# Why does the g (acceleration) go in the opposite direction?

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Thread starter 4 years ago
#1
4:31 of the video
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Thread starter 4 years ago
#2
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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4 years ago
#3
(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?
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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Thread starter 4 years ago
#4
(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.
Oh i see so the same would apply for (lets say q had a mass of 4kg) then if the string became slack on p, then there would be no tension on q so its resultant force would be: 4a=0-4g and we would still get get a = -g. So bassically the reason why gravity is the only force acting on the particle is because the string has no tension as it isnt taut aymore?
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Thread starter 4 years ago
#5
(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.
in the last question in this video:

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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