Question

Pick the right answer?
A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it.
A slight pull P is exerted on the crate, not enough to move it. A force of friction now acts:
a) which is ... less than, equal to, OR greater than ... P?
b) Net force on the crate is ... zero, OR greater than zero?
---
My answers:
Greater than; greater than zero (CORRECTION: zero. THX to @AllTehMaffs for correcting!!)

Answer 1

@AllTehMaffs I take you are looking at this? >.>

Answer 2

hallo - was afk, sorry.

Answer 3

Your first one is right.

For the second one:
If the crate isn't moving, what is the acceleration of the crate?

Answer 4

Meh -_-

Answer 5

stupid comment lag.
Zero.

Answer 6

and if the acceleration of the crate is zero, what's the net force on the crate?

Answer 7

Zero?
How, though?

Answer 8

The net force is defined as the acceleration of the crate times is mass - everything that you sum up in between the
\[F_net=................=ma\]
is what equals the acceleration of the crate - and if we know for sure that the create isn't moving, we know that all that junk in the middle cancels each other out.

We already know that the force of friction is greater than the force of the pull; and the friction can't pull back any *harder than the pull. So that means the pull gets you nowhere, and the friction gets you nowhere, so you stay stationary.

Answer 9

~
\[F_{net}\]

Answer 10

I got it, but if f doesn't equal Fa how does it make a net force of 0?

Answer 11

It does equal it - it equals **exactly** what the force of the pull is while

\[\mu_{static} N ≥ Fa\]
And while that inequality remaines true, the box is stationary

Answer 12

Unless there's a Wx in it, which makes no sense since it's on a flat plane and the pull isn't at an angle either...
Wait, what?
\[\mu_s N \ge F_a\]I don't get it. We never use that sign o_o;;

Answer 13

≥ means "greater than or equal to"

And correct - there is no x-component of the weight; and if there were, the force of static friction would not equal the force of the pull.

Answer 14

So it's greater than 0, but still works out?

Answer 15

Fp is greater than zero, and f is less than zero (because it's pulling in the opposite direction)

Answer 16

the inequality is talking about their magnitudes, sorry

Answer 17

You can basically treat the friction like glue

As long as you don't pull hard enough to break the bond the glue makes between the floor and the crate, the net force on the crate is zero, because it's not moving.

Answer 18

I get that \[F_a > 0\]\[f < 0\]but are they both equal & opposite? c:

Answer 19

yup yup!!!

Answer 20

o_o that contradicts the first answer

Answer 21

\[\text{max force of static friction }= \mu_s N\]

So if the force of the pull is less than mu_s N, the magnitude of the force of friction is just equal to the magnitude of the force of the pull; which means that they cancel each other out.

Answer 22

The force of static friction doesn't exist without the pull.

Answer 23

Wait, is it less because one is positive and one is negative?

Answer 24

What do you mean by "is it less?"

Answer 25

The pull/Fa, I mean.

Answer 26

I mean, the best explanation I can give is that if the box isn't moving and it's on a flat surface, the force of **static friction cancels out the force of the pull - if the box is moving, then the force of **kinetic friction slows down the acceleration of the box caused by the pull.

Answer 27

Ah, so there's two types of friction... WAIT no, there's only one..?

Answer 28

I mean - there's one per type of free-body

Answer 29

yah yah = ** kinetic friction is what acts on the box when it is moving

**static friction is what acts on the box when it's not moving.

Answer 30

the coefficient of kinetic friction is less than the coefficient of static friction

\[\mu_k<\mu_s\]

Answer 31

Okay so how do they cancel out when f is greater than P, per the first answer?

Answer 32

How? The box sticks to the ground, pretty much literally.

Answer 33

wat .-.
I don't get it.

Answer 34

Well, I know it sticks to the ground b/c N and W equal each other, but that doesn't explain how f > P and yet net force is 0...

Answer 35

\[\mu_s N\] is the maximum value for static friction.

If you have a box that weights 1000N (g=10m/s ; m=100kg) that has a coefficient of static friction =.5, and a coefficient of kinetic friction of .3
\[f_{static \ max}=\mu_s N = 500N\]
\[f_{kinetic}= \mu_k N = 300N\]
If you pull on it with a force of 10N, the force of static friction is 10N, and the box doesn't move. If you pull on it with a force of 50N, the force of static friction is 50N, and the box doesn't move. If you pull on it with a force of 437 N, the force of static friction is 437 N, and the box doesn't move. If you pull on it with a force of 500N, the force of static friction is 500N, and the box doesn't move.

If you pull on it with a force of 500.1N, the box is no longer stationary; it has overcome the force of **static friction, and now has a net force on it that equals
\[ F_p - f_{kinetic}=ma\]
\[500.1N - 300N = ma\]
\[ 200.1N = ma\]
so experiences an acceleration of
\[a=\frac{200.1N}{100kg}=2.001 m/s^2\]

Answer 36

eh, ok.
Holdup while I process info

Answer 37

I gots to go snooze. If some of that doesn't make sense I'll be back on later in the morn ^_^