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How do you calculate the force of friction on an inclined plane

  1. Jan 24, 2005 #1
    How do you calculate the force of friction on an inclined plane(ramp) a certain number of degrees from the horizontal?(Which formula is used)

    Mass of block = 10 Kg
    Angle of incline = 30 degrees
    Coefficient of friction = 0.25
    Gravity = 9.8m/s2

    What is the acceleration of the blockdown the ramp?
    (I already worked out the force of gravity using F = mg sin theta, or should I have used cosine theta?)
     
    Last edited: Jan 24, 2005
  2. jcsd
  3. Jan 25, 2005 #2

    quasar987

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    You shouldn't (musn't) try to 'guess' wheter it's mg cos or mg sin. If you don't remember how the sides of a triangle and its angles are related, go look it up. In a right triangle, [itex]cos \theta[/itex] = hypothenuse / adjacent side, [itex]sin \theta[/itex] = hypothenuse / opposite side, [itex]tan \theta[/itex] = opposite side / adjacent side.

    In an inclined plane problem, both components of the force of gravity are important. The component parallel to the plane, mg sin30° indicates you that the block is pushed downward the plane, and thus will be subject to a force of friction with the plane.

    The component of weight perpendicular to the surface of the plane, mg cos30°, indicates you that the normal force N exerted by the plane on the block is, according to Newton's 3rd law, mg cos30°.

    Now, how are the force of friction and the normal force related?
     
  4. Jan 25, 2005 #3
    and....

    and can you determine a relationship between the coefficient of friction and the acceleration of the block down the ramp...

    next, if the object were, say, a cylinder instead of a block, how would the coefficient of friction affect the cylinder's acceleration down the ramp.

    finally, i think there's an assumption here about the coefficient of friction between the block and the ramp. there are two different "coefficients of friction" between the block and the ramp. what are they, how do they differ, and how would that affect the actual answer to your question of how fast does the block accelerate down the ramp?
    :smile:
    +af
     
  5. Jan 25, 2005 #4
    Well, the block is just placed on the incline so it implies that the force is mg cos theta, but there is also a frctional force acting on the block, so it can also impky rhe block is moving, so the force would then be mg sin theta.

    If its the latter is the frictional force calculated using sin theta or cos theta?
    If its the first, which one is it for the frictional force?
     
  6. Jan 25, 2005 #5

    Gokul43201

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    Gogsey : you are confused about vecors and their resolution.

    Draw a free body diagram.

    Then resolve all the forces in 2 directions : (1) along the ramp , and (2) perpendicular to the ramp. If you don't remember how to do this, go over your vectors basics.

    Write Newton's second law for each of these two directions.

    You now have 2 equations. You can hence solve for up to 2 unknowns.
     
  7. Jan 26, 2005 #6
    The forces along the other planes are not required, the force along the inclined plane is all thats needed. The angle is already given to you and so is the mass.

    So, wieght of the block sitting still is mg sin theta, and the frictional force of the inclined plane on the block is mu mg sin theta.

    But if the block is moving down the incline, the gravitational force is mg cos theta, and the frictional force is mu mg cos theta?

    Although, if the block is placed on the ramp, doesn't it imply that the force of gravity is greater than that of the frictional force, so the block should be moving down the ramp?
     
  8. Jan 26, 2005 #7

    quasar987

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    Would you mind writing down the word-for-word formulation of the problem please?
     
  9. Jan 26, 2005 #8
    Sure

    A ramp is inclined at an angle of 30 degrees to the horizontal. A 10 kg block is placed on the ramp. The coefficient of frictionbetween the block and the ramp is 0.25.

    a) What is the force of gravity acting on the block?

    b) What is the frictional force acting on the block?

    c) What is the acceleration of the block down the ramp?
     
  10. Jan 26, 2005 #9

    quasar987

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    Not knowing whether 0.25 stands for the coefficient of static or kinetic friction, you cannot answer to b) and c). :grumpy:

    Alright, maybe you can, assuming that 0.25 stands for the coefficient of static friction. But under this assumption, at c), you cannot give the acceleration of the block "down the incline", but only it's acceleration the very instant you let go of the block...

    I'll look at what you said in your last post now..
     
    Last edited: Jan 26, 2005
  11. Jan 26, 2005 #10

    quasar987

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    I'm not sure what you mean by that, but you are contradicting yourself a few lines later when you mention the quantity mg cos theta. That is precisely the component of weight in a direction perpendicular to the plane.

    I'd like to mention the distinction betwen the weight and it's components. The weight of the block is a vector. It is a vector directed downward and perpendicular to the surface of the earth. Its magnitude is mg. It can be divided as a sum of 2 vectors: one perpendicular to the surface of the plane, the other parrallel to the surface of the plane.

    No. I asked you at the end of my first post to write what the expression of the force of friction is, in terms of the normal force. Had you done this, you would have written

    [tex]f_s_{max} = \mu_s N[/tex]

    Now, what is the expression of the normal force exerted by the plane on the block?

    ?!?!?

