physics help aaaaaaaaah

[[REDACTED]]

"One block rests upon a horizontal surface with an identical second block resting on top of the first block. The coefficient of static friction between the blocks equals the coefficient of static friction between the lower block and the surface. A horizontal force is applied to the top block and when the force reaches 47.0 N the upper block begins to slide. The force is then removed and the blocks are put back in their original positions. What is the magnitude of the horizontal force that should be applied to the lower block so it just begins to slide out from under the upper block?"
so 47 N=mu*mg for the top block
that's all i know
f_{s bottom}=mu*N_bottom+mu*N_top , because we have to account for the force the top box will exert on the bottom box to oppose its motion by newton's third law?
i still don't understand the reasoning behind solving this problem someone please help

[Travis]

[spoiler]who cares[/spoiler]

[[REDACTED]]

[spoiler]who cares[/spoiler]

you

[Socks]

i only need one thrust to enter her vagina, overcoming friction from tightness

[Daddy]

whoa a school related thread

[Hippopo]

Uuuuhhhhhh....  This one is hard.  

Okay, so we know that the force to make the block move is equal to the maximum static friction force which is equal to the coefficient of friction by the normal force.

F = (fs)max = Us * Normal

F = 47.0 N

We must keep in mind that the bottom block has two surfaces of friction since it seems the top block is going to remain stationary while it moves.

Free body diagramming is always good.  Here's the bottom block:


                                          ^  N on bottom by desk
                                           |
                                           |
     f on b by t    <---    ---------
                                    |           |
                                    |           |    --------> F on bottom by hand
    f on b by d   <-----    ---------
                                      |       |
          F on b by T      v       v   W on bottom by Earth


Notice the two friction forces.  The top friction force is smaller than the bottom.  This is because the top friction force is only the coefficient for friction by the weight of the block (it's not actually the WEIGHT of the block since it's a contact force.  This is actually the normal force caused by Newton's 3rd law pair.  For simplicity I'll refer to it as a weight).

(fs)top = Us * W of top

[W of top = Normal force of top]

(fs)top = Us * Normal force

[Us * Normal force = 47.0N]

(fs)top = 47.0 N

The bottom friction is equal to the coefficient of friction by the normal force of the bottom block.  The normal force of the bottom block is equal to the W of the top block and the W of the bottom block.  Since they're identical, it's equal to 2 W or 2 Normal top.

(fs)bottom = Us * Normal bottom

(fs)bottom = Us * 2 * Normal top

(fs)bottom = 2 * (Us * N)

(fs)bottom = 2 * (47.0 N)

(fs)bottom = 94 N

We have to add the two friction forces.

94 N + 47 N = 141 N

So the answer is 141 N.  


....

Of course I'm really bad at physics, so this is probably wrong.

[the shortest route to the sea]

I totally forgot how to do force stuff, and I need to worry about what happens when this happens at .866c, so I'm afraid I can't help you.

Hippo's solution looks alright.

[[REDACTED]]

Uuuuhhhhhh....  This one is hard. 

Okay, so we know that the force to make the block move is equal to the maximum static friction force which is equal to the coefficient of friction by the normal force.

F = (fs)max = Us * Normal

F = 47.0 N

We must keep in mind that the bottom block has two surfaces of friction since it seems the top block is going to remain stationary while it moves.

Free body diagramming is always good.  Here's the bottom block:


                                           ^  N on bottom by desk
                                            |
                                            |
      f on b by t    <---    ---------
                                     |           |
                                     |           |    --------> F on bottom by hand
    f on b by d   <-----    ---------
                                       |       |
           F on b by T      v       v   W on bottom by Earth


Notice the two friction forces.  The top friction force is smaller than the bottom.  This is because the top friction force is only the coefficient for friction by the weight of the block (it's not actually the WEIGHT of the block since it's a contact force.  This is actually the normal force caused by Newton's 3rd law pair.  For simplicity I'll refer to it as a weight).

(fs)top = Us * W of top

[W of top = Normal force of top]

(fs)top = Us * Normal force

[Us * Normal force = 47.0N]

(fs)top = 47.0 N

The bottom friction is equal to the coefficient of friction by the normal force of the bottom block.  The normal force of the bottom block is equal to the W of the top block and the W of the bottom block.  Since they're identical, it's equal to 2 W or 2 Normal top.

(fs)bottom = Us * Normal bottom

(fs)bottom = Us * 2 * Normal top

(fs)bottom = 2 * (Us * N)

(fs)bottom = 2 * (47.0 N)

(fs)bottom = 94 N

We have to add the two friction forces.

94 N + 47 N = 141 N

So the answer is 141 N. 


....

Of course I'm really bad at physics, so this is probably wrong.

yeah hippo, it's the correct answer.
i just need an explanation of all the forces you drew on your diagram :x because "f on b by d" confused me

[Travis]

why does math elude me so

[Hippopo]

yeah hippo, it's the correct answer.
i just need an explanation of all the forces you drew on your diagram :x because "f on b by d" confused me

Oh, my diagram is a bit confusing.  

"f on b by d" means "friction on bottom block by desk."

N = Normal force
f = friction
b = bottom block
T or t = top block
F = force


Sorry about that. :P

Remember that the net force of the bottom block is going to equal zero.  That means both the friction forces will equal the applied force.  The weight of the bottom block and the force exerted downward by the top block added together will equal the normal force of the bottom block.

Do you get it? D:

[Walter]

I'll give you a hint. You can ignore mu.

[spoiler]Better hint: hippo is right. All you need to think about is the combined forces of friction, and the one between the top and bottom is already given as 47. Thats like half the problem done for you right there. Then we know they are identical mass and identical mu, so we know it will take twice the amount of force between the floor and bottom box. 2 times 47 is 94. 94 + 47 = 141.

Now if you wanted to move both of the boxes without them sliding apart, the force applied to the bottom box would have to be greater than or equal to 94, but less than 141.[/spoiler]

[The artist formally known]

newton's 2nd bitches

[Minus;]

Im alright in physics don't want to work that out though...
Plus i have my own physics work that i have to do that i don't want to.

I have to find the moment of intertia of the neuton star at the center of the crab nebula.

[The artist formally known]

Im alright in physics don't want to work that out though...
Plus i have my own physics work that i have to do that i don't want to.

I have to find the moment of intertia of the neuton star at the center of the crab nebula.

seems pretty easy, just finding the measurements for that star and the crab nebula is pretty shitty.

working on gaussian proofs on my end. my teacher makes us do it all from scratch, just use gauss' law and some other shit.

[Minus;]

we can't use measurements.
they gave us how much power the star outputs...
and how fast its spinning and how fast it descreases per second in seconds per second haha.
and we have to find its moment of intertia with that
but i am already almost completely done, just needa plug it in to the calc, but i left it at school so ill do that tomorrow