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Finding the kinetic coefficient and acceleration

  1. Aug 3, 2009 #1
    A boy drags his 60N sled at a constant speed up a 15.0* hill. He does so by pulling with a 25N force on a rope attached to the sled. If the rope is inclined at 35* to the horizontal, (a) what is the coefficient of kinetic friction between sled and snow? (b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?

    So we know:
    (tension of rope) T=25N
    (weight of sled) mg=60N
    (m of sled) m=6.12kg
    (normal force) n=56.36N
    (force of gravity) Fg=20.51N

    So,

    [tex]\sum[/tex]Fx=T*cos[tex]\Theta[/tex]-mg*sin[tex]\Theta[/tex]-fk=max=ma

    [tex]\sum[/tex]Fy=T*sin[tex]\Theta[/tex]+n-mg*cos[tex]\Theta[/tex]=may=0?

    Then,

    [tex]\sum[/tex]Fx=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax

    ax=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax/6.12kg

    2.98-(uk*56.36N)/6.12kg=ax

    So, since I have two unknowns I am lost now. And I haven't even begun part (b).

    Could anyone help me and teach me to find these unknowns?

    Thank you!
     
    Last edited: Aug 3, 2009
  2. jcsd
  3. Aug 3, 2009 #2
    Hey, you must remember that the sled is moving constantly uphill ! So, the acceleration must be 0. Enlightened ?
     
  4. Aug 3, 2009 #3
    What, really? Both components too? Not just Fy either by n? Hmph... why is this?

    So theoretically speaking I could use this 0a circumstance to find the kinetic friction coefficient and use it in the sled + kid situation to find the downwards acceleration? Yeah, okay, that definately seems plausible.

    Thank you, prob_solv. :approve:
     
  5. Aug 3, 2009 #4
    It is because the speed is constant . If the speed is constant, it means that there's is no acceleration.
     
  6. Aug 3, 2009 #5
    Oh, okay. I see.

    But:

    When I solve the problem I hit a wall.

    ma=T*cos-mg*sin-fk

    Then,

    6.12kg*0=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)

    So,

    that makes the entire force 0... is that correct?
     
  7. Aug 3, 2009 #6
    2.98N-(uk*56.36N)=max=0?

    But, I get:

    uk= 0.05

    But the problem says that uk= 0.161

    :confused:
     
  8. Aug 3, 2009 #7

    rl.bhat

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    Homework Helper

    You have taken the angle of inclination as 20 degrees, whereas in the problem angle is given as 35 degrees. How is that?
     
  9. Aug 3, 2009 #8
    And even when I insert uk= 0.161 in my probelm I still get:

    ma=T*cos-mg*sin-fk
    ma=-4.59

    which seems very unlikely?

    Maybe it has to do with either the order of my force components or the fact that I took the angle rope minus the angle of the hill 35*-15* since it said that the rope was 35* at the horizontal. Yet, I have no idea. Anything catching an err would be appreciated. :tongue:
     
  10. Aug 3, 2009 #9
    It's wrong. mg cos theta and mg sin theta is wrong. You must use mg sin 15 or mg cos 15 ( because theta is 20 ). Got it correct ?
     
  11. Aug 3, 2009 #10
    So,

    what your saying is that:


    Fx=Tcos35-mgsin15-fk=max=ma=0

    Fy=Tsin35+n-mgcos15=may=0


    Well I think that the tension theta of 35 and the hill theta of 15 should be seperated and not subtracted by themselves?
     
  12. Aug 3, 2009 #11
    When I do how you said:

    Fx=Tcos(20)-mgsin(15)-ukn

    Fy=Tsin(20)+n-mgcos(15)

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

    I obtain uk=0.141

    Which is (0.020) off but something is still missing?
     
  13. Aug 3, 2009 #12
    When,

    Fx=Tcos(20)-mgsin(15)-ukn

    Since the uk=0.161

    If I insert this into the above I obtain Fx= -1.11

    So there is still something wrong with the angles I belive...
     
  14. Aug 3, 2009 #13
    I didn't think that this problem would pose such a problem. Is it hard or just boring?

    All it is, is deriving the acceleration and the kinetic coefficient on an inclinded plane with tension and frictional force... then again, I can not do it so... Ah well.
     
  15. Aug 3, 2009 #14

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    It's neither hard nor boring (well, I guess that latter is a matter of opinion). The angle of the incline is 15 degress. let's call this [itex] \theta [/itex]. The angle of the rope is 35 degrees (relative to the horizontal for those people like rl.bhat who weren't paying attention). Let's call this [itex] \phi [/itex]. Then the angle of inclination of the rope relative to the incline is [itex] \phi - \theta [/itex] = 20 degrees. I want to emphasize that there are two angles we're working with in this problem. Let's set up our coordinate system so that the x direction is along the incline, and the y direction is perpendicular to the incline. Then, we have two force balance equations. In the y direction, the normal force balances the other vertical forces, which include a contribution in the positive y direction (from the rope) and in the negative y direction (from the weight of the sled):

    [tex] \sum F_y = T_y - W_y + N = 0 = T\sin(\phi - \theta) - W\cos{\theta} + N [/tex] ​

    [tex] -N = (\textrm{25 N})\sin(20^{\circ}) -(\textrm{60 N})\cos(15^{\circ}) [/tex] ​

    [tex] = \textrm{-49.405046 N}[/tex] ​

    In the x direction, the net force is supposed to be zero and has a positive contribution from the rope and a negative contribution from the weight of the sled and from friction:

    [tex] \sum F_x = T_x - W_x - f = 0 [/tex] ​

    [tex] f = T_x - W_x = T\cos(\phi - \theta) - W\sin{\theta} [/tex] ​

    [tex] f = (\textrm{25 N})\cos(20^{\circ}) - (\textrm{60 N})\sin({15^{\circ} ) [/tex] ​

    [tex] f = \textrm{7.96317281 N} [/tex]

    The result is that:

    [tex] \mu_k = \frac{f}{N} = 0.161181366 [/tex]​
     
  16. Aug 3, 2009 #15
    I'm sorry, I've never taken a physics class before... I wish it were as easy for me, I suppose this is simply another step!

    Thank you cepheid, I appreciate it. :redface:
     
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