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Sliding Block Problem (We must solve together)

  1. Oct 4, 2006 #1
    Sliding Block Problem (Please Help!)


    The drawing shows a large cube(mass=25 kg) being accelerated accross a frictionless surface by a horizontal force P. A small cube(mass=4.0 kg) is in contact with the front surface of the large cube and will slide down unless P is sufficiently large. The coefficient of static friction between the cubes in 0.71. WHat is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

    What i did:

    4.0 x 9.8 = 39.2 x 0.71 = 27.832 N
    39.2 - 27.832 = 4.0a
    a = 2.842 m/s/s

    I do not know what the acceleration in the x direction must be in order to make the acceleration upward match 2.842.
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2

    Doc Al

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    Staff: Mentor

    If it doesn't slide down, what does that say about the vertical acceleration?
  4. Oct 4, 2006 #3


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    Homework Helper

    Draw the direction of the force of friction, and draw the direction of gravity acting on the small cube. Further on, what does the force of friction equal? In what relation must the force of friction be with gravity acting on the small cube? You can retrieve the minimal value of P from this inequality easily.
  5. Oct 4, 2006 #4
    the force of friction is 27.832 N upwards but it must be 39.2 N in order to keep the cube up. This means an extra 11.368 N must be applied upwards. I don't know what to do after this...
  6. Oct 4, 2006 #5
    does the force of gravity change from straight down to an angle?
  7. Oct 4, 2006 #6

    Doc Al

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    How did you determine that?

    There are only two forces acting vertically: friction (up) and gravitational force (down). They must balance if the block is not to slide down.

    Nope. It points straight down as always.

    Hint: Your first goal is to determine the normal force between the cubes that will provide sufficient friction.
  8. Oct 4, 2006 #7
    I got 27.832 N up by this (4.0 x 9.8 = 39.2 x 0.71 = 27.832 N) Doesn't that mean that 39.2 Netwons are directed downwards, and 27.832 N is directed upwards? Does this mean you need an extra 11.368 N upwards in order to keep the block from sliding?
  9. Oct 4, 2006 #8

    Doc Al

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    Ah... I see what you did. (Please be careful how you write things. That looks like an equation, but you really meant it in two steps, right?)

    (1) You found the weight of the small cube: mg = 39.2 N (OK!)

    (2) Then, for some reason, you multiplied by the coefficient of friction. (Not OK!)

    To find the force of friction, you need to multiply the normal force by mu. (The normal force is not the weight! The normal force acts horizontally in this case; the weight acts vertically.)

    You have to solve for the normal force.
  10. Oct 4, 2006 #9
    I looked on the internet and found that normal force is equal to mg - Tsin(theta), but how can I solve this without a degree in the question? I learn better if I can see the solution, because I memorize concepts very well.
  11. Oct 5, 2006 #10

    Doc Al

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    What makes you think that that expression is relevant to this situation? (T probably stands for tension!)

    Instead, do this: Identify all the forces acting on the small cube. (Hint: There are three such forces.) Tell me everything you know about those forces. For each force, tell me the magnitude if you know it (use symbols not numbers) and the direction. Then we will apply Newton's 2nd law to those forces.
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