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    I recognise that they both have the same magnitude but they just act in different directions.

    I have always wondered why the acceleration always acts in the opposite direction to the 'ma' force.

    This is best shown in circular motion where \ddot{\theta} = r\omega ^{2}. The acceleration is inwards but the force \left| m\ddot{\theta} \right| = mr\omega ^{2} is outwards in order to restore against the acceleration inwards. So in this case the ma force is both inwards and outwards. But isn't the ma a restoring force to ensure equilibrium as a result of acceleration?

    Please tell me if this is just a load of hot air I'm talking.
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    In the case of circular motion, there is a continuous inwards acceleration which is in the same direction as the central force. If the force disappears then the body will continue on a straight line course as Newton's first law says.

    Acceleration is in the same direction as force. In the case of a diagram, it is convenient to balance acceleration in against force out but this is a fiction.

    I do not know why you think ma is opposing acceleration, perhaps you could post a diagram or example.
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    In circular motion the radial acceleration is of magnitude r\omega ^{2} and is directed inwards as you stated. Therefore in the ma assumption the force should also be inwards
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    In circular motion the radial acceleration is of magnitude r\omega ^{2} and is directed inwards as you stated. Therefore in the ma assumption the force should also act inwards and be of magnitude mr\omega ^{2}.

    Let us take an example of a particle of mass m attached a light, inextensible string of length r and that the particle travels with angular velocity \omega in a horizontal circle on a smooth surface. The particle accelerates inwards with an acceleration r\omega ^{2}, however when resolving the forces on the ball the mr\omega ^{2} force acts outwards and the tension in the string acts inwards with equal magnitude.

    In this example the 'ma' force is not in the direction of the acceleration.

    I have found this article on the matter if it helps. This idea is called the d'Alembert force.
    http://www.britannica.com/science/dAlemberts-principle

    Another example is this. If two horizontal forces F and R act upon a particle of mass m that is at rest on a smooth horizontal surface in opposite directions, where F>R. Then the following equation is obtained:
    F-R=ma
    Therefore,
    F=ma+R
    In this instance the ma force must be acting in the opposite direction to maintain equilibrium as
    F-ma-R=0
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    (Original post by Cryptokyo)
    In circular motion the radial acceleration is of magnitude r\omega ^{2} and is directed inwards as you stated. Therefore in the ma assumption the force should also be inwards
    Correct, as I stated although the ma inwards is not an assumption, it is Newton's first law.
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    (Original post by Cryptokyo)
    In circular motion the radial acceleration is of magnitude r\omega ^{2} and is directed inwards as you stated. Therefore in the ma assumption the force should also act inwards and be of magnitude mr\omega ^{2}.

    Let us take an example of a particle of mass m attached a light, inextensible string of length r and that the particle travels with angular velocity \omega in a horizontal circle on a smooth surface. The particle accelerates inwards with an acceleration r\omega ^{2}, however when resolving the forces on the ball the mr\omega ^{2} force acts outwards and the tension in the string acts inwards with equa magnitude.

    In this example the 'ma' force is not in the direction of the acceleration.

    I have found this article on the matter if it helps. This idea is called the d'Alembert force.
    http://www.britannica.com/science/dAlemberts-principle

    Another example is this. If two horizontal forces F and R act upon a particle of mass m that is at rest on a smooth horizontal surface in opposite directions, where F>R. Then the following equation is obtained:
    F-R=ma
    Therefore,
    F=ma+R
    In this instance the ma force must be acting in the opposite direction to maintain equilibrium as
    F-ma-R=0
    The force is inward and the acceleration is inward which is fine.

    If you set up the situation as a balance of forces as you have & presumably EB has), this is a convenient fiction that leads to the correct answer.

    The Physics of the situation is that the Tension provides an inwards force that accelerated the mass inwards.

    You can dress the situation up in all sorts of frames of reference but to be honest, they serve to confuse rather than illuminate.
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    (Original post by nerak99)
    The force is inward and the acceleration is inward which is fine.

    If you set up the situation as a balance of forces as you have & presumably EB has), this is a convenient fiction that leads to the correct answer.

    The Physics of the situation is that the Tension provides an inwards force that accelerated the mass inwards.

    You can dress the situation up in all sorts of frames of reference but to be honest, they serve to confuse rather than illuminate.
    This is the part of M3 that I found rather bonkers.
    Thank you very much for your help.
    Cryptokyo.
 
 
 
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