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Real life problem - pushing object up incline plane

  1. Sep 8, 2010 #1
    This is not a homework question but rather a real life problem I need to work through. Sorry if I have used incorrect terms but I have not done any physics in 10 years and have forgotten everything so I was hoping someone would be able to help.

    1. The problem statement, all variables and given/known data

    How much effort would it take to move a) 600kg, b) 900kg object (on wheels) up a 1 / 100 incline plane (pretending friction = zero). Could 1-2 persons push it up? Or would I need mechanical assistance to get it up. The total distance to be moved is 2m.

    The object is a 6ft long fish tank. The 900kg weight will be filled, and the 600kg is partially emptied. I'm not sure how much friction - the floor surface is tiled. Yes, I'm aware the water will exert force on the glass but I'm hoping it can be moved slow enough for this to not be a major drama (especially if the tank is partially emptied).

    2. Relevant equations


    3. The attempt at a solution

    1/100 slope = x degrees
    tan x = 1/100
    x = 0.57 degrees ???

    F = W . sin x
    = 900 x 9.8 x sin 0.57
    = 88.2 N

    I'm not sure if this is right and I have no idea whether a person can push 88 N + the effects of friction...

    Can anyone help with this odd request? thanks
     
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2

    cepheid

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    Welcome to PF:

    First of all, here's how to solve for the angle of the incline in degrees. It's basic trigonometry. Viewed from the side, your ramp makes a right-angled triangle. If the vertical height of the ramp at some point along it is "x" metres (or feet if you prefer), then the distance along the base up to that point is 100*x metres (or feet). This just comes simply from the information you gave me about the gradient.

    So we can use trigonometry to solve for the angle of inclination. What I just said can be summarized as follows. You have the side of the triangle that is opposite to the inclination angle (the height), and the side of the triangle that is adjacent to the inclination angle. The RATIO of these two sides is 1/100 = 0.01. This ratio is also known as the TANGENT of the inclination angle. We can use the tangent of the inclination angle to solve for the angle itself since:

    tan(angle) = 0.01

    ==> arctan(0.01) = angle

    where the arctangent is the inverse operation to the tangent. Get out your calculator and you'll see that the angle is about 0.6 degrees.
     
  4. Sep 8, 2010 #3
    hi cepheid, thanks for your reply

    You've confirmed that I managed to calculate the angle correctly (after I thought about it for a while :) )

    I've had a go at solving the force - are you able to check my equation?

    thanks
     
  5. Sep 8, 2010 #4

    cepheid

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    If by "effort" you mean force, the the answer is that you will have to push with a force slightly higher than the component of the tank's weight that is trying to pull it down the ramp in order to get it moving. Once it is moving, it is sufficient to push with a force equal to that component. Think of the weight as a force vector (arrow) that points straight downward from the object. This vector can be considered to be the sum of two components, one that acts parallel to the ramp, and one that acts perpendicular to the ramp. Draw a picture. So, once again we have a right-angled triangle, and simple trigonometry tells us that the force pulling along the ramp is:

    weight*sin(0.57 deg).

    EDIT: I see you've edited your post and done this computation. In that case, to get some intuition for this amount of force, convert it into pounds or whatever units are used to measure weight in your country:

    http://www.google.ca/search?hl=en&s...newtons+in+pounds&aq=f&aqi=&aql=&oq=&gs_rfai=

    This will help because many people have a sense of what fraction of their weight they can exert in a pushing motion using their upper body. It doesn't seem like very much force is required to me.

    Another issue is whether you'll have enough traction to walk up the ramp. Recall Newton's third law: if you push on the tank, the tank pushes on you. The only thing preventing you from slipping is friction between your shoes and the ramp surface.
     
  6. Sep 8, 2010 #5
    Thanks again for your response

    So I calculated 88.2N to be equal to 9kg x accelleration, which sounds like it could be pushed by a single person quite easily. Is that the formula you mean to use F = ma ?

    Having traction on the tiles with shoes should not be a problem as they are outdoor rated with some grip. But this same traction will be causing friction problems on the wheels on the tank.

    I'm a little confused about the force required to get the tank moving in the first place though. Would pushing with 88N be enough to make the tank stop moving downwards and stop still. And then I would need to exert a further 88N to get it moving upwards? Or just any extra N over the 88 would get it moving upwards?
     
  7. Sep 8, 2010 #6

    cepheid

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    Actually I just meant convert 88 N into pounds or whatever unit of force/weight you're most familiar with. It works out to about 20 lbs, which I could easily exert, and I'm a not a big guy.

    That's true, but rolling friction will be much less than sliding friction. That's the big advantage of wheels of course. You'll probably have to push more than the idealized calculation would indicate, but not by a whole lot. It should still be okay.

    No. It would be enough to make it stop accelerating, but it would continue down the ramp at a constant velocity (this will be explained below). To make it stop, you'd have to push with greater than 88 N.

    No.

    Yes. Say, for example, the cart is at rest on the ramp near the bottom. To get it moving, you have to push with > 88 N. The reason for this is that in Newton's second law F = ma, the "F" refers to the NET force on the object. If you push with only 88 N, the net force will be zero. You have to push with more than 88 N to get a non-zero net force and hence a non-zero acceleration.

    Of course, in practice, you'll probably start pushing the tank when it's flat on the floor (before reaching the ramp) in which case you have negligible opposing forces, and ANY force you want is sufficient to get it moving. Of course, once you get to the ramp, you'll have to push harder (with at least 88 N) in order to maintain whatever velocity you have when you reach the ramp..
     
  8. Sep 8, 2010 #7
    Thank you so much for your help

    I can now put the fish tank in place without worrying about too much about the fact that we will have to move it in a couple of months.
     
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