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Blocks on an Incline

  • Thread starter lollikey
  • Start date
33
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1. 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?


2. Homework Equations
sin = o/h
cos= a/h
tan= o/a

3. 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 !!!!!!!
 

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Last edited:

tiny-tim

Science Advisor
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hi lollikey! :smile:
3. 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 :redface:

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? :smile:
 
33
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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

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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 :confused:
 
33
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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
 
Last edited:

tiny-tim

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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 :smile:

(m will cancel out in the end)
 
33
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Fg = 9.8m

Fn= (9.8m)cos(theta)

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

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

would these be right?
 
Last edited:

tiny-tim

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Fst= (9.8m)cos(theta) * .7
no, Fst (9.8m)cos(theta) * .7 :wink:

ok, so what is the maximum angle at which there can be static equilibrium?
 
33
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would it be

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

then solve for theta?
 

tiny-tim

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would it be

(9.8m)cos(theta)*.5 ≤ (9.8m)cos(theta)*.7
but that's obviously true for any theta, isn't it? :redface:

do a static equilibrium equation​
 
33
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i don't know what that is .....
 

tiny-tim

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an equation that tells you why nothing is moving! :rolleyes:

(and i'm off to bed :zzz:)
 
33
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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!!!!)
 
435
13
You don't need anything more than those equations...
 
33
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then why did tiny-tim say i did?
 
435
13
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?
 
33
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mass*acceleration = sum of forces
 
435
13
good. when the object is not moving what is the force?
 
33
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zero
 
435
13
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.
 
33
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Fg +Fn = 0

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

cos(theta) = 1 ???
 
435
13
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.
 
33
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are any of my equations in post #7 right?
 
435
13
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.
 
33
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so would it be

Fgx = 9.8m(cos theta)

Fgy = 9.8m(sin theta)
 

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