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Why does the g (acceleration) go in the opposite direction?
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OH is it because by the time the particle q moves up, it will be under the influence of gravity as the velocity will be 0 mid air?
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(Original post by shohaib712)
OH is it because by the time the particle q moves up, it will be under the influence of gravity as the velocity will be 0 mid air?
OH is it because by the time the particle q moves up, it will be under the influence of gravity as the velocity will be 0 mid air?
Until P reaches the ground, Q is accelerating upwards with a +ve acceleration that satisfies
.When P reaches the ground, the string becomes slack since there's nothing to drag the string down anymore, hence tension becomes 0. For Q this means there is no more tension acting up and only the one force is acting down, namely the gravitational force. SO in this process of motion, you get that
for it.
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(Original post by RDKGames)
Gravity is ALWAYS acting down. If you choose upwards to be +ve then your gravity acceleration is always -ve.
Until P reaches the ground, Q is accelerating upwards with a +ve acceleration that satisfies.
When P reaches the ground, the string becomes slack since there's nothing to drag the string down anymore, hence tension becomes 0. For Q this means there is no more tension acting up and only the one force is acting down, namely the gravitational force. SO in this process of motion, you get that for it.
Gravity is ALWAYS acting down. If you choose upwards to be +ve then your gravity acceleration is always -ve.
Until P reaches the ground, Q is accelerating upwards with a +ve acceleration that satisfies
.When P reaches the ground, the string becomes slack since there's nothing to drag the string down anymore, hence tension becomes 0. For Q this means there is no more tension acting up and only the one force is acting down, namely the gravitational force. SO in this process of motion, you get that
for it.
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(Original post by RDKGames)
Gravity is ALWAYS acting down. If you choose upwards to be +ve then your gravity acceleration is always -ve.
Until P reaches the ground, Q is accelerating upwards with a +ve acceleration that satisfies.
When P reaches the ground, the string becomes slack since there's nothing to drag the string down anymore, hence tension becomes 0. For Q this means there is no more tension acting up and only the one force is acting down, namely the gravitational force. SO in this process of motion, you get that for it.
Gravity is ALWAYS acting down. If you choose upwards to be +ve then your gravity acceleration is always -ve.
Until P reaches the ground, Q is accelerating upwards with a +ve acceleration that satisfies
.When P reaches the ground, the string becomes slack since there's nothing to drag the string down anymore, hence tension becomes 0. For Q this means there is no more tension acting up and only the one force is acting down, namely the gravitational force. SO in this process of motion, you get that
for it.
https://www.youtube.com/watch?v=SGEscXwQUFw
Why does he make F=2T? Shouldnt the pully also be affected by the mass of the particles and doesnt the T cancel out as they are affecting the string in opposite directions?
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