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Homework Help: F=ma pulley problem

  1. Nov 12, 2008 #1
    1. A 28.0 kg block is connected to an empty 1.35 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.450 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move. (a) Calculate the mass of sand added to the bucket. (b) Calculate the acceleration of the system

    2. F=ma Ffr = mumg

    3. I think I got part a.

    For a)

    mu(static)mg + mu (kinetic)mg =mg
    (.45)(28)(9.8) + (.32)(28)(9.8)= m (9.8)
    211.288 = m (9.8)
    m= 21.58
    subtracting 1.35 (mass of bucket)
    mass of sand = 20.21

    for b) You look at only the horizontal forces on the block, because vertical forces are going to cause any acceleration?

    So, the sum of the forces on the box (1) is

    m1a= Tension - Ffr

    And on box 2

    m2a = T - m2g

    the acceleration is going to be the same because it is all attached, correct?
    If my equations are correct, I'm just kind of confused about how to put them together.

    help is really appreciated! thank you!
  2. jcsd
  3. Nov 13, 2008 #2
    Hi, sorry but i don't really understand your working for part b. I thought there is only 1 box attached to 1 bucket via a pulley? why are there boxes 1 and 2?

    if there is only 1 box and 1 bucket, then for the part a, you should only consider the situation at which the box just begins to move. in other words, if you let x be the mass of sand added to the bucket, then let the static frictional force acting on box be equal to the tension which is equivalent to the weight of the bucket plus sand.
    hence, static frictional force = (0.45)(28)(g) = (1.35 + x)(g)... then you can solve for x.. in your solution, you used both the static frictional force and the kinetic frictional force in the same equation. however, since we are considering the situation in which the box is initially stationary, you shouldn't use the kinetic frictional force, since the box isn't moving initially.

    for the 2nd part, since we are assuming that the cord is inelastic, the acceleration of the block would be equal to the acceleration of the bucket, so we only need to consider the forces acting on the block.
    then, force acting on block = 28(a) = (1.35 + x)(g) - (0.32)(28)(g) ... then you can solve for a..

    hope this helps..
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