General

Thank you for the help, everyone~ (@ayduw, @Haishiro, @xVolcomStone, @henryumeash, @DrHye, @rockboarder, @achyif, and @Rianael)
I expected a lot less people to help so I appreciate it!

Ok, first you start with a free body diagram.
Then you do the sum of forces.
Sum of forces in the y direction = N - mg + F*sin(30) = 0

solve for N = mg+F*sin(30)

Friction = mu*N = mu*[mg+F*sin(30)]

Basically, none of the options is correct, so H

[quote=Haishiro]@Rianael: You're forgetting that the applied force has a vertical component, so the normal force is not just mg.[/quote]

Woops. Flew over my head. Let me redo the problem. So H? Okie.

[quote=Rianael]I'm in AP Physics C, but if I make mistakes or you have questions please point them out.

There is friction. Coefficient is mu. This coefficient has nothing to do with MASS. K? K. Friction = mu * F-normal force.

From the beginning. Normal force equals F-gravity so there is no y-component to those forces.

The box is being dragged however there is friction. There are only two forces acting upon it. And one(F-applied) is not directly in the x-direction, so you must use F-app*cos(30) to find the x-component. With that you can formulate the Fnet formula. F = ma.

NET FORCES = MASS * ACCELERATION.
[F*sin(30)] - [mu * F-normal force] = ma
[F*sin(30)] - [mu * mass * g] = ma

It asks for friction so lets see...

[b]It's not A[/b]. Friction is the coefficient of static/kinetic friction times the weight, F-normal force = F-FR.

[b]It's not B[/b]. The applied force overpowers the friction therefore it is able to be dragged. If it was equal it wouldn't be moving.

[b]It's definitely C[/b]. We just found friction. mu * mass * g. [b]WEIGHT IS MASS TIMES GRAVITY[/b]. It says weight not mass.

[b]Not D[/b]. Look at the Fnet formula we just made. What does mass times acceleration give you? Not friction. It gives you the F-applied minus friction! Which is the overall net of forces.

[b]It is also E[/b]. mu * m * g. MU is coefficient of friction. M*G is the weigh of the box. Multiplied together gives you friction.

[b]G doesnt even make sense.[/b] You can ask me why it doesn't if you don't get it.

[b]Not H..[/b] because we found answers.

Hope that helped or someone else already answered your question.[/quote]

C. and E. are what I thought at the beginning too. But the normal force [i]lowers[/i] when the applied force is at an angle. You can't simply multiply the mass by gravity if there's a force opposing gravity. The net force downward is what you should multiply mass by. @achyif already pointed this out in the first page.

I'm in AP Physics C, but if I make mistakes or you have questions please point them out.

There is friction. Coefficient is mu. This coefficient has nothing to do with MASS. K? K. Friction = mu * F-normal force.

From the beginning. Normal force equals F-gravity so there is no y-component to those forces.

The box is being dragged however there is friction. There are only two forces acting upon it. And one(F-applied) is not directly in the x-direction, so you must use F-app*cos(30) to find the x-component. With that you can formulate the Fnet formula. F = ma.

NET FORCES = MASS * ACCELERATION.
[F*sin(30)] - [mu * F-normal force] = ma
[F*sin(30)] - [mu * mass * g] = ma

It asks for friction so lets see...

[b]It's not A[/b]. Friction is the coefficient of static/kinetic friction times the weight, F-normal force = F-FR.

[b]It's not B[/b]. The applied force overpowers the friction therefore it is able to be dragged. If it was equal it wouldn't be moving.

[b]It's definitely C[/b]. We just found friction. mu * mass * g. [b]WEIGHT IS MASS TIMES GRAVITY[/b]. It says weight not mass.

[b]Not D[/b]. Look at the Fnet formula we just made. What does mass times acceleration give you? Not friction. It gives you the F-applied minus friction! Which is the overall net of forces.

[b]It is also E[/b]. mu * m * g. MU is coefficient of friction. M*G is the weigh of the box. Multiplied together gives you friction.

[b]G doesnt even make sense.[/b] You can ask me why it doesn't if you don't get it.

[b]Not H..[/b] because we found answers.

Hope that helped or someone else already answered your question.

EDIT: All that mess above is if there wasnt a vertical component to F-applied force... but there is LOL.

Anyway, the horizontal component of F app is F app * cos30. This adds on to the force of Fn. Fn overpowers gravity, therefore the box is lifted. And as a result everything on the list is false except H.

@henryumeash: Oh yeah, I completely forgot about what the answer said as I was writing my reply

[quote=DrHye]Because wouldn't the equation be Friction force = mu * mass * acceleration, where mu would be 0?

I could be totally off, which is why I'm expecting to be told I'm wrong (I haven't done this in a while)[/quote]
@DrHye
Well the question only says "the mass of the box multiplied by the acceleration of the box". Your friction calculations are correct, but d is still wrong since it never brought mu into the mix o.o
@DrHye, actually on second thought, Friction force shouldn't equal to mu*mass*acceleration.

Consider a 1 kg box accelerating horizontally at 9000m/s^2. If mu=0.5, then Ff=0.5*1*9.81.
The friction force is not dependent on acceleration, as achyif has said.

[quote=DrHye]Because wouldn't the equation be Friction force = mu * mass * acceleration, where mu would be 0?

I could be totally off, which is why I'm expecting to be told I'm wrong (I haven't done this in a while)[/quote]
nah. Friction force is calculated by multiplying mu by the normal force, which is not necessarily dependent on the acceleration.

[quote=henryumeash]Consider a puck on a frictionless ice surface. Applying a force accelerates this mass, but how can [friction=mass*acceleration] if [friction=0]?[/quote]

Because wouldn't the equation be Friction force = mu * mass * acceleration, where mu would be 0?

I could be totally off, which is why I'm expecting to be told I'm wrong (I haven't done this in a while)

[quote=DrHye]Can someone say why it's not D?

I think G is using the vertical component of the applied force, but you want the horizontal component, so it's not helpful. Correct me if I'm wrong[/quote]
Consider a puck on a frictionless ice surface. Applying a force accelerates this mass, but how can [friction=mass*acceleration] if [friction=0]?

not a
not b
not c
not d
not e
not f
not g
I guess h.....
the force of friction should equal to [b]mu*F(norm)[/b] and [b]F(norm)=F(grav)-F(app)*sin(30)[/b] so the force of friction is equal to [b]mu*(m*g-F(app)*sin(30))[/b]

hopefully from here you can work back and find out how to get the answer, but if you still can't just ask.

Can you quote me answers or the work?

Can someone say why it's not D?

I think G is using the vertical component of the applied force, but you want the horizontal component, so it's not helpful. Correct me if I'm wrong

Nevermind, c. and e. are talking about something else. Guess (h.) is correct.

Edit: For c. and e., I don't think it matters what angle the box is being pulled at. A higher angle = lower friction, but it also results in a lower normal force (these should balance out).

I can't say for sure, but (g) definitely looks the best, I haven't done this stuff in a while.

Just remember that you have to keep in mind that when you push on any sort of angle, that's gonna affect the force of friction. In this case when you push at an upward angle, it'll reduce it.