can someone show me how to do this problem?

[PV Audio 5,000+ posts]

i'm dying over here.

to move a large crate across a rough floor, you push down on it at an angle of 21 degrees. find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is .57

this is my work so far

EFy = ma

Fn - mg sin theta = 0

Fn - (32)(9.8) sin 21 = 0

Fn = 262.38

friciton work...

fs = .57(262.38)

fs = 144.55

i don't know what else to do... i need to find the force needed to overcome static friction at 21 degrees theta. heeeelp!

-Dave

no retarded posts here please.

Click to read more...

 

[Gauntlet 5,000+ posts]

Fu*k it in the mouth.

 

[maldecido33 10+ year member]

beginning HS physics? we started forces w/ friction today.

however, i have a retarded teacher, so i don't know what to do because he can't teach.

 

[Acidburn 5,000+ posts]

find the force necesary if you were just pushing it horizontally, and then use trig to find how much force is needed at 21*

 

[mattf 5,000+ posts]

I could have told you last year while I was in physics, but now...nothing.

 

[bigluis 10+ year member]

im learning that and im in 8th grade lol

 

[sumone 10+ year member]

I could have told you last year while I was in physics, but now...nothing.

x2...

But let me take a stab at it (if it's not right, whatever):

Ey = Normal - Force_of_Object - Vertical_Applied_Force = ma => 0 since a = 0

Ey = F_Normal - mg - F * sin(21) = ma = 0

Ey = F_Normal - (32 * 9.8) - F * sin(21) = 0

F_Normal = (32 * 9.8) + F * sin(21)

F_Normal = 313.6 + F * sin(21)

Ex = Horizontal_Applied_Force - Friction*Normal = ma => 0 since a = 0

Ex = F * cos(21) - u * F_Normal = ma = 0

Ex = F * cos(21) - 0.57 ( F_Normal ) = 0

Use Ex & Ey as a system of equations to find F but substituting F_Normal into Ex:

Ex = F * cos(21) - 0.57 ( 313.6 + F * sin(21)) = 0

F * cos(21) - 178.52 - 0.57 * F * sin(21) = 0

F (cos(21) - 0.57 * sin(21)) = 178.52

F = 178.52 / (cos(21) - 0.57*sin(21))

F = 244.779

So that's my guess...244.779 Newtons

 

[infamousZ71 10+ year member]

I wish I could help you. All I do in Physics is start at the board and laugh at my teacher's diagrams. Im passing with a B somehow so until I drop a lettergrade, Im sticking with this strategy.

 

[sumone 10+ year member]

im learning that and im in 8th grade lol

Well, why don't you offer some input....

 

[Decado 10+ year member]

find the force necesary if you were just pushing it horizontally, and then use trig to find how much force is needed at 21*

Ohh dear god...trig //content.invisioncic.com/y282845/emoticons/hide.gif.2d479cfd917eedfe201353b91522ceab.gif . I never could understand trig.

 

[JAZN 5,000+ posts]

x2...
But let me take a stab at it (if it's not right, whatever):

Ey = Normal - Force_of_Object - Vertical_Applied_Force = ma => 0 since a = 0

Ey = F_Normal - mg - F * sin(21) = ma = 0

Ey = F_Normal - (32 * 9.8) - F * sin(21) = 0

F_Normal = (32 * 9.8) + F * sin(21)

F_Normal = 313.6 + F * sin(21)

Ex = Horizontal_Applied_Force - Friction*Normal = ma => 0 since a = 0

Ex = F * cos(21) - u * F_Normal = ma = 0

Ex = F * cos(21) - 0.57 ( F_Normal ) = 0

Use Ex & Ey as a system of equations to find F but substituting F_Normal into Ex:

Ex = F * cos(21) - 0.57 ( 313.6 + F * sin(21)) = 0

F * cos(21) - 178.52 - 0.57 * F * sin(21) = 0

F (cos(21) - 0.57 * sin(21)) = 178.52

F = 178.52 / (cos(21) - 0.57*sin(21))

F = 244.779

So that's my guess...244.779 Newtons

That's what I got.

I'm actually learning this same exact shit as a 3rd year in college right now //content.invisioncic.com/y282845/emoticons/tongue.gif.6130eb82179565f6db8d26d6001dcd24.gif

 

[Beat_Dominator 5,000+ posts]

Oh! Now do it in 3 dimensions. Let's say the ramp is tilted at an angle of 3 degrees to the left so the crate doesn't slide down the middle.

Time to start using gradient vectors and Stokes Theorem!!! //content.invisioncic.com/y282845/emoticons/veryhappy.gif.fec4fed33b4a1279cf10bdd45a039dae.gif

 

[JAZN 5,000+ posts]

There are no such things as tilted ramps. Thats devil talk.

 

[DiamondFanatic 10+ year member]

Hooray for AP Physics (thats what I'm taking this year too) Yeah Lemans, for that problem all you need really is trig once you find the normal force/ friction force

 

[Beat_Dominator 5,000+ posts]

There is no ramp? Is this the matrix!?