Friction on an inclined plane

  • #1
5
0

Homework Statement


Given an inclined angle of θ = 30° and 10m long. The coefficient of friction is 0,1 and the mas of the block is 1kg.
a) Find the velocity needed for the block to climb the plane and its velocity be 0 in the top.
b) Find the time it took to reach the top.
c) Find the speed of the block when it reaches the bottom after its descent.

I solved a) and c) already (10,7 m/s and 9m/s)

Homework Equations



For a) and c) : Wr + Ec1 + Ep1 = Ec2 + Ep2 ===> u.m.g.cos θ.x + 1/2 m v1^2 + mgh1 = 1/2 m v2^2 + mgh2
For b) : Fr = uN ; Fr = ma ; fv = iv + a.t

The Attempt at a Solution



I think i've got to find the acceleration first. But i'm probably messing the equation. Supposedly the friction force (u.N) is equal to mass + acceleration, but im having second thoughts about if i have to add the weight projected on the x plane (m.g.cos ) ?

∑F = m.a
Fr = m.a
u.m.g.cos210 = m.a
u.g.cos210 = a
-0.84m/s^2 = a <===== Thats wrong i guess.

What am i missing?

Then i should use :

fv = iv + a.t
0 = 10.7m/s + a.t
(-10.7m/s)/a = t
 

Answers and Replies

  • #2
907
88
Why do you guess it's wrong?
 
  • #3
5
0
We are given the answers, and the time for this item is 1,87s . A number i cant arrive.
 
  • #4
569
85
I can get 9m/s for part c), but why isn't the answer to part a) the same?

Edit: If you want to get 1.87s, solve a.t=10.7m/s and 10m=0.5a.t^2.
 
Last edited:
  • #5
nrqed
Science Advisor
Homework Helper
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279

Homework Statement


Given an inclined angle of θ = 30° and 10m long. The coefficient of friction is 0,1 and the mas of the block is 1kg.
a) Find the velocity needed for the block to climb the plane and its velocity be 0 in the top.
b) Find the time it took to reach the top.
c) Find the speed of the block when it reaches the bottom after its descent.

I solved a) and c) already (10,7 m/s and 9m/s)

Homework Equations



For a) and c) : Wr + Ec1 + Ep1 = Ec2 + Ep2 ===> u.m.g.cos θ.x + 1/2 m v1^2 + mgh1 = 1/2 m v2^2 + mgh2
For b) : Fr = uN ; Fr = ma ; fv = iv + a.t

The Attempt at a Solution



I think i've got to find the acceleration first. But i'm probably messing the equation. Supposedly the friction force (u.N) is equal to mass + acceleration, but im having second thoughts about if i have to add the weight projected on the x plane (m.g.cos ) ?

∑F = m.a
Fr = m.a
u.m.g.cos210 = m.a
u.g.cos210 = a
-0.84m/s^2 = a <===== Thats wrong i guess.

What am i missing?

Then i should use :

fv = iv + a.t
0 = 10.7m/s + a.t
(-10.7m/s)/a = t
Draw a free body diagram. It should be clear that, as you guessed, the friction force is not the only force along the inclined. There is a component of the force of gravity (equal to mg sin(theta) if theta is the angle of the inclined with respect to the horizontal).
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement


Given an inclined angle of θ = 30° and 10m long. The coefficient of friction is 0,1 and the mas of the block is 1kg.
a) Find the velocity needed for the block to climb the plane and its velocity be 0 in the top.
b) Find the time it took to reach the top.
c) Find the speed of the block when it reaches the bottom after its descent.

I solved a) and c) already (10,7 m/s and 9m/s)

Homework Equations



For a) and c) : Wr + Ec1 + Ep1 = Ec2 + Ep2 ===> u.m.g.cos θ.x + 1/2 m v1^2 + mgh1 = 1/2 m v2^2 + mgh2
Regarding the above line:
It would help you to get the solution if you can explain what all those terms mean and especially explain how the second equation comes about.

For b) : Fr = uN ; Fr = ma ; fv = iv + a.t

The Attempt at a Solution



I think i've got to find the acceleration first. But i'm probably messing the equation. Supposedly the friction force (u.N) is equal to mass + acceleration, but im having second thoughts about if i have to add the weight projected on the x plane (m.g.cos ) ?

∑F = m.a
Fr = m.a
u.m.g.cos210 = m.a
u.g.cos210 = a
-0.84m/s^2 = a <===== That's wrong i guess.

What am i missing?

Then i should use :

fv = iv + a.t
0 = 10.7m/s + a.t
(-10.7m/s)/a = t
Did you find the acceleration for going up the ramp or for coming down the ramp?
 
  • #7
SammyS
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I can get 9m/s for part c), but why isn't the answer to part a) the same?

Edit: If you want to get 1.87s, solve a.t=10.7m/s and 10m=0.5a.t^2.
@insightful ,

Why should answers (a) and (c) be the same?

What are you using for acceleration?
 
  • #8
569
85
Ooops, never mind. I forgot the friction force is in the opposite direction when going uphill (red face).
 
  • #9
5
0
Draw a free body diagram. It should be clear that, as you guessed, the friction force is not the only force along the inclined. There is a component of the force of gravity (equal to mg sin(theta) if theta is the angle of the inclined with respect to the horizontal).
Thanks nrqed, you won a cookie. The acceleration was solved by doing:

Fr + Wx = m.a
u.m.g.sin240 + m.g.cos240 = m.a ==> cancelling all the m
0,1 . 9,8 . (-0.86) + 9,8 . (-0.5) = a
-5.74 = a


Then :

fv = iv + a.t
0 = 10.7 + (-5.74) . t
-1.86 = t

Regarding the above line:
It would help you to get the solution if you can explain what all those terms mean and especially explain how the second equation comes about.


Did you find the acceleration for going up the ramp or for coming down the ramp?
That second equation is the law of conservation of energy. English is not my first language so maybe im not translating into the right standars you guys use.

The acceleration im working with is always the one that goes up.


I can get 9m/s for part c), but why isn't the answer to part a) the same?
I think its because of the friction that you need a higher inicial velocity.





Thanks everyone.
 
  • #10
nrqed
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Thanks nrqed, you won a cookie. The acceleration was solved by doing:

Fr + Wx = m.a
u.m.g.sin240 + m.g.cos240 = m.a ==> cancelling all the m
0,1 . 9,8 . (-0.86) + 9,8 . (-0.5) = a
-5.74 = a


Then :

fv = iv + a.t
0 = 10.7 + (-5.74) . t
-1.86 = t
You are welcome :-)

Watch out, solving for t you get t= 1.86 second (not -1.86 s).
 

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