Question

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

Answer 1

\[\huge F_{\text{static friction}}=\mu_{s} F_{\text{normal}}= \mu_s mg\]
\[\huge F_{\text{kinetic friction}}= \mu_{k} F_{\text{normal}}= \mu_k mg\]
You'll notice that the frictional force only depends on the object's weight and not on the force being applied

Answer 2

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

Answer 3

I'm confused

Answer 4

@matt101

Answer 5

`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.

Answer 6

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?

Answer 7

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

Answer 8

*is stronger

Answer 9

Yep! What about kinetic?

Answer 10

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.

Answer 11

Right

Answer 12

So N would just be the static force?

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

For which scenario?

Answer 18

The first one still

Answer 19

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.`

Answer 20

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

Answer 21

@ganeshie8

Answer 22

Thats right!

Answer 23

Okay so what about the second part?

Answer 24

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".

Answer 25

So it's still 20.9 for t=0?

Answer 26

Yes

Answer 27

Okay so what about the t>0

Answer 28

It obviously has to be greater than statics max.

Answer 29

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

Answer 30

\(F_{k} = \mu_k mg\)

simply plugin the given numbers and see

Answer 31

Fk=(.255)(3.840*9.81)

Answer 32

@ganeshie8

Answer 33

Should it be set up like that?

Answer 34

Yep

Answer 35

Fk=9.61

Answer 36

Looks good!

Answer 37

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

Answer 38

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

Answer 39

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..

Answer 40

That's what I thought too.

Answer 41

@ganeshie8

Answer 42

@kropot72

Answer 43

@matt101

Answer 44

does it say the second part is right ?

Answer 45

Yes

Answer 46

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

12.9

Answer 47

Awesome. It took it.

Answer 48

Do you get why 20.9 is wrong ?

Answer 49

I think so

Answer 50

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

Answer 51

please explain, id like to know how you reason it

Answer 52

Thats it! friction adjusts itself to the applied force.

20.9N is the "maximum" force before it gives up

Answer 53

Good

Answer 54

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

\[F_{\text{max static}} = \mu_s N\]

Answer 55

Right. Thanks for the help!

Answer 56

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