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Newton's 3 Laws Problem

  1. Oct 7, 2014 #1
    1. The problem statement, all variables and given/known data
    The 20.0 kg box is being pushed with a 45.0N Force acting at an angle of 25.0*. The coefficient of kinetic friction on this surface is 0.150. Find the acceleration

    2. Relevant equations
    F=ma
    Fy=0
    Fx=0

    3. The attempt at a solution

    So I attempted this so far

    (sum)Fy=0
    Py+Fn-Fg=0
    Fn=Fg-Py
    Fn=(20.0kgx9.81m/s^2)-45.0Nsin335*
    Fn=215.217822N

    I tried to go and do sum of all the forces on the x axis but there's no point because we already have Ffk and Px.......... So now do I go ahead and put all of the forces in for Sum of F in
    (sum)F=ma and solve for a?
    as in would I do
    Fx-Fg+Fn-Ffk=ma???
     
  2. jcsd
  3. Oct 7, 2014 #2
    Well, actually, as my physics professor always says, do not put in the numbers, if not when you've arrived to the last formula.

    So...The box is moving on the x axis, so we know that the sum of forces on the y axis is 0.

    We have |Fp| = |N| +|Fy| With fp being the force exerted by gravity on the box, N the reaction that the table/ground is exerting on the box and Fy the y component of F.

    We get that N = |Fp|-|Fy|

    By definition we know that the friction force, let's call it Fat, is N*μ

    |Fat| = [mg-F*sin(25)] * μ

    In this case, checking the forces acting on y, is just needed to get the friction force right.

    The block is moving on the x axis, so it has forces acting on it, and an acceleration in that case, the forces acting are Fx (x component of F), and the force of friction.

    Fcos(25)-([mg-F*sin(25)]μ) = m*a

    The acceleration a is equal to (Fcos(25)-([mg-Fsin(25)]μ))/ (m) [m/s^2]

    At this point plug in all the numbers on the calculator and you get what you're searching for, with an human precision.
     
  4. Oct 7, 2014 #3
    Wouldn't the angle being used be -25* rather than positive??
     
  5. Oct 7, 2014 #4
    Why should it be -25?

    Isn't it like this

    https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/10726541_838742206166089_1666516233_n.jpg?oh=83223884bd9c63131013f875e990f880&oe=543720A0&__gda__=1412890032_b43d01452f1154c45db8a98041885387
     
  6. Oct 7, 2014 #5
    Using your formula I got .568 m/s^2....
    Does that sound right?
     
  7. Oct 7, 2014 #6
    Well if you've got the solution for that we would know :p

    Well, it can be anything, we don't know prior...We just know it isn't negative, as the friction force can block it, not drag it.
     
  8. Oct 8, 2014 #7

    ehild

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    Homework Helper
    Gold Member

    @Bedeirmuir: You must not give (almost) full solution.



    In the problem text, the box is pushed at the angle of 25°. So it is like in the following picture.

    ehild
     
  9. Oct 8, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Is that angle enclosed with the horizontal or with the vertical?
    I see, that 25°was counted from the horizontal. You got Fn rigth.

    The acceleration is a vector and its x component satisfies the equation ##ma_x=\sum{Fi_x}## So you need Px and Ffk. No sense to add x components with y components.

    ehild
     
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