Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Spehss _]

Well, you're substantially overthinking this. There are only 3 forces at play: gravity, the force pushing it up the slope, and the normal force. Instead of trying to change the coordinates to be relative to the ground, why not have them be relative to the force pushing the box? In that case, gravity becomes two forces- the force pushing it into the slope (the normal force) and the force pushing against the force pushing the box up the slope. M*g*cos(60) = F, M*g*sin(60) = N

In various example problems involving ramps, the diagram was always drawn relative to the angle of the ramp, so that the ramp is the x plane. Both the examples given by my professor and the (few) examples in the textbook do this. So I blame that.

I don't see how gravity "becomes two forces" in your scenario. If F is treated as the x-axis, then W (m*g) is pulling directly downwards. But Fn isn't pushing directly up, because Fn is the force of the ramp pushing back on the box as the box pushes down on the ramp, so Fn is at an incline since the ramp is at an incline.

Where do you get 60 degrees from? The angle of Fn relative to the x-axis?

[hops]

Only the vertical component of Fn push vertically, that's why things slide down ramps.

[Spehss _]

Well, it's because we're taking Fn to be "up." So, gravity's coming down at an angle. It's just some simple trig after that.

...that sounds like what I was doing in my first post. By setting the x plane to be parallel to the ramp, force of gravity is at an angle, and the force F is at an angle. Use trig to break the two forces into x and y components. They components of gravity and F are both pushing down the Y axis, so Fn, which is directed up the y-axis in response to Wy and Fy pushing down, is equal in magnitude to Fy+Wy.

The problem is I don't know what F is and therefore don't know what the components of F is. I thought that F would have to be equal to the weight of the box in order for the net force to become zero but according to grading that's wrong.

[Spehss _]

I mentioned F was horizontal in the first line that was setting the problem up. Probably should've specified it was perpendicular to gravity, yeah. Or include a picture. But posting an image would be a hassle.

Fnet has to be zero for the speed to be constant and acceleration to be zero, right? So magnitude of F has to be equal to the force applied by gravity, right?

[Amperzand]

₁₀P_r

Dammit I was taught this in class but it's gone now halp.

[TheBiggerFish]

What is that even supposed to be?
(10^P)r, or something else, and what are you trying to do with it?
Do the variables stand for anything?

[Amperzand]

I don't know what the name for it is. Has to do with factorials and permutations. The equation for it is ₁₀P_r = 10!/(10-r)!

While I can calculate them without the variables involved, I seem to have shenanigan'd myself with business over the weekend, and forgotten how you handle variables, since I seem to be intended to solve down to an integer, and the textbook says nothing.

[TheBiggerFish]

Ohhhh.


I see.


_{I have no idea what you're doing.}

[frostshotgg]

That seems like a specific case of permutations. General form for permutations is nPr, where n is the number of choices you have and r is the number you choose from it.

What questions do you have about it?

[Amperzand]

Largely standard, with some division and stuff. This one just didn't provide r. Having done some research, it apparently just wants 10!/(10-r)!, which is weird and I swear that's not what the teacher wanted, but whatever.

_{I have no idea what you're doing.}

Precalculus. :V

[TheBiggerFish]

Buhuhuwha?  How is that precalc I don't know what it is.

[Amperzand]

College precalc? Factorials?

[TheBiggerFish]

I mean I know what factorials are, but I haven't had a factorial with a variable in it ever...

[Amperzand]

Apparently the variable was weirdness and it went away. I'm equally confused.

[TheBiggerFish]

Okay then.

*still confused, but less so*