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  1. #1
    pilot in training..u mad? dadonkey's Avatar
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    Physics Help - Tension/Others (REPS)

    Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.


    Problem 1)

    The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.



    A) What is the acceleration of the system in meters/seconds^2?

    B) What is the tension at the top of the heavy rope in Newtons?

    C) What is the tension at the midpoint of the rope in Newtons?

    -----------------------------------------------------------------------------------




    Last Problem

    The position of a 2.75 * 10^5 N training helicopter under test is given by
    r = (0.02 m/s^3)*t^3 ihat + (2.2 m/s)*t jhat - (0.060 m/s^2)*t^2 khat

    Find the net force on the helicopter at t =5.0 s.
    Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures. In newtons.
    Last edited by dadonkey; 02-01-2009 at 09:45 PM.
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  2. #2
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  3. #3
    pilot in training..u mad? dadonkey's Avatar
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    Anybody know either of these two problems?
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  4. #4
    Registered User Sledgebat's Avatar
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    Alright, well F=ma, right? I can't see the picture in the first one, so I can't really go anywhere from there, but I guess you have to find the total force applied to the rope and then use F=ma to find the acceleration. Though I suppose you probably already knew that.

    I can't say I remember how to do the second one at all, I haven't done that stuff in three years. If I find my formula sheet from physics, I'll try to help you out.
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  5. #5
    Registered User PunkGoesSwoll's Avatar
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    Originally Posted by dadonkey View Post
    Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.


    Problem 1)

    The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.



    A) What is the acceleration of the system in meters/seconds^2?

    B) What is the tension at the top of the heavy rope in Newtons?

    C) What is the tension at the midpoint of the rope in Newtons?

    -----------------------------------------------------------------------------------
    I got a B in Phys last semester so Ill give it a try

    A. Total mass of the whole system: 15kg
    Total force by gravity applied on system: 15kg * 9.8 = 147N
    Net force on the system: 200N - 147N = 53N going up
    Force = acceleration x mass
    Acceleration = Force / mass = 53N/15kg = 3.5 m/sec^2
    B. Tension @ top is total weight of the rope and the block below = (4kg+5kg) x 9.8 = 88.2N
    C. Tension @ midpoint is weight the block below and half of the rope= (5kg+2kg) x 9.8 = 68.6N
    Last edited by PunkGoesSwoll; 02-01-2009 at 10:03 PM.
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  6. #6
    Banned Klutch89's Avatar
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    I think the acceleration is 3.5m/s^2 is that right?
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  7. #7
    pilot in training..u mad? dadonkey's Avatar
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    Originally Posted by PunkGoesSwoll View Post
    I got a B in Phys last semester so Ill give it a try

    A. Total mass of the whole system: 15kg
    Total force by gravity applied on system: 15kg * 9.8 = 147N
    Net force on the system: 200N - 147N = 53N going up
    Force = acceleration x mass
    Acceleration = Force / mass = 53N/15kg = 3.5 m/sec^2
    B. Tension @ top is total weight of the rope and the block below = (4kg+5kg) x 9.8 = 88.2N
    C. Tension @ midpoint is weight the block below and half of the rope= (5kg+2kg) x 9.8 = 68.6N
    You got A right. B and C are wrong ( I already tried putting in 68.6, didn't work). I also tried 31.77 for B which didn't work. (4 + 5) * 3.53.
    Last edited by dadonkey; 02-01-2009 at 10:11 PM.
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  8. #8
    Registered User PunkGoesSwoll's Avatar
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    B. 31.5
    C. 24.5

    ?

    Edit: oh... i never got tension problems right... haha
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  9. #9
    Registered User PunkGoesSwoll's Avatar
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    try...

    B. 56.7
    C. 44.1
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  10. #10
    pilot in training..u mad? dadonkey's Avatar
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    Originally Posted by PunkGoesSwoll View Post
    B. 31.5
    C. 24.5

    ?

    Edit: oh... i never got tension problems right... haha
    Nope, not 24.5
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  11. #11
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    try

    b) 53N
    c) 53N

    edit: nvm, system is moving
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  12. #12
    Registered User PunkGoesSwoll's Avatar
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    B. 141.2n
    C. 121.6
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  13. #13
    pilot in training..u mad? dadonkey's Avatar
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    Originally Posted by PunkGoesSwoll View Post
    B. 141.2n
    C. 121.6
    nope
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  14. #14
    u misc wat i did der? amerrocks's Avatar
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    Originally Posted by dadonkey View Post
    Thanks for any help guys. Definitely will rep you a lot. Explanations would be nice too, I want to learn how to do these, especially the tension problems.


