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Earth exerts a downward gravitational force

  1. Jul 17, 2007 #1
    1. The problem statement, all variables and given/known data
    Earth exerts a downward gravitational force of 8.9 N on a cake that is resting on a plate. The plate exerts a force of 11.0 N upward on the cake. [ add the rest of the problem: and a knife exerts a downward force of 2.1 N on the cake. Draw a free-body diagram of the cake.]

    3. The attempt at a solution
    My question here is that i know 8.9 N is the cake's weight, and this force is pointing downward, but why then the force from plate pointing upward is 11.0N not 8.9 N equally?
    thank you.
    Last edited: Jul 17, 2007
  2. jcsd
  3. Jul 17, 2007 #2
    The normal force (by plate on cake) only needs to equal the cake's weight if the force of gravity is the only other force acting on the cake, and neither the cake nor the plate is not accelerating.

    edit: please post the entire problem.
  4. Jul 17, 2007 #3
    sorry, i am still not so clear. why it's 11N? why it's larger than the cake's weight 8.9 N.
  5. Jul 17, 2007 #4
    Can you please post the whole problem?
    edit: sorry, just read the rest of your post. hang on.
  6. Jul 17, 2007 #5
    Did you draw a free body diagram?
  7. Jul 17, 2007 #6
    no, because i was actually getting confused about why the upward force from plate was not equal to the cake's weight 8.9 N but 11 N
  8. Jul 17, 2007 #7
    Drawing a free-body diagram will help you answer your question.
    Also, remember that [tex]\Sigma F=ma[/tex].
  9. Jul 17, 2007 #8
    there were only 3 forces on my diagram which wasnt correct.
  10. Jul 17, 2007 #9
    Three forces is good.

    Which force(s) point down?
    Which force(s) point up?

    Do you want up to positive or down to be positive?

    What's the acceleration of the cake?
  11. Jul 17, 2007 #10
    i want up to be positive, 11.0 up which is normal force i guess, 8.9 is down, 2.1 is down. but still how come is 11.0 not 8.9? btw, does the acceleration of a cake have something to do with gravity 9.81? or we just use net force/m. i was confused about this , too.
  12. Jul 17, 2007 #11
    OK. Sorry about the acceleration of the cake. What I meant is that the cake is not accelerating (it's just sitting on the plate), so a=0 so [tex]\Sigma F=ma=m(0)=0[/tex].

    Using your free-body diagram as a guide, write what [tex]\Sigma F[/tex] is, then make it equal 0. Yes, this will answer your question about why the normal force on the cake is greater than the gravitational force on the cake.
  13. Jul 17, 2007 #12
    oh, ok ,i get what you said, thanks. btw, i was confused about the gravity 9.81 and the formula that i just said to you. for example, if there are two objects dropped from the top of the building, one is heavier, the other one is lighter. lighter one has more acceleration than the heavy one i think if we know Newton's second law formula. but why their accelerations are not both equal to 9.81 from the earth?
  14. Jul 17, 2007 #13
    No, they have the same acceleration: a=9.81 m/s^2

    Say the heavy object has mass M and the light object has mass m, and they're both dropped off of a building.

    for the heavy object:

    for the light object:

    The FORCE acting on the heavy object is greater than the FORCE acting on the light object, but the actual accelerations are the same (which is why they both land at the same time [assuming no air resistance]).
  15. Jul 17, 2007 #14
    oh yes, i was just about to change my question since i just figured out that they have the same accelerations, but thanks for the explanation, it's good. but the weird thing was today in class, a question was like someone dropped a physics book, and it soon landed on the floor, actually physics book and the earth both moved to each other, but just the book's acceleration 9.81m/s^2 is greater than the earth's acceleration. so earth's acceleration is not 9.81? but why my teacher said earth and physics book have the same weight? they have different mass i know, but why the weights are same?
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