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    Hi, I was doing some mechanics questions on moments and I came over this one (Q8b on this paper) I don't understand why for moments- in this example- why the particle m doesn't have a normal reaction on the rod? Thanks
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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    (Original post by examstudy)
    Hi, I was doing some mechanics questions on moments and I came over this one (Q8b on this paper) I don't understand why for moments- in this example- why the particle m doesn't have a normal reaction on the rod? Thanks
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    In moments, particles only have weight, a downwards force. According to Newton's third law there must be an equal and opposite reaction, this is exerted by the supports in an upwards direction (unless the question specifies that the force exerted by the support is ftdownwards.) The total upward forces would then equalise the total downward forces when in equilibrium.
    Hope it helps.
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    (Original post by Thachi3)
    In moments, particles only have weight, a downwards force. According to Newton's third law there must be an equal and opposite reaction, this is exerted by the supports in an upwards direction (unless the question specifies that the force exerted by the support is ftdownwards.)
    Just one slight clarification here (not one that is necessary to help solve the problem) - the weight of a particle and the reaction from the beam/support/etc/ on the particle may be equal in magnitude and opposite in direction, but they are not Newton's third law partner forces. Such partner forces must be of the same type, and that isn't the case here; the weight of a particle is a gravitiational force, while the normal reaction is a contact force. The Earth pulls on the particle, causing a gravitational force on the particle (it's weight), and in return, the particle pulls on the Earth, with a force that is equal in magnitude and opposite in direction. Both forces are gravitational. Now, because the particle is in contact with a beam, this weight causes the particle to push down on the beam, exerting a contact force on it, and in accordance with Newton's third law, the beam pushes back upwards on the particle with another contact force, the normal reaction. These two forces are the Newton's third law partner forces.

    A subtle distinction, which makes no practical difference to the solution of problems like these, but one that is very important to people taking Physics, where questions like this are often set to trap the unwary!
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    (Original post by Pangol)
    Just one slight clarification here (not one that is necessary to help solve the problem) - the weight of a particle and the reaction from the beam/support/etc/ on the particle may be equal in magnitude and opposite in direction, but they are not Newton's third law partner forces. Such partner forces must be of the same type, and that isn't the case here; the weight of a particle is a gravitiational force, while the normal reaction is a contact force. The Earth pulls on the particle, causing a gravitational force on the particle (it's weight), and in return, the particle pulls on the Earth, with a force that is equal in magnitude and opposite in direction. Both forces are gravitational. Now, because the particle is in contact with a beam, this weight causes the particle to push down on the beam, exerting a contact force on it, and in accordance with Newton's third law, the beam pushes back upwards on the particle with another contact force, the normal reaction. These two forces are the Newton's third law partner forces.

    A subtle distinction, which makes no practical difference to the solution of problems like these, but one that is very important to people taking Physics, where questions like this are often set to trap the unwary!
    That is correct, apologies for referring to the law in a wrong way.... and am aware of the traps in physics.... thanks for the reminder...
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