Max Angle of Incline for Box Sliding w/ Friction on Ramps

[lollikey]

Homework Statement

Consider the following two scenarios with the same box sliding up two differently inclined ramps. The coefficient of static friction is 0.7, the coefficient of kinetic friction 0.5. Eventually, the box will stop. Up to what maximum angle of the incline (in degrees) would it not slide back down?

Homework Equations

sin = o/h
cos= a/h
tan= o/a

The Attempt at a Solution

cos-1 (.7/.5) = error
cos-1 (.5/.7)= 44.41 = wrong
cos-1 (.4) = 66.42= wrong
cos-1 (1) = 0 so I thought it could also equal 360 so I added 360+66.42 = 426.42 = wrong

sin-1 (.7/.5)= error
sin-1 (.5/.7) =45.58 (I think its wrong)

tan-1 (.7/.5) = 54.46 = wrong
tan-1 (.5/.7)= 35.53 = scared to try to see if its wrong

HELP PLEASE !

 

[tiny-tim]

hi lollikey!

The Attempt at a Solution

cos-1 (.7/.5) = error
cos-1 (.5/.7)= 44.41 = wrong
cos-1 (.4) = 66.42= wrong
cos-1 (1) = 0 so I thought it could also equal 360 so I added 360+66.42 = 426.42 = wrong

sin-1 (.7/.5)= error
sin-1 (.5/.7) =45.58 (I think its wrong)

tan-1 (.7/.5) = 54.46 = wrong
tan-1 (.5/.7)= 35.53 = scared to try to see if its wrong

this is silly

you're not even trying to work out the correct answer, you're just trying every answer you can think of

that's no way to learn physics, and it's no way to pass your exams!

start again …

what equation(s) do you think relevant? ​

 

[lollikey]

I KNOW! but I'm on the last question and this unit has been confusing and torturous I just want it to end :(

I first though that the reverent equations would be the kinetic friction and the static friction equation but then that doesn't really matter because there are no number

 

[tiny-tim]

I first though that the reverent equations would be the kinetic friction and the static friction equation but then that doesn't really matter because there are no number

but the numbers are .7 and .5

 

[lollikey]

yeah but that's the coefficient so i thought if i could find the normal force and all the other forces i could use them but the i saw that that was wrong so i just tried doing trig functions as you can see i have no idea what I'm doing at this point

 

[tiny-tim]

yeah but that's the coefficient so i thought if i could find the normal force and all the other forces i could use them …

yes … so call the mass "m", and carry on from there

(m will cancel out in the end)

 

[lollikey]

Fg = 9.8m

Fn= (9.8m)cos(theta)

Fk= (9.8m)cos(theta) * .5

Fst= (9.8m)cos(theta) * .7

would these be right?

 

[tiny-tim]

Fst= (9.8m)cos(theta) * .7

no, Fst ≤ (9.8m)cos(theta) * .7

ok, so what is the maximum angle at which there can be static equilibrium?

 

[lollikey]

would it be

(9.8m)cos(theta)*.5 ≤ (9.8m)cos(theta)*.7

then solve for theta?

 

[tiny-tim]

would it be

(9.8m)cos(theta)*.5 ≤ (9.8m)cos(theta)*.7

but that's obviously true for any theta, isn't it?

do a static equilibrium equation​

 

[lollikey]

i don't know what that is ...

 

[tiny-tim]

an equation that tells you why nothing is moving!

(and I'm off to bed :zzz:)

 

[lollikey]

the only equations i know that he taught us for this unit is Newtons second law, static friction and kinetic friction!(sooo regretting taking ap physics now!)

 

[Panphobia]

You don't need anything more than those equations...

 

[lollikey]

then why did tiny-tim say i did?

 

[Panphobia]

so the force of static friction is only on an object while it is not in motion, the force acted upon an object needs to be greater than this static friction in order for it to move. What does Newtons second law equation say?

 

[lollikey]

mass*acceleration = sum of forces

 

[Panphobia]

good. when the object is not moving what is the force?

 

[lollikey]

zero

 

[Panphobia]

Yes that is good, so sum up all the forces in the x direction, and keep in mind that m*a = 0 when the object is not in motion.

 

[lollikey]

Fg +Fn = 0

9.8m +(9.8m)cos(theta)= 0

cos(theta) = 1 ?

 

[Panphobia]

Not exactly, you don't add the normal force, and you haven't incorporated static friction yet. Also you can't just add mg to the forces in the x direction, because mg is neither in the x nor y direction with respect to the surface of the plane, you have to work with components.

 

[lollikey]

are any of my equations in post #7 right?

 

[Panphobia]

2,3, and 4 are correct, though #1 is correct...it is more appropriate for this question to split it into components of Fgx and Fgy.

 

[lollikey]

so would it be

Fgx = 9.8m(cos theta)

Fgy = 9.8m(sin theta)

 

[Panphobia]

think about it, if the normal force = mgcos(theta) how can Fgx = mgcos(theta)?

 

[lollikey]

i thought the fg would break up into fg in and fg out and the fg in would match up fn. I am looking at a past problem we've done but we solved for acceleration

 

[Panphobia]

ok so think about it like a triangle, the force collinear to the plane is Fgx, and the force orthogonal is Fgy, and Fn = Fgy.

 

[lollikey]

okay so fgy = 9.8mcos(theta) and fgx = 9.8msintheta)

 

[Panphobia]

correct, ok so now that you know that, sum up all the forces in the x direction again.

 

[lollikey]

the forces in the x direction are fgx and fst

9.8msin(theta) + (9.8m)cos(theta) * .7

 

[Panphobia]

fgx and fst are opposing forces, when you say "sum up the forces" you don't add the magnitudes. You have to set one direction to be positive and the other negative, so when they oppose forces one is negatively acting on the object and the other is positively acting on the object.

 

[lollikey]

this is sooo confusing! thank you soooooo much for helping me !

so would it be 9.8msin(theta) - (9.8m)cos(theta) * .7

 

[Panphobia]

yes it would be, now that you know what the sum of forces is recall what they are equal to from post 19-20.

 

[lollikey]

9.8msin(theta) - (9.8m)cos(theta) * .7 = 0

9.8msin(theta) = (9.8m)cos(theta) * .7

9.8msin(theta) /(9.8m)cos(theta)= .7

sin(theta)/cos(theta) =.7

tan(theta) = .7