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Forces with Friction help please

  1. Nov 9, 2004 #1
    ->Forces with Friction help<- ASAP please

    I have two problems that i didnt quite understand, i would appreciate any help i could get, my work follows the questions

    1.) A block with a mass of .2 kg is slid across a patch of ice with an initial velocity of 20 m/s. After Traveling 70 m, the velocity is 10 m/s. Find the following:
    a.) The acceleration of the puck a = acceleration
    Vf^2=Vi^2 + 2a(Xf-Xi)
    Vf^2 / a = Vi^2 +2(Xf-Xi)
    10^2 / a = 20^2 +2(70-0)
    100 / a = 400 + 140
    100 / a = 540
    a = 100/540
    a = .19 m/s/s
    b.) The Force of Friction - Ff = Force of Friction
    Ff = ma
    Ff = .2 kg(.19 m/s/s)
    Ff = .04N
    c.) Coefficient of Friction - U = Coefficient of Friction, FN = Force Normal
    U = Ff / FN FN = mg = .2 (9.8) = 1.96N
    U = .04 / 1.96
    U = .02

    2.) A 40 kg box is being pulled by a rope to the right with a tension of (T = 300 N) which is at a 37 degree angle to the horizontal. Another force of 50N is applied to the box pulling left perfectly horizontally. The coefficient of friction between the box and the floor is .4 The box is moving.

    a.) Find the Force with which the floor pushes up on the box
    Force Gravity = Force Pushing Up
    F = ma
    F = 40kg(9.8m/s/s)
    F = 392N = 390N with sigfigs

    b.) Find the acceleration of the box - Fr = Resisting force
    Tx = cos(Theta)T Ty = sin(Theta)T
    Tx = cos(37)(300) Ty = sin(37)(300)
    Tx = 240N Ty = 180.5N
    Fnet = Tx - 50N - Ff = ma
    Fnet = 240N - 50N - Ff = ma
    Fnet = 190N - FnUk = ma
    Fnet = 190N - 210N(.4) = ma
    Fnet = 190N - 84N = ma
    Fnet = 106N / m = a
    Fnet = 106 / 40 = a
    Fnet = 2.65 m/s/s = a


    Any help i could get on either of those two problems would be greatly appreciated...if i got them right, plz let me know, because i think I'm starting to get it. Thank You
     
    Last edited: Nov 9, 2004
  2. jcsd
  3. Nov 9, 2004 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You started out with the right formula, then messed up:
    Vf^2=Vi^2 + 2a(Xf-Xi)
    vf^2 - Vi^2 = 2a(Xf-Xi)
    a = [vf^2 - Vi^2]/[2(Xf-Xi)]
    a = [10^2 - 20^2]/[2(70)]
    No. You found the weight of the box, but you need to find the normal force between the floor and the box. Since the box is being pulling upward (consider the vertical component of the 300 N force), the normal force does not simply equal the weight.
     
  4. Nov 9, 2004 #3
    Re: Help

    Ok, so I put in the numbers for problem number one, and here is what I got.

    (Vf^2 - Vi^2) / [2(70)] = a
    (10^2-20^2) / [140] = a
    (-300) / [140] = a
    -2.14 = a = 2.14 m/s/s


    For the second problem, I think i found my mistake, correct me if im wrong.

    Ff = ma - Ty
    sin(Theta)T = Ty
    sin(37)(300) = Ty
    180N = Ty
    Ff = 40(9.8) - 180
    Ff = 390 - 180
    Ff = 210N
    So the upward force is 210 Newtons, not 390 Newtons

    I hope I got this right, and I also came across another challenging problem, devising of an equation for Acceleration on an inclined plane.

    Q: A toy car is released from rest at the top of a ramp going down to the right. Assuming that friction IS present derive an equation that could determine the acceleration of the car.

    SFy - sum of forces in the y direction
    SFx - sum of forces in the x direction
    Fn - Force Natural
    Fgy - Force Gravity in the Y-direction
    Fgx - Force Gravity in the X direction
    m - Mass
    a - Acceleration
    U - Coefficient of Friction
    Fg - Force of Gracity
    Ff - Force of Friction

    Here is my work, though I think i went wrong with the sin and cos values

    SFy = Fn - Fgy = 0
    Fn = Fgy
    Fn = cos(Theta)Fg
    SFx = Fgx - Ff = ma
    SFx = Fgx - FnU = ma
    SFx = Fgx = ma + FnU
    SFx = sin(Theta)Fg = ma +cos(Theta)Fg
    SFx = sin(Theta)Fg / cos(Theta)Fg = ma + U
    SFx = tan(Theta) = ma + U
    SFx = tan(Theta) - U = ma
    SFx = [tan(Theta) - U] / m = a

    Once again help would be appreciated..thankz for all help and corrections
     
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