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Help Appreciated

  1. Feb 27, 2004 #1
    A mass of 10kg lies on a smooth plane which is inclined at to the horizontal. The mass is 5m from top measured along the plane. One end of a light inextensible string is attached to the mass. The string passes up the line of greatest slope and over a smooth pulley fixed at the top of the plane. The other end is attached to a freely suspended 15kg mass. This mass is 4m above the floor. The system is released from rest 1 and 3/7 seconds later

    Find the volume of

    ~~~~~~~~~~~

    The "1 and 3/7 seconds" refers to one second and then 3 sevenths of a second --- i couldn't do a fraction.




    Thanks for all you help,

    Conor
     
  2. jcsd
  3. Feb 27, 2004 #2

    Integral

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    That's easy the volume is zero!

    Now if you would post the rest of the problem I'd give you a serious answer. What happens 1 3/7 s later?
     
  4. Feb 27, 2004 #3
    That's all it said, I think -- the system was released

    **I'll try to find out if there was anymore
     
  5. Feb 28, 2004 #4
    Sorry you're rite - there is another bit


    A mass of 10kg lies on a smooth plane which is inclined at xº to the horizontal. The mass is 5m from top measured along the plane. One end of a light inextensible string is attached to the mass. The string passes up the line of greatest slope and over a smooth pulley fixed at the top of the plane. The other end is attached to a freely suspended 15kg mass. This mass is 4m above the floor. The system is released from rest and the string first goes slack 1 and 3/7 seconds later

    I need to find the angle at which the plane was inclined

    The bit in italics is the edited bit

    Thanks,
    Conor
     
  6. Feb 28, 2004 #5

    HallsofIvy

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    "Find the angle" is a heckuva lot different from "find the volume"!

    The string will "go slack", of course, when the 15kg mass hits the floor. That is, when it has gone down 4 m. What you need to do is find the "net" force on the 15kg (Its weight minus the upward force due to the 10 kg mass on the incline). The upward force will be the same as the component of 10 kg mass's weight down the slope and that will depend on the angle. Calculate the acceleration necessary to go 4 m in 1 and 3/7 seconds. Use "F= ma" to find the net force. Use that to find the upward force and, finally, use that to find the angle.
     
  7. Feb 28, 2004 #6
    In my first post, i ment to say "size of the angle"

    I don't really understand your "theory" of it. I posted this in the general board and it got moved here.

    I'm only 13 and our maths teacher set us this challenge - if we can work this out from Monday then we get off homework for the rest of Lent.

    If possible, could I have the answer and how to do it.


    As I said, this isn't homework
     
  8. Feb 28, 2004 #7
    OK - I think the answer is 25º

    Can someone confirm if this is correct?
     
  9. Feb 29, 2004 #8
    As said before I have never done a problem of this sort:

    I think the answer is one of the following:

    a) 25º
    b) 8.21º
    c) 53º
    d) 26.74º


    Could anyone tell me the correct answers as I am at a loss and this is very important
     
  10. Feb 29, 2004 #9

    Doc Al

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    Halls told you how to do it. If you can't find the acceleration, and you don't know how to apply F=ma, then you can't do this problem.
    Sure it is!

    By the way, none of the four answers you gave is correct. (Of course, I may have made a mistake.)
     
  11. Feb 29, 2004 #10
    Yeah - my teacher asked us to find out and try to explain it -- we haven't done anyof this before.

    I'm trying to learn but am a bit above my depth.


    I will try my best to work it out myself - if someone would kindly explain how to calculate acceleration and how to apply F=MA


    **By TheWay, i'm not a 17 yr old posing as a 13yr old so you can do my homework - i said i'll try to work it out if someone told me how to calculate acceleration and how to apply F=MA (If this was really homework, then I would know how to do acceleration and F=MA**

    Thanks :smile:

    Conor
     
  12. Feb 29, 2004 #11
    We know the weight fell 4m in 1 and 3/7 seconds. From this we can calculate the acceleration, since acceleration is distance travelled (d) divided by time (t) squared. This gives us the following:

    4m
    _____________
    (1 3/7 secs)^2 2.040816 sec^2

    =

    4m
    _____________
    (1.42857sec)^2

    =

    4m
    _____________
    2.040816 sec^2

    approx 1.96m/sec^2 = acceleration=a

    Our mass is given at 15 kilograms. Now we can calculate F.

    F = ma

    F = (15kg)(1.96m/sec^2)

    F = 29.4 kg-m/sec^2

    This is the force with which the weight fell, aka the force the 15kg weight exerted on the 10kg weight on the smooth plane as they were connected with an inelastic string. Of course the 10kg weight was also acting upon the 15kg weight, slowing its acceleration in comparison to a free falling body, which would have the normal Earth acceleration (aka gravity). Now we need to find out this force in order to find out the angle of the force that resisted the free fall of the 15kg weight.


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


    I've figured this out?? Is it rite??

    And now, how do I change the force (29.4N) into the angle?
     
  13. Feb 29, 2004 #12
    Please, this is really important -- I just need some person to explain to me how to get from N(ewtons) --> º

    Thanks
     
  14. Feb 29, 2004 #13
    OK, well thanks for trying every1
     
  15. Feb 29, 2004 #14
    Would the upward force be

    117.6 N?
     
  16. Feb 29, 2004 #15

    Doc Al

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    Almost. D = 1/2 a t2
    No. If you had used the correct acceleration, this F would be the net force on the 15 kg mass.

    The way to do this problem is to apply F=ma to both masses. Note that each mass has two forces acting on it: the force of gravity (or a component of that force) and the tension in the string. Find the net force on each mass and then apply F=ma. You'll get two equations with two unknowns: the tension in the string and the angle. Solve for the angle.
     
  17. Feb 29, 2004 #16
    The correct acceleration?! I'm not sure i understand that bit of it?

    do you mean

    D = 1/2 a t^2?? If so, then does this make sense as we already have D?




    So I need to do

    F=ma for the suspended weight and agin for the weight on the plane when I do this I will have two values --> the forces of the two masses.

    Then I use these to get the angle? Using what formula?


    THanks,
    Conor
     
  18. Feb 29, 2004 #17

    Doc Al

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    Solve for a, of course. This is the correct relationship between distance, acceleration, and time. What you were doing was incorrect.
    No. Applying Newton's 2nd law (Fnet = ma) to each mass will give you two equations. Start by finding the forces on each mass and write these equations.
     
  19. Feb 29, 2004 #18
    Nah, I'm lost - I don't understand it at all

    We'll just do the homework.


    Thanks for all your help :smile:


    Conor
     
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