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Homework Help: Painter and a Chair - Force

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data

    A house painter uses the chair and pulley arrangement of the figure to lift himself up the side of a house. The painter's mass is 74.5 kg and the chair's mass is 10.3 kg. With what force must he pull down on the rope in order to accelerate upward at 0.21 m/s^2.


    what is the magnitude of the force on the painter by the chair?

    2. Relevant equations

    newtons 2nd law

    3. The attempt at a solution

    I found the first part. The force he must pull down with is 424 N and that is correct.

    I cannot find the question that is in bold about "The magnitude of the force on the painter by the chair"

    This is what i tried but im not sure if it is correct.

    w = 74.5*(9.8+.21) = 746 N

    does that look right?

    thank you
  2. jcsd
  3. Feb 14, 2010 #2


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    I think your answer is right, although I'm not sure whether you understand why. If the painter were in equilibrium, the chair would be pushing upward on him with just enough force to balance his weight. However, since he is accelerating upwards, this suggests that the chair is pushing up on him with a force greater than his weight (just like you feel heavier when you're in an elevator and it first starts to move upward). To figure out with how much force the chair is pushing up on the man, just use Newton's Second Law, but this time only for the painter, in order to figure out what the net force on him must be if he is to accelerate upwards at 0.21 m/s2. Once you know the net force, you can use that plus the weight of the painter to figure out the upward force on the painter.
  4. Feb 14, 2010 #3
    thank you that makes a little more sense the way you explained it.

    except i just tried that answer and it said it was wrong?

  5. Feb 14, 2010 #4
    is it maybe half of that answer because he is pulling himself up with the rope?
  6. Feb 14, 2010 #5


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    Interesting! That seems to be how you get the 424.84 N in the first part as well, although to be honest I'm not 100% sure I understand it myself.

    The only thing I can think of is that maybe the following reasoning applies:

    If you look at a free body diagram of the painter, there are actually three forces acting on him, not two. There's the force upwards due to the tension in the part of the rope that attaches to the chair (EDIT: No, this is a free body diagram for the painter only, so the upward force technically comes from the seat pushing up on him), there's gravity downwards, and then there is the reaction force on him at the point where his hand makes contact with the rope.

    By Newton's third law, the force with which which he pulls down on the rope is equal in strength and opposite in direction to the force which with the rope pulls up on him at the contact point. Hence, this reaction force should be upward, and there's actually therefore [an additional upward force on the painter - EDITED].

    Does that make any sense? Again, I'm not 100% sure of it...
    Last edited: Feb 14, 2010
  7. Feb 14, 2010 #6
    so since his force he is pulling down with is counteracted by the force pushing him up then those two forces would cancel out correct?

    or since its twice the tension would you add more to the 745?

    im really confused now haha. im just not sure what the answer is.

    do you think it is half of the 746? so 373 N
  8. Feb 14, 2010 #7


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    No, the comment about 'twice the tension' was applicable to the system (chair + painter) as a whole and was just meant to explain why the answer was 424 N in the first part and not 848 N (which is the answer you would get by considering the force balance argument I gave in post #2 alone).

    I've since edited my post (#5). So, re-read it, and draw a free body diagram for the painter. Three forces act on him:

    1. gravity downwards

    2. The force from the seat upwards (which is what you are trying to calculate)

    3. The reaction force from the rope on his hand (which is equal and opposite to his applied pulling force by Newton's third law, and hence is also upward).

    How must these forces balance (the two upward, and the one downward) in order for the net force ma for the painter to be given by: (74.5 kg)(0.21 m/s2) upward?

    Based on that force balance, you can figure out what the unknown force that you are trying to calculate must be.
  9. Feb 14, 2010 #8
    so you have 730 N down due to gravity

    424 N up due to the reaction force from the rope on his hands

    so subtracting those two you get 306 N up

    but for him to have a net force of 15.6 up wouldnt you have to also consider this into the 306 N.

    so would you go 306 N + 15.6 N?

    thank you for all of the explanations
  10. Feb 14, 2010 #9


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    I didn't check your numbers, but in short the answer to your question is 'yes'. If The net force is 15.6 N upward, then clearly all of the upward forces must add up to a number that is 15.6 N higher than what all of the downward forces add up to. That's what 'net' means: the total, taking everything into account.

    Now, the big question is, does Mastering Physics agree that 322 N is the right answer? ;)
  11. Feb 14, 2010 #10
    It sure does agree :)

    Thank you so much for all the help. it really makes perfect sense now. thank you!
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