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Monkey Business Please Confirm!

  1. Oct 23, 2005 #1
    A 5.0 kilogram monkey hangs initially at rest from two vines, A and B. Each vine is 10 meters in length with negligible mass.

    Vine A is on the left with an angle of 30 degrees from the horizontal from the monkey and B is at 60 degrees from the horizontal of the angle next to the monkey.

    1. Determine the tension of B while it is at rest
    2. Determine the speed of the monkey.
    3. Determine the tension of vine B at the lowest point through the swing.

    Here is what i got:
    1. T=Wcos(theta)
    =(5)(9.81)(cos 60)
    =24.525N

    2. PE(i)=KE(f)
    mgh=.5(m)(v^2)
    (5)(9.81)(10-10sin60)=(.5)(5)(v^2)
    v=5.127

    3. T=mg+F(centripital)
    =(5)(9.81)+((5*5.127^2)/10m)
    =62.192 N

    Can anybody confirm this for me? I have been stuck on this problem for like 2 hours, Im confused if i calculated the tension in part 1 right or if i accidentlly did the tension for vine A instead. This is my worst part of physics, so any help would be appreciated!
     
    Last edited by a moderator: Oct 24, 2005
  2. jcsd
  3. Oct 23, 2005 #2
    What a notice is that angles of A and B are complements, so if i messed up using sine and cosine i would have done the opposite vine. I cant figure out if I'm using the right equations and solving them the right way.
     
  4. Oct 24, 2005 #3
    Consider this when solving for tension in the vines.
    First set up system of force vectors for both the x and y axis.
    (X) [tex]T_{a}\cos{\alpha}-T_{b}\cos{\beta}=0[/tex]
    (Y) [tex]T_{a}\sin{\alpha}+T_{b}\sin{\beta}-Mg=0[/tex]
    You can then solve for the tension of both vines. Then back subtitute and you should be left with an equation like this. I'll give you one, but not both.
    [tex]T_{a}=\frac{Mg}{\sin{\alpha}+\cos{\alpha}\tan{\beta}}[/tex]
    I hope this has been some help :smile:
     
  5. Oct 24, 2005 #4
    ok first i solved the x equation to be: T(a)=T(b)(cosB/CosA)

    Then i substituted it into the y equation. I moved mg to the right side and factored out T(b) from the numbers. Then i solved for what was left on the right side (the sines and cosines) to recieve T(b)=42.4785N

    Does this sound about right?
     
  6. Oct 24, 2005 #5
    I think part 2 of the problem is right, since it doesnt deal with the tension.

    I am still confused about part 3 though also. According to my book: At the bottom of a vertical circle:
    F(centripital)=F(normal)-mg

    I assume that F(normal)=T (thats what i always thought, but im not sure)

    So F(centripital)+mg=F(normal)

    Does this seem about right?
     
  7. Oct 24, 2005 #6
    Normal Force is used to describe the force perpendicular to an object in contact with another object. A book on the table is good example, or a tire on the road.

    In this instance the tension is the centripetal force + gravitational force. So, you are almost on the right track, in that your equation is right for this example.

    However it is not correct though to think of tension in the string as normal force. Consider how the forces would behave if the monkey was at a right angle to the force of gravity. Then the tension would be equal to the Centripetal Force without any influence of gravity.
    Why?
    Tension is the Centripetal Force! - It is the net-force component perpendicular to the direction of motion.
     
  8. Oct 24, 2005 #7
    Ahh i get it. Ok im off on that part, but it works since the object is being pull downed towards the earth so it would add to the tension. Anywhere besides straight down though, my calculations would be totally off. Except at the exact top of the circle, where the tension would be F(c) minus the weight since the weight is subtracting or pushing towards the center of the cirlce...which lessens the tension.

    Thanks for the help
     
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