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a) and b) $w=539N$
c) $w=717N$
d) $w=361N$
e) $w=0N$

Giancoli 6th Edition, Chapter 5, Problem 53


Chapter 5, Problem 53 is solved.

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Transcript for this Giancoli solution

For parts a and b when there's no acceleration that means that the normal force is going to equal to the weight on this woman in the elevator, so let’s draw the woman and the forces on her. We have the normal force up and the normal is going to be equal in magnitude to the perceived weight, and so we have normal force up and gravity down. Since the normal equals gravity we can say that it’s equal to ‘m’ times ‘g’ so we have fifty five kilograms times nine point eight and that gives us a spring scale reading of five hundred and thirty nine newtons for parts a and b. For part c it’s a different picture because she’s accelerating upwards so we’ll draw the normal force longer than the gravity because there’s acceleration upwards, there's an unbalanced force. So in this case we have the normal force positive because it’s up minus the gravity which is negative because it’s down equals mass times acceleration. Solving for normal force: mass times acceleration plus gravity which is ‘m’ times ‘g’ so we end up getting the acceleration is: zero point three three times ’g’ plus ‘m’ times ‘g’, factoring out ‘m’ times ‘g’ we have ‘m’ times ‘g’ times zero point three three plus one which equals one point three three times ‘m’ times ‘g’. ‘m’ times ‘g’ we calculated in part a so we’ll substitute that in and we have: one point three three times five hundred and thirty nine newtons giving us an answer of seven hundred and seventeen newtons which is that answer for part c. In part d she’s accelerating downwards so we’ll draw the normal force up but we’ll draw it fairly short this time because she’s accelerating downwards so we have: normal force minus gravity equals mass times acceleration and this is going to b negative. So we’ll solve for the normal force and it’s going to be: ‘Fg’ minus ‘ma’ which is ‘m’ times ‘g’ minus ‘m’ times zero point three three times ‘g’ which is ‘m’ times ‘g’ minus zero point three three which is sixty seven percent of ‘m’ times ‘g’ which we got in part ‘a’, so we have zero point six seven times five hundred and thirty nine newtons and this gives us three hundred and sixty one newtons which is the spring scale reading when she’s accelerating downwards. For part e since it’s a freefall there’s no normal force because there's only one force exerted on this lady at this point, gravity, normal force being the perceived weight so that’ll be zero. For the 5th Edition we have that the weight is fifty eight kilograms instead of fifty five so that changes the answer for a and b to five hundred and sixty eight newtons. So substituting this gives a 5th Edition answer of seven hundred and fifty six newtons for part c and three hundred and eighty one newtons for part d and part e has the same answer, no normal force so no apparent weight at all.


where is the "1" coming from in this problem?

in part C?

Hi LeeCReynolds, thanks for the comment. The "one" in part C is a result of factoring out the common factor "mg".

$0.33mg + mg = mg(0.33 + 1)$

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