    You place the block on the ramp. It is not moving when you let go of it. Whether it starts moving or not is a matter of wheter the total force acting on it is greater than zero or not. This is what the equation F = ma it telling you. Therfor, your goal in this problem is to determine all the force vectors acting on your block, and see if the sum of all those vector forces is different than 0. If it is, then F = ma, written as F/m = a, give you the acceleration of the block.
     
  12. Jan 26, 2005 #11
    You asked for the expression of the force of friction, which i didn't answer, sorry.

    The frictional force of the incline on the block is

    F = mu N cos theta


    Doesn't FN = FG, or mg = mg or is FN = 9.8 and FG = 9.8 times the mass.

    I'm never sure that the normal force is mg or just the same acceleration due to gravity.

    In otherwords, FG = mg, so is FN = mg or just g.

    I know this sounds extremely stupid, and simple, but i'm getting a little confused about the value of FN .
     
    Last edited: Jan 26, 2005
  13. Jan 26, 2005 #12
    There are three force vectors acting on the block, right?

    1) Gravity at 30 degrees from the vertical

    2) Normal force, 30 degrees from the vertical

    3) Frictional force acting up the incline, 30 dgrees from the horizontal


    Also, part c) asks what is the acceleration of the block down the ramp.
    Doesn't this imply the vector sum is not zero, therfore the block is accelerating down the ramp, so friction is less that gravity on the incline.

    P.S. is ther e not a force of garvity acting down the incline directly against the force of frection?
     
  14. Jan 26, 2005 #13
    Sorry, there are four vectors, the three i mentioned and the component of gravity down the incline.

    I already know that the two vertical vectors of gravity and the normal are equal. I also know that the force of gravity down the incline is greater than the frictional force against it.

    so part c) is ok if i calculate a) and b) correctly, becuaes for c) its the NET force divided by the mass.

    But, its the angles that throw me off.

    All i'm not sure about is whether to use cos or sin for the angles of the forces acting up and down the incline
     
  15. Jan 26, 2005 #14

    quasar987

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    It is neither of the thousand possibilities you listed. Normal is a synonim of "perpendicular". It is a vector force of response. Always directed perpendicularly to the rigid surface. Its magnitude is equal to the component of force directed towards and perpendicularly to it.

    Here, examining the graphical representation of the block resting on the incline and drawing the vector weight, we can see that this vector has a component parralel to the plane, and another perpendicular to it. It is only to the perpendicular component of weight that the incline will react. How it reacts is governed by Newton's 3rd law: by exerting on the block a force equal in magnitude, and opposite in direction.

    We have found previously that this perpendicular component of the weight is mg cos theta. According to what I have just said about the behavior and nature of the normal force, this means that the normal force as direction opposite to the perpendicular component of weight, and of magnitude mg cos theta also.
     
  16. Jan 26, 2005 #15

    quasar987

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    Right.

    But this all follows directly from the definitions of sin, cos and tan as I wrote them in my first post. Starting from the fact that the inclined plane makes an angle theta with the horizontal, and drawing your vectors correctly, you can use all the theorems you know about right triangles (the sum of the angles is 180°), complementary angles, opposite angles, the sine law, and cosine law (although you can always get away without making use of this one) to figure out what angles each vector makes with one another.
     
  17. Jan 27, 2005 #16

    VietDao29

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    Hi,
    The definition of sin, cos, and tan in your first post is WRONG. Please re-check it, quasar987:
    [itex]\cos{\theta}[/itex] = adjacent side / hypothenuse, [itex]\sin{\theta}[/itex] = opposite side / hypothenuse, [itex]\tan{\theta}[/itex] = opposite side / adjacent side.
    Am I right??
    Hope yes,
    Viet Dao,
     
  18. Jan 27, 2005 #17
    Looks correct to me.
     
  19. Jan 27, 2005 #18

    quasar987

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    Teehee yea thanks for pointing that out VietDao29 :blushing:
     
  20. Jan 27, 2005 #19
    Looking at the angles position(betwenn the Hyp and Adj), i would say, that, the force of gravity down the incline, sgould be cos theta, theta being the angle being 39 degrees from the horizontal.

    So, if this is true, the force of friction should be at the same angle, cos theta, directed up the ramp, opposite in dirction of the force of gravity down the ramp.

    FF = mu N cos theta
    So, FF = mu mg cos theta
     
  21. Feb 1, 2005 #20
    trying to confuse me?

    who snuck in the 39 degrees thing???? :tongue:

    the slope is 30 degrees; the mass is m. the component of gravitational force on the mass acting down the ramp is m*g*sin(30) = 0.5*m*g, yes?

    the friction force opposing the motion of the mass is equal to the normal force times the coeffecient of friction, or mu*m*g*cos(30), or about 0.25*m*g*0.866, or (approx= .22*m*g)

    so, when you put the block on the ramp, it's gonna movel, since the frictional force up the ramp is about 28% less than the force of gravity pulling the mass down the ramp.


    so, if the angle were 45 degrees, what would the coefficient of static friction have to be for the mass to not slide down the ramp?

    izzat closer?
    :devil:
     
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