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Static friction forces on ramp

  1. Oct 1, 2007 #1
    An 55.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 30.0° to the ground.

    I sketched the problem using a popular paint program. I seem to be having problems visualizing components of the normal force for some reason. I know by rote that 55cos(30) will give me the normal force but I do not know why. Could anyone help me see it?

    By luck I was able to answer " What is the minimum possible value of the coefficient of static friction" by dividing 55cos(30) by 55sin(30).

    The last part of the problem asks:

    "The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?"

    Does anyone know what question wants me to do?

    EDIT: I answered this problem correctly by asserting the magnitude and direction of the contact force is 55N and 90 degrees respectively. It was a lucky guess. Anyone know what this question is getting at?


    Last edited: Oct 1, 2007
  2. jcsd
  3. Oct 2, 2007 #2
    First of all, I want to applause at the perfect problem description - especially with the FBD included *Sniff*. Users like you make us proud.

    As to answer your first question, I'll show you why using the FBD you included


    First of all, you know you can find that angle of 60 degrees from geometry, where all triangles have 180 degrees, and you have a right triangle so

    30 + 90 + x = 180

    x = 60 degrees

    Then, you know the normal force is perpendicular to the surface - that is the definition. Taking this to an account, the angle that makes with the normal force and the inner triangle has to equal 90 degrees. So to find that 30 degrees located right by 60 degrees,

    60 + x = 90

    x = 30 degrees.

    Understand so far? Then rest comes from trigonometry.

    You are resolving your weight into two components strictly - the normal force, and the force parallel to that, as the "Square root of Sum of the squares" has to equal to your weight.

    Therefore, you know the other component of W comes from the perpendicular direction of normal force. Just think of this as X and Y, tilted slightly to the left (I usually call them X', Y')

    A lot of students makes the following mistake:


    Which is clearly wrong, because you've tilted your coordinate system for Y', but you aren't tilting your X at all, so [tex](F_N^2 + X'^2)^(0.5)[/tex] does not equal your weight.

    Moving on to the actual calculation,

    You have a right triangle, you've given one length and you know an angle. What can you use to find the normal force, that is, the adjacent from the 30 degrees angle? Turns out it's cosine function as cosine is equal to adjacent over hypotenuse therefore.
    [tex]\cos(30) = \frac{F_N}{W}[/tex]
    [tex]F_N = W\cos(30)[/tex]

    As for your second question, it doesn't make much sense to me either. The question is poorly written. I was thinking maybe it was asking to find every forces on the surface of the block, but since answer was 55N, this cannot be it.
    Last edited: Oct 2, 2007
  4. Oct 3, 2007 #3
    Thanks. I was indeed tiltling my cordinate system.
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