What magnitude of force must the worker apply?

    A factory worker pushes a 30.0-kg crate a distance of 4.5. along a level floor at constant speed by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply?

    2. Relevant equations

    summation of F x forces = 0

    frictional force = coefficient of fric * normal force

    3. The attempt at a solution

    Hello, my solution is this

    since the frictional force and the force of the worker in the x direction are equal,

    fric force = 0.25 * 9.8 * 30 kg

    fric force = 73.5 NEWTONS

    so to find the magnitude of the force that the worker apply, we simply use trigonometric ratios.

    cos (30) = 73.5 / r

    r = 84.87 newtons

    my answer is 84.87, but the answer in the book is 99.2 newtons. can you tell me where did i go wrong? tnx
  2. jcsd
  3. why cos(30) ? you never mentioned in your formulation of the problem that there are any angles other than 90 degree.
  4. Hello, im referreing to question 6.4a, (the one in black pen)


    tnx. ive tried my best but where did i go wrong? whats ur answer?
    Last edited by a moderator: Oct 12, 2011
  5. Redbelly98

    Redbelly98 12,038
    Staff Emeritus
    Science Advisor
    Homework Helper

    This is confusing because you posted Question 6.3(a), but show your work for 6.4 which is different because of the 30 degree downward angle of the worker's force.

    The normal force is not simply the weight of the crate here. To work through a problem like this, you need to:

    1. Draw a diagram showing all forces on the crate. Remember that the force from the worker is at an angle, not along the horizontal as it was in problem 6.3.

    2. Set up two equations, using ƩFx=0 and ƩFy=0. Remember to separate the worker's force into components using sin and cos.

    See if you can at least get that far, and if you're still stuck then post your work here for us to look at.
  6. Indeed, when the worker is pushing downwards, he is increasing the weight of the crate and so the friction force as well.

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