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Motion and Force

  1. Sep 24, 2010 #1
    Attachments: Questions and Answers

    Could someone please help me take a look at my work?
    Especially Question 5, I am not sure what 5a is asking, and I don't know if I could calculate "Us" using the way I did.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 24, 2010 #2

    Lok

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    Damn it, I forgot...
    What is the international measuring unit for weight and Force? Put the values in their right place... (5 Q answer) ... It is not a necessity but think of the children who read it like that and turn evil.

    Answer for 5 seems ok but I cannot find a 5a question?
     
  4. Sep 24, 2010 #3
    Sorry, I mean 5c.
     
  5. Sep 24, 2010 #4

    jhae2.718

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    For 5c, what force does the conveyor belt exert on the package? Look at your free body diagram...
     
  6. Sep 24, 2010 #5
    I calculate the "x" and "y", do I just add them together?
     
  7. Sep 24, 2010 #6

    jhae2.718

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    Edit: Okay, so you're basically calling the frictional force y and the normal force x.

    These forces act in reaction to the motion of the package, so the total reaction force [tex]\vec{R}[/tex] would be the vector sum of x and y, or [tex]\vec{R}=x\hat{i}+y\hat{j}[/tex]. The magnitude of a vector, [tex]|\vec{R}|=R=\sqrt{R_x^2+R_y^2}[/tex]. (Are you familiar with unit vector notation?)

    However, based on the problem statement, I think they just want the reaction force in the [tex]\hat{j}[/tex] direction, which would be...?
     
    Last edited: Sep 24, 2010
  8. Sep 24, 2010 #7
    Does that mean Fn is the answer for this question?
     
  9. Sep 24, 2010 #8
    Also, did I do the other questions correctly?
     
  10. Sep 24, 2010 #9

    jhae2.718

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    That's what I think they're after; based on the equation it seems like it is trying to reinforce the relationship between the normal force and friction.

    (I.e. first you find the normal force, which you then use, along with ∑F=ma to find μ.)

    What kind of physics course is this?

    Edit: I only briefly looked through your work, but you seemed to do everything correctly.
     
  11. Sep 24, 2010 #10
    The way that I find "Us" is not correct?
    There is no velocity and time, how can I find acceleration?

    The course name is Physical principles of food strucuture and functionality. I have no idea why we are doing physics!
     
  12. Sep 24, 2010 #11

    jhae2.718

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    No, you found μs correctly.

    Since the package is at rest, the net acceleration is equal to zero, so ∑F=ma=0 along the incline.

    If we have ∑Fx=mgsinθ-Fs=0 and ∑Fy=N-mgcosθ=0, we can solve them separetly (as you did) to get:
    mgsinθ=Fs
    N=mgcosθ

    If we assume that static friction is at its maximum, then FssN.

    Using these three equations, we find that μs=(mgsinθ)/(mgcosθ)=tanθ
     
    Last edited: Sep 24, 2010
  13. Sep 24, 2010 #12
    What about other questions? Did I do them correctly?
     
  14. Sep 24, 2010 #13

    jhae2.718

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    Briefly reading through it, you seemed to have done everything correctly.

    Edit: Actually, on 2, you just found the vector sum of the two forces given. You need to find a third vector whose magnitude should be equal to the magnitude of the first two, but opposite to the direction of the sum of the first two.

    You have three forces, F1, F2, and F3. You know F1 and F2, and you know that ∑F=0, so you need to break the forces into their component form and find what F3 needs to be to make ∑Fx=0 and ∑Fy=0.
     
  15. Sep 24, 2010 #14
    Thank you so much!

    By the way, the course is under Nutrition and Food Science program....
     
  16. Sep 24, 2010 #15

    jhae2.718

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    See my edited post; I missed a mistake you made on 2.
     
  17. Sep 24, 2010 #16
  18. Sep 24, 2010 #17
    So the answer will be -25N with an angle to the horizontal of 1.5 degree?
     
  19. Sep 24, 2010 #18

    jhae2.718

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    You have the resultant force of the first two forces already , by the way. The third force is this force with the components negated.

    I just want to make sure you understand the process of summing the forces to find the unknown force.

    The magnitude is always positive. (The magnitude of a vector [tex]\vec{R}=\langle r_x,r_y \rangle[/tex] is defined as [tex]R=\sqrt{r_x^2+r_y^2}[/tex]. This quantity is always positive for real numbers. The individual components of the third force's vector are both negative. The magnitude is positive.

    Your angle should come out to what you had before, but this force is in the fourth quadrant, so you need to add 180 deg. to the result of the arctangent.
     
  20. Sep 24, 2010 #19
    o~ I got it! Thanks alot!
     
  21. Sep 24, 2010 #20

    jhae2.718

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    So the third force should be 25N at 181.5 degrees measured from the x-axis.

    No problem.
     
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