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Basic Forces Question from 9th grader - Hard to put in words

  1. Jan 25, 2016 #1
    1. The problem statement, all variables and given/known data
    I have here a problem asking me to find the tension force on the lower end of a rod with mass of 0.6 kg. The rod connects two blocks, the first / upper one has a mass of 5.0 kg while other/lower one has a mass of 4.0 kg. There is an upward force of 150 N applied on the whole system.

    2. Relevant equations
    Given:
    F = 150 N
    Mass of whole system : 9.6 kg (mass of upper block+mass of rod + mass of lower)
    a = 150 N/9.6 kg
    a= 15.625 m/s^2

    3. The attempt at a solution
    Knowing that the system has acceleration:
    Fnet= ma
    Fnet = Tb (tension force on point b/ lower end of rod) - (Weight of rod + Weight of lower block)
    Tb - (Wrod + Wlowrblck) = ma
    HERE IS MY QUESTION (only capitalised for emphasis)
    Tb - (5.88 N + 39.2 N) = ma
    Tb - 45.08 N = m (15.625 m/s^2)
    At this juncture I was having 2nd thoughts on what mass to use but since this problem is in multiple choice, I had to try out some things:
    Tb - 45.08 = (4.6 kg)(15.625 m/s^2) -------> the mass I used is the sum of the masses of rod and lower block
    Tb = 116.96 N -------> This answer was too large and was not part of the choices.
    I had an alternative:
    Tb - 45.08 = (0.6 kg)(15.625 m/s^2) ---------> this time I plugged in only the mass of the rod
    Tb = 54.455 or approximately 54 N -------> This answer of mine was part of the choices so decided to stick with it.
    If my 2nd solution is correct, why should I only consider the mass of the rod and not include the mass of the lower block?
    Thank You very much
     
    Last edited by a moderator: Jan 25, 2016
  2. jcsd
  3. Jan 25, 2016 #2
    Welcome to Physics Forums.

    It's very hard to picture this. Can you provide a diagram?

    I assume the force is applied to the bottom of the stack? Have you drawn a free body diagram on the lower block and shown the forces acting on it?

    Chet
     
  4. Jan 25, 2016 #3
    I would expect that since there is a reference to tension, vis-a vis compression, which to me implies a stretching of the connecting rod that the force is applied to the upper mass.
     
  5. Jan 25, 2016 #4
    OK. So, again, have you drawn a free body diagram on the lower mass and shown the forces acting on it? Based on this free body diagram, what is the force balance on the lower mass?

    Chet
     
  6. Jan 27, 2016 #5
    Mod note: Copied diagram from Word doc and added it as an attachment.
    Here is the diagram of the problem. Thank you for replying.
    FBD.png
     

    Attached Files:

    Last edited by a moderator: Jan 27, 2016
  7. Jan 27, 2016 #6
    the answer is 62.5 N my friend
    Who's gonna calculate gravity as an external force(as the 150 N force acts upwards,gravity acts on the whole system downwards)
    still make the question a bit clear,would you?
     
  8. Jan 27, 2016 #7
    My friend,it's obvious gravity's there
    Please redo your question and i'll tell you something called the centre of mass method to solve such problems(:))
     
  9. Jan 27, 2016 #8

    Ray Vickson

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    It is not at all obvious that gravity is present; it was not mentioned in the original problem. The whole thing could be here on earth in the presence of gravity, or it could be in outer space where there is no significant gravitational force on such a small object.

    Certainly the OP should clarify this issue!
     
  10. Jan 27, 2016 #9
    Assuming, as UchichaClan13 suggests, that gravity is present.

    You already determined that the mass of the combined assembly is 9.6 kg. Using the equation w = mg, what is the weight of the combined assembly?
    From your free body diagram, there are two external forces acting on the assembly: 150 N and the weight (gravitational force) mg. What is the net force? What is the acceleration?

    Next, you are going to have to do a free body diagram on just the lower block so that you can determine the force of the rod acting on this block. From Newton's 3rd law, how would this force compare with the tension force that the lower block applies to the rod?

    Chet
     
  11. Jan 27, 2016 #10

    SammyS

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  12. Jan 28, 2016 #11
    that's true but my friend,surely the question would have been elaborated further by the OP wouldn't it???
    It's normally assumed that when we say that the 150 n force pulls the system upwards ,there's gravity acting downwards!
     
  13. Jan 28, 2016 #12

    Ray Vickson

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    Normally, yes, but who knows if that is the case here? I refuse to assume; I would rather have the OP state clearly all the conditions of the problem. I know, I know... if one speaks of upward and downward that typically refers to the direction of gravity, but again, I am unwilling to make an assumption. Much better, I think, would be to force the OP to deal with the issue!
     
  14. Jan 28, 2016 #13

    jbriggs444

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    Suppose that you add gravity to the free body diagram based on some unknown local gravitational field strength. Does that field strength change the final answer?
     
  15. Jan 28, 2016 #14
    $$150-9.6g=9.6a\tag{overall force balance}$$
    where a is the acceleration of the assembly. So,
    $$a=\frac{150}{9.6}-g$$
    $$T-4g=4a\tag{force balance on lower mass}$$
    where T is the tension exerted by the rod on the lower mass, and the tension exerted by the lower mass on the rod.
    So,
    $$g+a=\frac{T}{4}=\frac{150}{9.6}$$
    So, $$T=62.5 N$$

    Yikes, you're right. The answer to the problem does not depend on whether gravity is included or not.
     
  16. Jan 29, 2016 #15
    See told you
    gravity assumption would yield the correct answer
    :)
     
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