How Does Elevator Physics Affect the Work Done by a Cable?

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titansarus
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Homework Statement


Question from fundamental of physics, Halliday Resnick Walker
In Figure below, a ##m=0.250## kg block of cheese lies on the floor of a ##M=900 kg## elevator cab that is being pulled upward by a cable through distance ##d1 =2.40 m## and then through distance ##d2 = 10.5 m##. (a) Through d1, if the normal force on the block from the floor has constant magnitude ##F_N## = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d2,if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of ##F_N##?
halliday phy.png

Homework Equations


##W = F d## (in 1 dimension)
##\Sigma F = m a##
##\vec F_{12} = - \vec F_{21}##

3. The Attempt at a Solution

I don't have problem at the solution itself, My problem is with the solution of the book.
Answer of the book:
answer halliday.png
My solution: ##F_N - mg = m a## (##*## equation) and we get ##a = 2.2 m/s^2## now for the whole system, I write ##F - (m+M)g = (m+M)a## (##**## equation) and by substituting, I get ##F = 10800 N##. The book also get the same approximate answer but It wrote ##F + F_N - (m+M) g = (m+M) a##. I think these ##F_N## part is wrong because when we are speaking about the whole system ##F_N## and its reaction (Newton 3rd Law) cancel each other. (i.e. they are Internal Forces for the system). Also for the second part, I think we must find F with ##W = F d## and then find ##a## using (##**## equation) and then find ##F_N## with (##*## equation). Is my solution correct?
 

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titansarus said:
I think these ##F_N## part is wrong because when we are speaking about the whole system ##F_N## and its reaction (Newton 3rd Law) cancel each other. (i.e. they are Internal Forces for the system). Also for the second part, I think we must find F with ##W = F d## and then find ##a## using (##**## equation) and then find ##F_N## with (##*## equation). Is my solution correct?
You are correct.
 
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