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Why does the maximum tension in the cable occur when it is accelerating? watch

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    Just for part ii) Why does the maximum tension in the cable occur when it is accelerating and not in any of the other phases? It says that in the mark scheme for ii. Many thanks
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    Well, given that all forces acting on the object are vectors with directions, you know that the forces upwards will counteract the forces downwards. Hence the equation:
    T - mg = ma (where mg is the weight of the object and ma is the resultant force in the upwards direction)

    notice that this is the equation used in the answer

    Rearranging this gives:

    T = mg + ma

    This means that if the object wasn't accelerating upwards, travelling at a constant speed, the tension would become:

    T = mg + m(0) = mg

    And if the object was decelerating, the tension would become:

    T = mg - deceleration

    Since tension is amplified by any force in the same direction, the phase at which it has the maximum tension will be when it has maximum acceleration.
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    (Original post by Maivnek)
    Well, given that all forces acting on the object are vectors with directions, you know that the forces upwards will counteract the forces downwards. Hence the equation:
    T - mg = ma (where mg is the weight of the object and ma is the resultant force in the upwards direction)

    notice that this is the equation used in the answer

    Rearranging this gives:

    T = mg + ma

    This means that if the object wasn't accelerating upwards, travelling at a constant speed, the tension would become:

    T = mg + m(0) = mg

    And if the object was decelerating, the tension would become:

    T = mg - deceleration

    Since tension is amplified by any force in the same direction, the phase at which it has the maximum tension will be when it has maximum acceleration.
    Thank you very much But what do you mean by T = mg - deceleration ?
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    (Original post by h26)
    Thank you very much But what do you mean by T = mg - deceleration ?
    Maybe I wrote it in a confusing way, sorry about that. Deceleration is actually just negative acceleration (or acceleration in the opposite direction). So if you have the equation:

    T = mg + ma (where a = 0.05 ms^-2)

    and replace acceleration with deceleration, you get:

    T = mg + m(-a) = mg - ma (where a = 0.1 ms^-2)

    What I meant was that for when the object is decelerating, tension is equal its weight minus the value of its deceleration times its mass.
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    (Original post by Maivnek)
    Maybe I wrote it in a confusing way, sorry about that. Deceleration is actually just negative acceleration (or acceleration in the opposite direction). So if you have the equation:

    T = mg + ma (where a = 0.05 ms^-2)

    and replace acceleration with deceleration, you get:

    T = mg + m(-a) = mg - ma (where a = 0.1 ms^-2)

    What I meant was that for when the object is decelerating, tension is equal its weight minus the value of its deceleration times its mass.
    I understand it now Thank you very much for the help! Really appreciate it So Tension is greatest when accelerating, then in the middle when stationary and lowest when decelerating, right?
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    When the acceleration is stationary and its travelling at a constant speed, then yes it will be in the middle out of all three phases.

    You got it!
    :congrats:
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    (Original post by Maivnek)
    When the acceleration is stationary and its travelling at a constant speed, then yes it will be in the middle out of all three phases.

    You got it!
    :congrats:
    Thanks ever so much!
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    (Original post by h26)
    Thanks ever so much!
    I now realise that whether the object is at rest (0 velocity) or moving at a constant speed, its tension will be the same. So it really didn't matter whether or not you meant stationary motion or stationary acceleration.

    All the best!
 
 
 
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