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Motion on Incline

  1. Dec 16, 2006 #1
    1. The problem statement, all variables and given/known data

    A 215 N box is placed on an inclined plane that makes a 35.0º angle with the horizontal.

    a) Draw a free body diagram which indicates all forces acting on the box.
    b) Calculate components of weight force parallel and perpendicular to the plane.
    c) If the crate is at rest, calculate the force of friction.
    d) Now assume that the create accelerates down the incline at 5 m/s^2. What must force of friction be?
    e) Find the coefficient of friction in part d

    2. Relevant equations

    http://www.algebralab.org/img/d0c27dd9-1bf4-48da-96bd-89f6397bc762.gif

    3. The attempt at a solution

    http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e7.gif

    I know how to draw the free body diagram. I think the components of the weight force is mgcos35 and mgsin35 which is 176 and 123 respectively. I don't know how to do questions c, d, or e.

    4. Sorry

    I know I sound stupid asking this amateur question. I really need to stop sleeping in physics class...
     
  2. jcsd
  3. Dec 16, 2006 #2

    cristo

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    Staff Emeritus
    Science Advisor

    "respectively" will depend on which component you are stating first! The perp component is mgcos35.
    For c, what can you say about the force of friction if the box is at rest? For d, use the equation F=ma, which you state on your web link, but remember that F is the resultant force! For e, do you know an equation involving the coefficient of friction?
     
  4. Dec 16, 2006 #3
    Your components are correct. Beware of the signs of these vector-components, but first things first:

    Following the picture you added in the link, we define the x-axis along the incline (positive direction go in UPWARD direction) and the y-axis is perpendicular to the incline.

    The gravity on the box gives us indeed :

    1) x direction : -mgsin35
    2) y direction : -mgcos35

    Watch your signs of the vectors. Both components are in the opposite direction of the positive directions of x and y axis, hence the -.

    Next questions :

    c) the box is at rest (read : not moving down). This means that friction is strong enough to avoid the box from coming down. In b) you calculated the gravitycomponent along the x axis, which is the force responsible for sliding the block down. Apply Newton's second law (in the x direction) with these data

    d) again, apply Newton's second law with the given data (both in x and y direction)

    good luck

    marlon
     
  5. Dec 16, 2006 #4
    yeah thanks guys i think i figured it out. Friction is the same as mgsin35 since its at rest. friction with the acceleration would be F=ma. coefficient is F=(mu)*N
     
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