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What is the force exerted on the liquid by the pot?

  1. Oct 31, 2006 #1
    I have a few questions on some homework I'm trying to do:

    1) A 3.5 kg flowerpot drops from a tall building. The initial speed of the pot is zero, and you may neglect air resistance.

    After the pot has fallen 33 m, it enters a viscous liquid, which brings it to rest over a distance of 1.5 m. Assuming constant deceleration over this distance, what is the magnitude of this deceleration?

    I found the force to be 215.56 (I know this is correct).

    Now, here's the part I can't figure out:

    What is the force exerted on the liquid by the pot?

    I did F = ma but that was not the correct answer. What did I do wrong?

    2) A block of mass M = 15 kg is suspended at rest by two strings attached to walls, as shown in the figure. The left string is horizontal and the right string makes an angle theta = 50° with the horizontal. What is the tension in the left string?

    3) A 0.6 kg sphere hangs on a string from the ceiling of an automobile. The car has a constant horizontal acceleration and the string is displaced an angle θ = 65.2° from the vertical. To a person sitting inside the car, it appears that the sphere is motionless (the angle between the sphere and the vertical is constant). What is the magnitude of the acceleration of the car?

    Thanks for any and all help.
     
  2. jcsd
  3. Oct 31, 2006 #2

    PhanthomJay

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    In problem 1, you found the acceleration to be -215.6 m/s/s. I think this is what you meant, always show your units. Then you can use F_net =ma to solve for F_net, bearing in mind that that F_net = N - mg, where mg is the weight and N is the average liquid force acting on the pot. Swap signage for the force of the pot on the liquid, per Newton III.
    In problems 2 and 3, show what you've come up with thus far.
     
  4. Oct 31, 2006 #3
    For question 2 resolive the tension of the cord into its components tsin50 and tcos50. Use Newtons laws. Obviously the sum of the forces are going to be zero in the x and y directions since the object is suspended and not moving, therefore no acceleration. I called up and to the right the positive directions. I am calling the left string T1 and the right T2.

    The most important thing to do is draw yourself a free body diagram once resolving all vectors into components.

    Fy=0 Fx=0
    T2sin50-mg=0 T2cos50-T1=0
    T2sin50=mg T2cos50=T1
    T2sin50=15kg(9.8m/s^2) 192Ncos50=T1
    T2sin50=147N 123.4=T1
    T2=147N/sin50
    T2= about 192N
     
  5. Oct 31, 2006 #4
    For question 2 resolive the tension of the cord into its components tsin50 and tcos50. Use Newtons laws. Obviously the sum of the forces are going to be zero in the x and y directions since the object is suspended and not moving, therefore no acceleration. I called up and to the right the positive directions. I am calling the left string T1 and the right T2.

    The most important thing to do is draw yourself a free body diagram once resolving all vectors into components.

    Fy=0
    T2sin50-mg=0
    T2sin50=mg
    T2sin50=15kg(9.8m/s^2)
    T2sin50=147N
    T2=147N/sin50
    T2= about 192N

    Fx=0
    T2cos50-T1=0
    T2cos50=T1
    192Ncos50=T1
    123.4=T1

    sorry about last post came out a little unclear
     
  6. Nov 1, 2006 #5
    For #2 I had something similar to what rmarkatos said, but I think I skipped a step and therefore got the wrong answer.

    For #3 I'm not really sure. We had a mass in an elevator where the rope could handle a certain amount of tension (60N) before it broke and we had to find the maximum acceleration, but I'm not sure if this is the same as what I'm doing for #3. Basically, the mass was 5kg and acceleration was a. So I used T^2 = (mg)^2 + (ma)^2 and got the right answer but I don't think that's what I'm supposed to use for this.
     
  7. Nov 1, 2006 #6
    I am still having trouble with #2 and #3, does anyone have any idea?
     
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