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Can't get correct sign for a simple force problem

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    A tension force of 2.50 N acts horizontally on a 2.00 kg block. The block accelerates at 0.750 m/s2. What is the force of kinetic friction?

    2. Relevant equations
    Fnet = ma = F1 + F2 + F3 + . . .

    3. The attempt at a solution
    I define the reference frame to be, in terms of an x axis, positive to the right and negative to the left. Thus, the force of tension is positive because I define it to be in the right direction, and the force of friction should be negative since it opposes the tension.

    ##F_{fx}## = force of friction
    ##F_{tx}## = tension

    ##F_{netx} = ma = F_{tx} + F_{fx}##
    ##F_{fx} = ma - F_{tx}##
    ##\left \| \vec{F_{f}} \right \|cos(180^{\circ}) = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##
    ##-\left \| \vec{F_{f}} \right \| = ma - \left \| \vec{F_{t}} \right \|cos(0^{\circ})##
    ##\left \| \vec{F_{f}} \right \| = \left \| \vec{F_{t}} \right \|cos(0^{\circ}) - ma##
    ##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (2.00 ~kg)(0.750 ~m/s^{2})##
    ##\left \| \vec{F_{f}} \right \| = (2.50 ~N) - (1.50~N)##
    ##\left \| \vec{F_{f}} \right \| = 1~N##

    As we can see, I get positive 1 N rather than -1 N. Why is this? What am I doing wrong?
     
  2. jcsd
  3. Feb 10, 2015 #2

    Nathanael

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    Homework Helper

    You don't need to say what is "left" and what is "right." For example, I would of set up the directions as: "the positive direction is the direction of tension and the negative direction is the direction opposite of tension." At any rate, this doesn't affect your answer.

    I'm not sure why you're doing this... But this step is why it turns out positive instead of negative. All you did in this step was multiply Ff by -1 (without multiplying the other side by -1) which is why your answer came out altered by a factor of -1

    Anyway, doesn't the ||...|| mean "the magnitude of ... "? The magnitude of -1N is 1N.
     
  4. Feb 10, 2015 #3

    lightgrav

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    your absolute value Force is positive ... what direction does it push? (The sign of a vector tells its direction)
    ... I suggest you NOT use absolute values, and not use cos(180), so you get the F_x component with sign still attached.
     
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