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I'm stuck on this problem involving Tension, Force, and Acceleration!

  1. Oct 1, 2012 #1
    Two packing crates of masses m1 = 10.0 kg and m2 = 4.80 kg are connected by a light string that passes over a frictionless pulley as in the figure. The 4.80-kg crate lies on a smooth incline of angle 41.0°.

    4-p-028alt.gif

    (a) Find the acceleration of the 4.80-kg crate.

    QUESTION 1: WHENEVER I'm given TWO things such as m1 and m2, that means I will be taking the SUM of something right in ORDER to find a variable such as "a" (acceleration) or something else, right?

    This is what I did...

    Mass 1:
    Fnet = T1 + (-m1g)
    -m1a = T - m1g

    Mass 2:
    Fnet = T2 + (-m2gsin θ)
    QUESTION 2: Why is sin θ being included in THIS equation for mass 2 but not in the equation for mass 1???
    m2a = T - m2gsin θ

    Subtract mass 2 from mass 1:

    Mass 2: m2a = T - m2g[/B]sin θ
    Mass 1: -(-m1a = T - m1g)
    –––––––––––––––––––––––––––––––––––––
    Equation: m2a + m1a = m1g - m2gsin θ

    a(m2 + m1) = m1g - m2gsin θ
    a = (m1g - m2gsin θ) / (m2 + m1)
    a = [ (10)(9.81) - (4.80)(9.81)(sin 41) ] / (4.80 + 10.00)
    a = (98.1 - 30.86) / 14.8
    a = 67.24 / 14.8
    a = 4.54 m/s2

    (b) Find the tension in the string.

    QUESTION 3: How would I go about solving for this one?
    Just plug in a in either of the original mass equations??


    Can you please, please, not be vague/abstract in your answer also?
    Thank you.
     
  2. jcsd
  3. Oct 1, 2012 #2
    Question 1:
    Typically, if you are given two masses, the sum of them will be involved somewhere.

    Question 2:
    Mass 1 is not on an inclined plane, and thus feels the full effect of gravity; however mass 2 is on the incline, and thus has gravity affecting it in both the x and y directions, but the y direction is what you are concerned with. Multiplying the total force of gravity by the sine of the angle gives the y component; likewise, taking the cosine would show the x component. You could also do this for mass 1, but since it makes a 90 degree angle with the ground, the sine of that angle would result in 1.

    Question 3:
    Yes, but you would have to derive an equation that deals with the total tension, taking both masses into account. You want to add T1 and T2, resulting in Ttotal.
     
  4. Oct 1, 2012 #3

    Doc Al

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    Staff: Mentor

    I don't know what you mean. When two masses are involved you'll usually have two force equations, which you'll solve simultaneously.
    Good.
    You want force components parallel to the incline, which is the direction of the acceleration for m2.
    OK.
    Sure.
     
  5. Oct 1, 2012 #4

    Doc Al

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    Staff: Mentor

    :bugeye: The rope has a single tension throughout its length.
     
  6. Oct 1, 2012 #5
    Good point. Throughout my tension unit, we only did problems with multiple strings so I got that stuck in my head. My fault, but now I'll remember. Thanks!
     
  7. Oct 1, 2012 #6
    Ok....I think I understand..

    Imma get back to y'all if I'm even more confused

    WAIIIT.
    I STILL don't get why the sin theta is being multiplied to -m2g...

    It literally makes no sense to me. I've read all of your responses and I'm trying to understand it but it just doesn't make any sense...

    The mass 2 is technically on the HYPOTENUSE of the triangle........ You only do mg cos theta OR mg sin theta when you are looking for 2 legs (??????)
     
    Last edited: Oct 1, 2012
  8. Oct 1, 2012 #7
    SO for my Question 2, you ESSENTIALLY multiply -mg (for whatever mass that's attached to the pulley) by the sin of WHATEVER angle THAT mass is in right?
     
  9. Oct 1, 2012 #8
    Correct, it's just like if you were wanting to find one component of any other force or value when working with vectors.
     
  10. Oct 1, 2012 #9
    OK!
    Thanks SO much everyone!!
     
  11. Oct 1, 2012 #10

    Doc Al

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    Staff: Mentor

    The surface of the incline is not the hypotenuse of the force triangle. (It's the hypotenuse of the incline plane triangle, but so what.) You are finding the components of the weight (mg) parallel and perpendicular to the incline, so the weight must be the hypotenuse of your force right triangle.

    Read this: Inclined Planes
     
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