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Homework Help: Determine the magnitude of force on a suspended object

  1. Mar 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A 5.0 kg mass is suspended by a rope. A horizontally directed force F is applied to the mass. What magnitude of force is needed to produce an angle of 65˚?

    mass = 5kg
    angle = 65˚

    2. Relevant equations

    F = mgsinx
    T = rFsinx

    3. The attempt at a solution

    I tried using the first formula:
    F = (5kg)(.8m/s^2)(sin65˚)
    F = 44N

    I'm not sure if this is correct though, I am worried I used the wrong formula, but since I don't know the length of the rope I can't use the formula for torque.
    Last edited: Mar 12, 2015
  2. jcsd
  3. Mar 12, 2015 #2


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    Hello MadMcB. Welcome to PF!

    Your answer is not correct. Be careful with pulling formulas out of your notes unless you are sure they apply to your new problem. You can solve this problem without using any physics formulas, just a little trig.

    When the object is being held at rest at 65o, how many forces act on the object? Draw a force diagram ("free-body diagram") and proceed from there.
  4. Mar 12, 2015 #3
    You shouldn't use torque. You know the magnitude of the net force and it's vertical component.

    (Although you calculated F incorrectly. Try drawing a diagram rather than using a formula, formulas don't always work because details may change)

    With this, what can you solve for?
  5. Mar 12, 2015 #4
    Hi TSny! Thanks for the advice. I've drawn a diagram, and I know that the vertical component of the normal force on the object is 49N. How do I find the vertical component?
  6. Mar 12, 2015 #5
    Thanks Brain. How do I know the net force? If I had the net force and the vertical component, I could solve for the horizontal component...
  7. Mar 12, 2015 #6
    Could you perhaps attach an img of your diagram so we could see what you have?
  8. Mar 12, 2015 #7
    Can I insert an image as a file from my computer??
  9. Mar 12, 2015 #8
    here you go! Apologies- It is sideways

    Attached Files:

  10. Mar 12, 2015 #9
    Okay. Just one question: is it 65 degrees from the horizontal or from the perpendicular?

    Once you figure that out (not sure if the problem makes it ambiguous), use trig to figure out the tension and then get the horizontal component from that
  11. Mar 12, 2015 #10
    65˚ from the horizontal. So the tension is sin65˚ = o/h = 49/h. So h = 49/sin65 = 54.07 N

    using pythagorean's theorem: x^2 + 49^2 = 54.07^2. So x^2 = 522.56, and x = 22.86 N

    So, F = 23N. Is this correct?
  12. Mar 12, 2015 #11
    Yup looks good.
  13. Mar 12, 2015 #12
    Thanks for your help.
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