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Net force holds object in equilibrium

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data
    "A 6.0 kg box is held at rest by two ropes that form 30-degree angles with the vertical. An external force F acts vertically downward on the box. The force exerted by each of the two ropes is denoted by T. . . . The magnitude of force F is 410 N. The magnitude of force T is closest to: . . ."

    m = 6.0 kg
    a = -9.8 m/s2
    vi = 0 m/s
    fF = 410 N
    fT = ?

    A figure is included in the practice example, but I am unable to upload it. Each rope is denoted by T--and this is key to the questions I have about the problem, noted below.



    2. Relevant equations
    w = mg

    Fnet = ma

    cos [theta] = adjacent/hypotenuse



    3. The attempt at a solution
    w = mg
    fW = mg = 6.0 kg * -9.8 m/s2 =-58.8 N

    Fnet = ma
    Fnet, negative-y-axis = -58.8 N + (-410 N) = -469 N

    cos [theta] = adjacent/hypotenuse = 469/T
    --> T = 469/cos30 = 542 N

    Concerns:
    First, I'm not sure if I determined the magnitude of T the correct way. I figured I could use the positive y-axis as the equal but opposite force of the net force of fw and fF; this would in turn serve as the "adjacent" aspect of the trig function calculation.
    Second, the problem refers to "force T" as if it is a singular force--but in the diagram (and explanation), there are two ropes denoted by T.
    Third, the answer is provided: it is 271 N--exactly half of what I thought the magnitude of a *single* T is. I'm guessing I may be on the right path but am missing something because of the double nature of the force T--just not sure how to convey it in my work. (This isn't something to turn in--I just want to understand how to correctly do such a problem.)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 20, 2011 #2

    rl.bhat

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    Homework Helper

    469 N force is balance by the components of 2 Ts in the vertical direction.
    So 2*T*cos(30) = 469.
     
  4. Jun 20, 2011 #3
    I think I understand what you mean by the balance of components, as the y-axis divides the two T forces, but . . . I'm still not following, exactly. Any other way to frame it? Would you have approached things similarly?
     
  5. Jun 21, 2011 #4

    rl.bhat

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    Homework Helper

    The title of the post says that Net force holds object in equilibrium
    Since point of action of two tensions and the weight of the box and the force are in equilibrium, the resultant of two tensions must be equal to the total downward force.
    The angle between tensions is 60 degrees. Find the resultant of the tensions and equate it to the total force. That gives you the magnitude of the tension.
     
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