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I've been staring at this newtons first law assignment for hours!

  1. May 25, 2014 #1
    And I'm starting to feel kind of dumb here...

    here is the picture to go along with the question


    Question: Draw a free-body diagram for Point A, showing all the forces acting at A.
    Resolve the vectors into components, where necessary.
    Using the first condition of equilibrium, solve for T1 and T2.

    Repeat for Point B, and solve for T3 and T4 .

    So what I've done is realize that the sum of forces on the y axis should equal zero, so I have written that T4Sin35 + T2Sin65 - 80.0N = 0

    What I've learned is that you have to substitute in a value for T4 or T2 so you can then solve for the other one. Problem is I'm not sure how to approach this since there are two separate strings on the x axis.

    I have it written that T2cos65 + T4cos35 - T1 - T3 = 0, but I'm not sure if that is right.

    A little direction here would be immensly appreciated.
    Thank you so much! (sorry it's not written in TeX! I plan on learning it but have been so busy with school and haven't found the time yet.)

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    Last edited by a moderator: May 26, 2014
  2. jcsd
  3. May 25, 2014 #2


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    You had the right idea but you can pretend that point B is connected to a wall (it doesn't matter if there's a bunch of strings holding it there, it only matters that it's stationary)
    So it would be two separate equations, the first one being
    T[itex]_{2}[/itex]*cos(65°) - T[itex]_{1}[/itex] = 0

    (Technically your equation you made is true, but it's not solvable unless you separate into two equations.)
  4. May 25, 2014 #3
    Thanks for the reply man!

    So Is this line of thinking correct then?

    because T2*cos(65°) - T1 = 0
    T2 = T1/cos(65°) ?

    Then you can sub this into the equation T2*sin(65°) - 80.0N = 0 to get
    T1/cos(65°) * Sin(65°) = 80.0N ? which gives you T1 = 37.3 N?
  5. May 25, 2014 #4


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    Yep, looks good.

    T2 was known from the start (from: T2*sin(65°)=80.0N) and then you can use T2*cos(65°)=T1 to find T1

    I got the same answer as you so I believe it is correct (T1 = 37.3N).
  6. May 25, 2014 #5
    Awesome! Thanks for the help. I really appreciate it.

    So does this then mean that that T4*Sin35 - 80 = 0 as well then? Or is it T4*Sin35 - 80 - T2 Sin65 = 0 ? (so therefore T4*Sin35 - 160 = 0 ?)
  7. May 26, 2014 #6


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    When you drew a free body diagram of joint A, T3 and T4 did not enter the picture. So now, draw a free body diagram of joint B. What forces act at joint B??
  8. May 26, 2014 #7
    So unless I am mistaken, the forces acting on point B are the tension force between point B and the left wall acting to the left, the tension force between the ceiling and point B acting at a 35 degree angle from horizontal, and then a force from T2 along the y axis and a force of acting on the x axis of T2 as well.

    So I ended up getting T2 = 88.3 N, T1 = 37.3 N, T4 = 139 N, T3 = 76.6 N. :)
    Last edited: May 26, 2014
  9. May 26, 2014 #8


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    Yes, T4*Sin(35°) - 80 = 0 because point b is "not concerned" with what is below. The vertical part of T4 only has to counteract the vertical part of T2 (which clearly only has to counteract the 80N weight, therefore, the vertical part of T4 only has to counteract the 80N weight.)

    I do definitely understand how it can be confusing, but here's a way of looking at it (that I personally find helpful).
    Let's look at a (vertical) rope hanging from a ceiling and holding an 80N weight. The tension is clearly 80N. Now, let's bisect the rope so now it's two ropes, one hanging from the ceiling, the other hanging from the first rope, and the weight hanging from the second rope.
    (Does this make sense? It's the same picture, but now I'm pretending it's two ropes, instead of one continuous rope.)
    Would the tension in the top rope be 80N or 160N? Well it's the same system, nothings changed except were pretending it's two ropes, so it should still be the same tension 80N, right?
    (If not, then it should logically follow that the tension in a rope will get larger as you move up, but we know it's constant throughout the rope.)

    I don't know if that helped but it's essentially the same question you asked, just in a simplified way (no angles or anything) and (in my opinion) it makes the intuition a little bit more clear. (In my mind, you can consider a single rope as a large number (billions) of connected ropes, and from that perspective it's quite clear that the tension from the weight is not going to be multiplied a billion times.)

    Sorry if this was long and unhelpful, I just wanted to try to share the intuition behind my answer.
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