1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    3. The attempt at a solution

    So I attempted this so far


    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
  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

  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


    User Avatar
    Homework Helper

    @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.

  9. Oct 8, 2014 #8


    User Avatar
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted