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Work done on an elevator cab

  1. Jun 27, 2014 #1
    The problem
    A [itex]0.250 \operatorname{kg}[/itex] block of cheese lies on the floor of a [itex]900 \operatorname{kg}[/itex] elevator cab that is being pulled upward by a cable through distance [itex] d_1 = 2.40 \operatorname{m}[/itex] and then through distance [itex]d_2=10.5 \operatorname{m}[/itex]. Through [itex]d_1[/itex], if the normal force on the block from the floor has constant magnitude [itex]F_N = 3.00\operatorname{N}[/itex], how much work is done on the cab by the force from the cable?
    From Fundamentals of Physics, 9th Editon, Problem 25, Chapter 7

    My solution manual
    Here is what my solution manual says:
    The net upward force is given by [itex]F + F_N-(m+M)g = (m+M)a[/itex] where [itex]m=0.250 \operatorname{kg}[/itex] is the mass of the cheese, [itex]M = 900 \operatorname{kg}[/itex] is the mass of the elevator cab, [itex]F[/itex] is the force from the cable, and [itex]F_N = 3.00 \operatorname{N}[/itex] is the normal force on the cheese. On the cheese alone, we have [itex]F_N - mg = ma[/itex], and etc..., the solution continues

    My question
    I do not see why the first equation is correct. To me, the force [itex]F_N[/itex] is internal between the cab and the block, so once one considers both as a system and applies the Newton's second law, the equation should read [itex]F - (m+M)g = (m+M)a[/itex].

    Is the solution manual wrong, or am I overlooking something?
    Thank you.
     
  2. jcsd
  3. Jun 27, 2014 #2

    Lok

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    It's about the notation as you could have done it your way, that is split the F into Fn(cheese) Fc(cab) and you would get the same result. In your notation you just consider F as the total force not just the force that acts on the cab. The question is how much work is done on the cab, and not cab + cheese.
     
  4. Jun 27, 2014 #3

    PhanthomJay

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    Science Advisor
    Homework Helper
    Gold Member

    Putting aside the issue of work for a moment, the solution manual for the calculation for the force in the cable is incorrect. Your equation is correct.
    Alternatively, if you want to calculate the cable force by first incorporating the normal force exerted by the cheese on the cab, then you must not include the cheese weight in the formula, because it is not part of the FBD for forces acting on the cab.

    So, use either
    [itex]F - (m+M)g = (m+M)a[/itex], or


    [itex]F - F_N - Mg = Ma[/itex].
     
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