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Math Help - Need help getting started..

  1. #1
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    Need help getting started..

    Sled of 196 N is being pulled across a horizontal surface at a constant velocity. So it's in equilibrium right??
    The pulling force(Tension force?) has a magnitude of 80 n and is direct at an angle 30degree above the horizontal. Determine the coefficient of the kinetic friction.

    I am stuck on drawing the FBD. I suck at those. I get them confused.
    Need help on getting started, thanks.

    So 30 degree angle from the 0degree, Force of 80 is perpendicular to that, so at 120 degrees?? Where do I put the Ff? to figure out the u
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by bradycat View Post
    Sled of 196 N is being pulled across a horizontal surface at a constant velocity. So it's in equilibrium right??
    The pulling force(Tension force?) has a magnitude of 80 n and is direct at an angle 30degree above the horizontal. Determine the coefficient of the kinetic friction.

    I am stuck on drawing the FBD. I suck at those. I get them confused.
    Need help on getting started, thanks.

    So 30 degree angle from the 0degree, Force of 80 is perpendicular to that, so at 120 degrees?? Where do I put the Ff? to figure out the u
    Here is the free body diagram


    Need help getting started..-freebody.jpg

    \displaystyle \sum F_x=80\cos(30^{\circ})-\mu_k\eta=ma_x=0


    \displaystyle \sum F_y=\eta+80\sin(30^{\circ})-196=ma_y=0
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  3. #3
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    Ok, so do I solve for Fy and sub that into Fx? What letter do you give the 80N just P for pull or T for tension?
    I solve for Fn for Fy and plug that into Fn in Fx?

    Thanks for the diagram I was off somewhat for sure.
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  4. #4
    Behold, the power of SARDINES!
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    Solve for eta (\eta) the normal force in the 2nd equation.

    Sub this into the first equation and solve for the coefficient of kinetic friction \mu_k
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  5. #5
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    OK that is what I meant, I typed it wrong. Thank you. I will study that.
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