    Problem 1)

    The two blocks in the figure (Intro 1 figure) are connected by a heavy uniform rope with a mass of 4.00 kg . An upward force of 200 N is applied as shown. The top block has mass 6 kg, the bottom 5 kg.



    A) What is the acceleration of the system in meters/seconds^2?

    B) What is the tension at the top of the heavy rope in Newtons?

    C) What is the tension at the midpoint of the rope in Newtons?

    -----------------------------------------------------------------------------------
    a) sum of forces =ma, so force up-force down=ma, force up is 200N, force down is gravity, which is 15lbs x 9.8, subtracting the two gives u about 50, set that = to ma, so 50=(15)a, a= about 3.33
    b) forces are the 200 N up, and the gravitational force on the specified part =(9kg)(9.8)= about 90 subtracting the two gives you 110 N
    c) same process as in b, except weight of rope is half, so 200-(7kg)(9.8)= about 130 N



    am i close?
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    u misc wat i did der? amerrocks's Avatar
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    Originally Posted by dadonkey View Post


    Last Problem

    The position of a 2.75 * 10^5 N training helicopter under test is given by
    r = (0.02 m/s^3)*t^3 ihat + (2.2 m/s)*t jhat - (0.060 m/s^2)*t^2 khat

    Find the net force on the helicopter at t =5.0 s.
    Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures. In newtons.
    so take second derivative of the position equation to give u derivative. Then plug in the time given.

    now convert weight of helicopter to kg's, then multiply the acceleration vector u got by the weight.....there u go
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  16. #16
    pilot in training..u mad? dadonkey's Avatar
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    Originally Posted by amerrocks View Post
    a) sum of forces =ma, so force up-force down=ma, force up is 200N, force down is gravity, which is 15lbs x 9.8, subtracting the two gives u about 50, set that = to ma, so 50=(15)a, a= about 3.33
    b) forces are the 200 N up, and the gravitational force on the specified part =(9kg)(9.8)= about 90 subtracting the two gives you 110 N
    c) same process as in b, except weight of rope is half, so 200-(7kg)(9.8)= about 130 N



    am i close?
    B is 120, don't know how that is it (got it off some forum that had that part done)

    For C i entered 131. That wasn't right.
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  17. #17
    pilot in training..u mad? dadonkey's Avatar
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    Originally Posted by amerrocks View Post
    so take second derivative of the position equation to give u derivative. Then plug in the time given.

    now convert weight of helicopter to kg's, then multiply the acceleration vector u got by the weight.....there u go
    I did. I got the weight as 28061.2 kg ( 2.75 x 10^5 / 9.8)

    Then the acceleration components = .12t ihat + 0 - .12

    then i multiplied each by 28061.2. still not it.
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  18. #18
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    Originally Posted by dadonkey View Post
    B is 120, don't know how that is it (got it off some forum that had that part done)

    For C i entered 131. That wasn't right.
    oh, ok that makes sense....acceleration is 3.333 everywhere, so Force up on the part is 9(3.333) which is about 30, force down is still abt 90. when I pictured it, i saw the block being attached to one of those spring scales u hang stuff from, and i realized that when you pull the scale up, the weight actually increases
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    u misc wat i did der? amerrocks's Avatar
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    Originally Posted by dadonkey View Post
    I did. I got the weight as 28061.2 kg ( 2.75 x 10^5 / 9.8)

    Then the acceleration components = .12t ihat + 0 - .12

    then i multiplied each by 28061.2. still not it.
    did u put the 5 seconds in, so your vector would be <.6, 0, -.12> then multiply that by weight?
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    Originally Posted by amerrocks View Post
    did u put the 5 seconds in, so your vector would be <.6, 0, -.12> then multiply that by weight?
    yeah, and got 16836.7 ihat, 0, 3367.34 khat.

    It asked for two sig figs (Express your answer in terms of \hat i, \hat j, \hat k. Express your coefficient using two significant figures.) so I put

    1.7 x 104 i hat, 0.0, 3.4 x 10^3 khat
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    Anyone? Please! haha I have to have this by tonight.
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    Just need the helicopter one now guys, thanks.
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