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March 28, 2024, 07:22:10 pm

Author Topic: Forces in an elevator  (Read 529 times)  Share 

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Adapt

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Forces in an elevator
« on: May 07, 2019, 08:21:20 pm »
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Hey guys,

Got one question that is confusing me. For part b, the solution says that the force is 18.9 N downwards, but I don't understand how to calculate that?
From my understanding, the sum of all forces on A would be 10.5N upwards. I'm assuming that the sum of all forces on A would just be its weight force plus normal force, and that the normal force A experiences would be the force that B exerts on A. Letting upwards be positive, I would then assume 10.5 = -29.4+x, where x is the normal force that A experiences, or the magnitude of the force that A exerts on B. However, this gives a final answer 39.9 N downwards, could someone please explain this?
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fun_jirachi

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Re: Forces in an elevator
« Reply #1 on: May 07, 2019, 09:22:05 pm »
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Hi there :)

For part b), the force due to gravity exerted on B by A is equal to (mass of A) x g ie. 3x9.8=29.4 N downwards. Now, when the elevator accelerates downwards, think of it as inducing a sort of reaction force equal to the mass of the object by the acceleration from F=ma, but in the opposite direction. (though this is oversimplified and not technically how it works). This gives you a force upwards of (mass of A) x acceleration, which is 3x3.5=10.5 N upwards. Add these up and you have the correct answer of 18.9 N down. You have the wrong answer because you're not resolving the vectors correctly ie. you've factored in the 'reaction' force, just in the wrong direction.

Hope this helps :)
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Adapt

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Re: Forces in an elevator
« Reply #2 on: May 07, 2019, 10:03:24 pm »
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oh my god, this whole time I thought it was accelerating upwards :( Thanks!
2018: Methods 45
2019: Physics 47 | Specialist 41 | Further 49 | English 42 | Religion & Society 41
ATAR: 99.55