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Basic Mechanics question I cant seem to solve

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data

    An adult swimmer is pulling to small children on a tobaggen over level snow. The total wieght including the children of the tobaggen is 47kg. The tobaggen rope makes an angel of 23dgrs with the horizontal. The coefficient of kinetic friction is 0.11. Calculate the magnitude of the tension in the rope needed to keep the sleigh moving at a constant velocity. (hint: normal force does not equal gravitational in magnitude).

    2. Relevant equations

    3. The attempt at a solution
    Okay so I drew a fbd of the 47kg object simplified as a box. for I calculated Fg. (9.8x47) = 460.6N. I drew the Fa force on a angle at 23 degrees.I used trig to and found that the ratio of the force applied going away at the horrizontal to the force applied going away at 23dgrs is 0.92:1
    Since uniform velocity means horrizontal forces are equal in magnitude opposite in direction I multiplied G by 0.92 to get Fn then by 0.11 to get Ff. Then I devided by 0.92 to get the magnitude if the force was 23 dgrs off of horizontal. I got 50.66N the answer says 53. Im not sure if im totally off, If the answer is correct we just rounded diffenerently or if the teacher used 10 for g instead of 9.8 I thnk I may be worng because I never used the weight of the tobaggen. Any help is much appreciated
  2. jcsd
  3. Sep 20, 2011 #2
    you may know this but remember to include all vectors in your calculations. I had this problem with homework too. the hint where the normal force is not equatl to the weight is key, because that is what scewd me up since I thought the other way around.

    so, sum of all forces in y direction=(Tension)sintheta - Weight + normal force=0 (think about it, how is the system accelerating in the y direction?..think about that)

    sum of all forces in x=Tensioncostheta - kinetic friction + applied force

    now think about this kinetic friction always equals normal force times the coefficient of friction.

    so, what would you do next? (Hint: the sum of the forces in x direction are too complexed to solve. So, think outside the box)
  4. Sep 20, 2011 #3
    wait maybe you should scratch the applied force part. I think that was wrong.
    Try to prove that it is, because I'm not too sure. Sorry about that.

    but now that I see the constant velocity part, tensioncostheta - kinetic friction=0
  5. Sep 20, 2011 #4
    thanks for the reply! So I understand the tension in the y direction plus the normal force has to equal the weight (aboslute values, I know that since they are vectors downwards will be defined as the negative direction) But you lost me in your second equation I thought the tension in the rope was the applied force as there will be no friction yet, so why do you have the two being added? and do you see something obviously wrong with my solution cause i was quite close and it made sense to me the way I solved it. Thanks alot for the tips already though
  6. Sep 20, 2011 #5
    ok yea i just did the problem at got 52N. That means applied force is out, prove it after youre done though!
  7. Sep 20, 2011 #6
    I don't understand the ratio part, but I don't think that is necessary to do this problem.

    Dude I have no idea how to prove that no applied force can be accounted for. It just isn't. You should ask your teacher or someone else about that, but I know the calculations are right.
  8. Sep 20, 2011 #7
    oh hint:find an equation that fits this scenario and equals the normal force.
  9. Sep 20, 2011 #8
    Ok so Really the ratio part I was saying is assuming the magnitude of the applied force at the offset angel from the horizontal has a magnitude n. the magnitude of the horrizontal componet will be 0.92n. With really is the same as your costheta i believe as cos theta = 0.92? Does that make sense? sorry im taking so long to get your point. And i think that it really is telling u your applied force inderictly. Its (Ffriction)/(cos23)?
  10. Sep 20, 2011 #9
    i think we have the same solutions just yours is conveyed much better, did you get exactly 52? because so far the answers my teacher has had are all pretty exact so im afraid im missing something! or maybe you have the right solution and Im wrong in thinking mine is the same as yours
  11. Sep 20, 2011 #10
    sigfigs like my professor said and i agree are just icing on the cake but unfortunetely actually affect the precision of your answers. Thats whose fault it is. Its the right answer, but sigfigs are off and to those out there who do, no need to b@@itch because you will be successful regaurdless of being that precise. You know what I'm saying? :)
  12. Sep 20, 2011 #11
    but as far as your intuitive idea that there is a ratio of .92n, even I have a trouble understanding what you mean and I don't want to think about that, because it will probably screw up my ideas that are usful enough, but thats awesome that you think that abstractly about these problems. Screw it, I'm going to absorb what you mean and sit down and try to understand. hold on.
  13. Sep 20, 2011 #12
    come to think of it, I think the applied force is the tension. I cant explain why but I think thats what it is. so, you dont need to include it.
  14. Sep 20, 2011 #13
    thats alright. Yesterday, I spent the whole day having the topic of relavtive velocity on my head and it was really hard to understand. I just really couldnt understand why certain factors occur.
  15. Sep 20, 2011 #14
    and Tension=friction/cos23

    so in a sense that I cant really prove, since tension seems to be applied force, applied force=friction/cos23
  16. Sep 20, 2011 #15
    This is how i came too it: We need the normal force to calculate the frictional force, since some upwards force is provided by the applied force that is why the gravitational force is more then the normal. so the normal force plus that upwards componet of the tension in the rope = -force of gravity. Then the horrizontal componet left would have to equal the frictional force. So the horrizontal componet of the applied force is in the expression cos23=tensionrope/horizontal componet. if u sub 1 in for tension of rope jsut to get a simple ratio you get 0.92 for the horizontal part. you multiple that times fg to get fn. and the other part of the positve y axis force is acounted for in the vertical componet of the applied force. then we have fn. We multiply by the coeficient of friction which is 0.11 to get Ff which i got to be 46.6 Then I used that as the horizontal componet of applied and did trig to turn that back to the angled componet (23 degres of of horizontal) sorry I know my explanaition is very conveluded. I apreeciate all the helpp tho!
  17. Sep 20, 2011 #16
    Yeah I agree the applied force is deffintly the tension of the rope.
  18. Sep 20, 2011 #17
    if you substitute 1 for the tension and find the ratio, the ratio will represent the constant you multiply by the tension to get the frictional force of the rope. if you know that there is a ratio to get tension of rope, however, then you may be in luck as far as finding the tension itself, but thats too compicated. the ratio in this case we'd call "an equation." But yea lets not go there.

    "normal force plus that upwards componet of the tension in the rope = -force of gravity. Then the horrizontal componet left would have to equal the frictional force"

    frictional force is equal to the normal force times the coefficcent of frictioon. now does that help a bit?
  19. Sep 20, 2011 #18
    you just want me to spill the beans and show you exactly how I did it? or do you think you got it?
  20. Sep 20, 2011 #19
    YEs thats exactly what i have done with the ratio I was taking about. Im glad someone else has came by it the same way! thanks for the help and i will post tommorrow after talking to me teacher and seeing his solution
  21. Sep 20, 2011 #20
    can I just show you my solution? haha
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