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When is gravitational value - and/or +?

  1. Feb 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Projectile motion questions, Kinematics questions: my physics teacher defined acceleration due to gravity as -9.81 m/s2. Specifically in questions regarding: dropping an object, throwing an object downwards, obliquely launching an object.

    Today we are studying a box on an inclined plane question, and for some reason Fg = mg and the gravitational (9.81) was NOT negative.

    Can someone please explain:
    Let me give you my attempt at the question

    3. The attempt at a solution
    a box is on the top of a ramp with an inclination of 35˙. It is released from rest and it begins to accelerate down the ramp. The mass of the box is 1.5kg and the force of friction between the box and the ramp is 2.6N. What is the acceleration of the train down the ramp?

    Step 1: Find the force of gravity on the train (Fg on the diagram)

    Fg=mg =(1.5kg)(9.81) =14.7N
    Fg=mg =(1.5kg)(-9.81) =-14.7N

    Step 2: Find the Fgx

    Fgx= Fgsin =(14.7N)(sin35o) = 8.43N
    Fgx= Fgsin =(-14.7N)(sin35o) = -8.43N

    Step 3: Calculate the force of train.
    Fnet = Fgx – Ffriction
    Fnet = 8.43N - 2.6N
    Fnet= 5.83N

    Fnet = -8.43N - 2.6N
    Fnet= -11.03N

    Step 4: Find the acceleration


    -11.03N = (1.5kg)(a)
    a=-7.35 m/s2 WHICH IS WRONG!

    Please help.
  2. jcsd
  3. Feb 6, 2017 #2


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    well it depends on which direction you assign as positive or negative. In your inclined plane example, if you define the component of gravity down the plane as negative, then is the friction positive or negative?
  4. Feb 6, 2017 #3


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    As PJ explains, it is arbitrary.

    In one example, they chose positive y values to be "up" - in another they may not have been specific, just providing a magnitude. Choose the value that makes sense in the given scenario. The only rule is that you must be consistent within that scenario.

    If I were doing a 1 dimensional car speed scenario, I might choose East as positive velocity, and therefore West is the negative velocity. Car accelerating Eastward is positive values, Car declerating (accelerating Westward) is negative values.

    I could just as soon reverse those signs, deciding that acceleration Eastward is negative and acceleration Westward is positive - if it made more sense for the problem.
  5. Feb 6, 2017 #4
    If I define gravity as "+" in this example, then I can define friction as "+" too?
    Therefore, if I define gravity as "-", then friction has to be "-" because it's opposite to the movement?

    Now I am more confused
  6. Feb 6, 2017 #5


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    In general, "g" is a positive constant. The sign that you assign to it depends upon your choice of coordinate system for a given problem.

    When you begin a problem, the first thing you should always do is choose your coordinate system (frame of reference). Everything else with direction (i.e. vector quantities) then get assigned their signs according to that convention. In projectile motion problems, for example, it is common to choose "up" to be the direction of the positive y (or sometimes z) axis, and the horizontal direction of the projectile to be the positive x axis. In such a situation g is taken to be negative since gravity causes things to accelerate downwards, along the -y direction, according to the coordinates chosen.

    It is also possible to choose the +y direction to be downwards if it convenient to the problem. So for example if your projectile were thrown downwards from a plane or off of a cliff, you might choose your coordinate origin to be the launch point and downwards to be positive. In that case you would give g a positive sign since it points in the direction of +y.
  7. Feb 6, 2017 #6


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    If you assign the direction of gravity as positive, why would you assign the direction of friction as positive, if you know it is in the opposite direction? Or if you assign the direction of gravity as negative, why would you assign the direction of friction as negative, if you know it is in the opposite direction?
  8. Feb 6, 2017 #7
    When it comes to assigning +/-, you have to be consistent, as others have said. For example, if you want "down the ramp" to be the positive direction, then every vector pointing down the ramp has to be positive. If you haven't yet, make a force diagram. Any arrows facing the same way need to have the same signs. Any arrows facing opposite directions need to have opposite signs. Friction opposes motion, so it should not have the same sign as the acceleration due to gravity in this example.
  9. Feb 6, 2017 #8


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    Much excellent advice in the responses so far, but perhaps none quite address this:
    Since you are taking gravity as negative, you are using the convention that up is positive for vertical forces. But what does that mean for components along slopes, especially if, as may happen, you are not yet sure which way the slope goes?

    When you apply the component rule, i.e. taking the cosine of the angle between the vector (force, say) of interest and the direction vector of interest, the angle should be that between the positive directions you have chosen for the two vectors.

    Let the slope be θ, drawn as acute.
    If you take up slope as positive in the diagram then the angle between that and a vertically upward-positive force is π/2-θ, giving a factor cos(π/2-θ)=sin(θ). So your -8.43N is correct, but that is with up slope as positive.
    If you take downslope in the diagram as positive then the angle between that and the upward-positive force is π-(π/2-θ)), so the factor is now cos(π-(π/2-θ))=cos(θ+π/2)=-sin(θ).

    Now, all this can be a bit complicated to work through formally. Much easier just to check what makes physical sense. Does it make sense that the downslope component of gravity would be negative?
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