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Confusion about tension

  1. Nov 8, 2015 #1
    For question about the falling object
    we will always take T=mg+ma
    so we take mg and ma positive

    for question about the object rolling down the inclined plane
    we will say that mg sin theta - ma=0
    although mg sin theta and ma is still on the same direction, we take mg sin theta positive but we take ma negative?

  2. jcsd
  3. Nov 8, 2015 #2
    If the object in the pulley system is falling, then your statement is wrong. We take mg-T=ma. This is with the convention of taking the downward direction positive. The second statement is correct.
  4. Nov 8, 2015 #3
    but i found the first statement from this textbook

    Attached Files:

  5. Nov 8, 2015 #4
    I think the difficulty is to understand that there is a qualitative difference between the term m⋅a ("inertia") and the rest of the terms (outer forces). I would recommend to wright the equation with m⋅a on one side and the rest on the other. Then you get:

    1) m⋅a = T - m⋅g
    2) m⋅a = m⋅g sinθ

    On the left side you now can find the inertia/acceleration of the mass, on the right side the sum of all forces which affect the mass (and accelerate it). Depending on how you defined the positive direction of the coordinate the signs of the forces of the right side can be changed.

    In case 1) the positive direction is upwards: If the tension in the rope is larger than the weight of the mass, the mass will accelerate up (in direction of the tension in the rope).
    In case 2) the positive direction is parallel to the ramp and downwards.
  6. Nov 8, 2015 #5
    A lot of confusion is generated by not stating your coordinate system. If you are writing Newton's 2nd law, then irrespective of the coordinate system:

    T + W = ma, where the tension, the weight and the acceleration are vectors. Note that there are no negative signs here. I wrote the vectors in bold font.
    Now, if you specify the coordinate system, and take the upward direction as positive, then the components of the above equation in the vertical direction give you
    T - mg = ma. In this equation, a is an algebraic quantity. If the tension is larger than the weight, then a is positive. If T is smaller than mg, then a is negative.

    For the inclined plane without friction, again, first write Newton's second law for the forces
    N + W = ma (no negative signs). N is the normal force, and W is the weight
    Take the x-axis down the inclined plane, and the y axis perpendicular to the plane, along the direction of the normal force. Take components, with θ the angle of the incline above horizontal.
    N - mg cos(θ) = 0 (y-components)
    mg sin(θ) = ma (x-components)
    The x-component equation is precisely what you wrote. Note that this does not mean a is negative. It is actually positive. Generally, everything with a minus sign in front of it is not necessarily negative. These are all algebraic quantities, and can stand for positive or negative quantities, with or without a negative sign in front of them.
  7. Nov 8, 2015 #6
    what if there is another force F against direction of mgsin(θ)?
    i got this equation from my textbook(the object is rolling down)
  8. Nov 8, 2015 #7
    For example friction?

    m⋅a = m⋅g sinθ - m⋅g⋅cosθ ⋅μ

    Now the friction "accelerates" the mass upwards, hence decreasing the acceleration due to the gravity. If the friction force is as large as m⋅g sinθ, the terms cancel out and the mass stays where it is.
    Last edited: Nov 8, 2015
  9. Nov 8, 2015 #8
    It does not matter. Follow the same steps every time, and you will not go wrong.
    0. Draw the free-body diagram with all forces.
    1. Write down Newton's 2nd (sum of all forces = mass x acceleration). Remember this is a vector equation
    2. Choose a coordinate system.
    3. Take components of Newton's 2nd.
    4. Solve for the unknown.

    If you like, try it and let me know if you need more help.
  10. Nov 9, 2015 #9
    The confusion arises because we do not distinguish between force "F" , which arises because of interaction of the object with some thing external to the object and "m*a", which also has the dimensions of force. In ma, a is the acceleration of the object as such. This is further compounded by our usual practice of taking gravitational force acting on the object as mg. This is conceptually completely true when the object is free falling; with acceleration g, which follows from the Newton's second law of motion.but at other times it means GMm/r^2, that is g is taken to be equal to GM/r^2.

    Your original statement has to be viewed in this perspective, you say mgsin theta - ma = 0 you say mgsin theta and -ma have opposite direction. But it actually means a has the same direction as mg sin theta. One should understand that ma in Newton's second law is numerically equal to force but is it self not force. It is mass times acceleration and Newton's second law states that it must be equal to net force acting on the object by external means.
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