Can anyone help?
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- anonymous

Can anyone help?
http://oi64.tinypic.com/300fh4x.jpg

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- 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

F static = .555(37.6704)=.555(9.81*3.840)
Fkinetic = .255(37.6704)=.255(9.81*3.840)

- anonymous

I'm confused

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## More answers

- anonymous

@matt101

- 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

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

The static is strong than the applied force so it's not going to move.

- anonymous

*is stronger

- anonymous

Yep! What about kinetic?

- 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

Right

- anonymous

So N would just be the static force?

- anonymous

N as in normal force or N as in Newtons? In your blank space for your answers, the N stands for Newtons.

- anonymous

But should the answer be the number for the static force?

- 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

So for the T>0 I take the number for Fkinetic which is 9.61

- anonymous

For which scenario?

- anonymous

The first one still

- 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

Okay so they both have the static friction as the answer t=20.9 and t>20.9

- anonymous

@ganeshie8

- ganeshie8

Thats right!

- anonymous

Okay so what about the second part?

- 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

So it's still 20.9 for t=0?

- ganeshie8

Yes

- anonymous

Okay so what about the t>0

- anonymous

It obviously has to be greater than statics max.

- 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

\(F_{k} = \mu_k mg\)
simply plugin the given numbers and see

- anonymous

Fk=(.255)(3.840*9.81)

- anonymous

@ganeshie8

- anonymous

Should it be set up like that?

- ganeshie8

Yep

- anonymous

Fk=9.61

- ganeshie8

Looks good!

- anonymous

So wait the 9.61 is the answer to the t>0? What about the added force of 25.9?

- anonymous

It's counting the first two wrong. Any idea why? I entered 20.9

- 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

That's what I thought too.

- anonymous

@ganeshie8

- anonymous

@kropot72

- anonymous

@matt101

- ganeshie8

does it say the second part is right ?

- anonymous

Yes

- ganeshie8

Ofcourse it has to be wrong. Try entering below for both t = 0 and t > 0 :
12.9

- anonymous

Awesome. It took it.

- ganeshie8

Do you get why 20.9 is wrong ?

- anonymous

I think so

- anonymous

It's not moving so the static force should equal push force

- ganeshie8

please explain, id like to know how you reason it

- ganeshie8

Thats it! friction adjusts itself to the applied force.
20.9N is the "maximum" force before it gives up

- anonymous

Good

- ganeshie8

So saying \(F_{\text{static}} = \mu_s N\) is wrong. we must say :
\[F_{\text{max static}} = \mu_s N\]

- anonymous

Right. Thanks for the help!

- anonymous

@ganeshie8 That makes sense! Can't believe I overlooked that

- ganeshie8

happens haha! I have overlooked that too...

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