Should it be set up like that?
ganeshie8
  • ganeshie8
Yep
anonymous
  • anonymous
Fk=9.61
ganeshie8
  • ganeshie8
Looks good!
anonymous
  • anonymous
So wait the 9.61 is the answer to the t>0? What about the added force of 25.9?
anonymous
  • anonymous
It's counting the first two wrong. Any idea why? I entered 20.9
anonymous
  • anonymous
I'm not really sure I understand why it would count the first two wrong. The friction by the table onto the box is followed by the equation that we stated from before. If we plug everything in, we seen that we should get 20.9, which is greater than the force being applied, which is 12.9. Therefore, the box will not move for t=0 and t>0. Both answers should be 20.9..
anonymous
  • anonymous
That's what I thought too.
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@kropot72
anonymous
  • anonymous
@matt101
ganeshie8
  • ganeshie8
does it say the second part is right ?
anonymous
  • anonymous
Yes
ganeshie8
  • ganeshie8
Ofcourse it has to be wrong. Try entering below for both t = 0 and t > 0 : 12.9
anonymous
  • anonymous
Awesome. It took it.
ganeshie8
  • ganeshie8
Do you get why 20.9 is wrong ?
anonymous
  • anonymous
I think so
anonymous
  • anonymous
It's not moving so the static force should equal push force
ganeshie8
  • ganeshie8
please explain, id like to know how you reason it
ganeshie8
  • ganeshie8
Thats it! friction adjusts itself to the applied force. 20.9N is the "maximum" force before it gives up
anonymous
  • anonymous
Good
ganeshie8
  • ganeshie8
So saying \(F_{\text{static}} = \mu_s N\) is wrong. we must say : \[F_{\text{max static}} = \mu_s N\]
anonymous
  • anonymous
Right. Thanks for the help!
anonymous
  • anonymous
@ganeshie8 That makes sense! Can't believe I overlooked that
ganeshie8
  • ganeshie8
happens haha! I have overlooked that too...

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