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Apparent Weight |
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| Nov7-07, 07:29 PM | #1 |
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Apparent Weight
1. The problem statement, all variables and given/known data
A man with a mass of 85.0 kg rides downward on an elevator.Initially the elevator accelerates downwards at 1.845 m/s^2 and eventually slows at 1.60 m/s^2 as it reaches the ground floor. Find the mans apparent weight as he starts down and as he reaches the ground floor. 2. Relevant equations fn=m(a-g) 3. The attempt at a solution fn= (85.0 kg) (1.85m/s^2--9.81m/s^2) =991.1 N fn=m(a-g) = (85.0 kg) (1.60 m/s^2--9.81 m/s^2) =969.85 N The question I have is in the second part of the question should th value for a be positive or negative? |
| Nov7-07, 07:46 PM | #2 |
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This is not quite right: you added instead of subtracted.
It might be clearer if you wrote it as: [tex]\Sigma F = F_n - mg = ma[/tex] So: [tex]F_n = m(a + g)[/tex] But note that the acceleration is downward and thus negative. |
| Nov7-07, 08:08 PM | #3 |
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So fn=85.0kg ( -1.85m/s^2 +(-9.81m/s^2)
=-991.1N fn=85.0kg (-1.60 m/s^2 + (-9.81 m/s^2) = -969.85N Is this correct? |
| Nov7-07, 08:21 PM | #4 |
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Apparent Weight
No, you're making the same mistake. In the expression "a + g", it's "a" (the acceleration of the elevator) that is negative, not g! (g is just a positive constant = 9.81 m/s^2.)
Always do a sanity check: If the elevator accelerates upward, you'd be squashed against the floor giving a greater apparent weight; if it accelerates downward, you'd be pulled away from the floor, reducing your apparent weight. |
| Nov7-07, 09:44 PM | #5 |
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fn =85kg (-1.85 m/s^2 +9.81m/s^2)
= 676 N fn = 85kg ( 1.6m/s^2 + 9.81m/s^2) = 969.85 N Is this correct? |
| Nov8-07, 05:22 AM | #6 |
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Excellent! The wording of the problem was a bit tricky. At first the elevator accelerates downward at a = -1.85 m/s^2. Then it slows down, which means it accelerates upward at +1.6 m/s^2.
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| Nov8-07, 07:07 AM | #7 |
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Thanks for the help !
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