Monkey Climbing a ChainDate: 09/22/1999 at 21:30:38 From: Huma Subject: Physics question A 20-lb. monkey is attached to a 50-ft. chain that weighs 0.5 lb. per (linear) foot. The other end of the chain is attached to the 40-ft.-high ceiling of the monkey's cage. Find the amount of work the monkey does in climbing up her chain to the ceiling. I don't even know where to start... help! Date: 10/14/1999 at 17:46:49 From: Doctor Fwg Subject: Re: Physics question Dear Huma, Thank you for sending such an interesting question. Here is one possible solution, but be sure to check my work to make sure I didn't make any foolish errors in solving the monkey/chain problem. Before attempting to solve this problem there are a few assumptions that might be made. For example, one should probably assume that the monkey is a "point" mass attached to the very end of the chain. This will avoid the problem of not knowing the exact position of the monkey's center of mass, either at the floor or at the ceiling. A second assumption might be that there is no "loop" in the chain, once it is lifted above the floor. This assumption allows one to let the total length of the chain always equal the exact sum of the distances between the monkey and the loop and the ceiling and the loop. This problem would be relatively easy to solve if one wanted to find the work done by an unchained monkey climbing up a 40-ft. chain. In this case, the work (W) would be the product of the force exerted by the monkey and the distance that the force was exerted through. Typically, if the force (F) that must exerted is constant, the work required is just the product of the force and the distance (X) that the force is exerted through, or W = FX. However, a more general method for calculating mechanical work, even when the force exerted changes with distance or time, involves the following differential expression: 1) dw = Fdx = Mgdx. In this case, integrating the expression above to get the total work performed by the monkey would involve using Mg in place of F, where M is the mass of the monkey (in slugs, this value of M = 20/32), and g is the acceleration due to gravity. F (or Mg) is the weight of the monkey and is also the minimum muscular force that the monkey must exert in order to overcome her weight and thus lift herself. M and g are both constants and the integration limits for dw are between 0 and W, and for dx between 0 and 40 ft. The integration step itself is trivial and the final result is: 2) W = Mg(40 - 0) = 40Mg = 40(20/32)(32) = 40(20) = 800 ft-lb where the mass of the monkey is in "slugs" (i.e., 20/32) and g = 32 ft/sec^2. This same result could have been obtained just by using W = FX = MgX, but illustrating how this equation is obtained, using integration, is the purpose of the preceding paragraphs. Now, in the present case, the mass that the monkey must lift increases as she climbs the chain because the monkey supports more and more of the chain's mass as she climbs. Furthermore, as soon as the monkey climbs above the height of 10 ft., the increase in chain mass supported by the monkey changes rather abruptly because the length of the chain is no longer the same as the monkey's height above the floor. So now one must account for the mass of the monkey as well as the increasing mass of the chain that must also be lifted by the monkey. One result of all of this is that the minimum work that must be performed by the monkey, in climbing as well as lifting part of her attached chain, must be greater than 800 ft-lb. For illustrative purposes, first consider the work that the monkey would have to do if the chain were only piled up on the floor and not attached to the ceiling at all. In this situation the integration limits for dw are also between 0 and W, and for dx between 0 and 40 ft, and the differential work expression would be: 3) dw = Fdx = (M + (0.5x/32))gdx, where the term (M + (0.5x/32)) gives the total mass that the monkey must lift (in slugs) as a function of "x," the height climbed. M is the monkey's mass and the (0.5x/32) term gives the additional mass of the chain supported by the monkey at height x. Performing the integration produces: 4) W = Mg(x - 0) + (0.5/32)(x^2 - 0)(g/2) = Mg(40 - 0) + (0.5/32)(40^2 - 0)(g/2) = 40(20/32)(32) + (0.5/32)(1600)(32/2), so: 5) W = 800 ft-lb + 400 ft-lb = 1,200 ft-lb. Now, this amount of "mechanical" work is more than the actual amount of work that would be required to lift a chain that is partially supported by a ceiling attachment, as in this case. So, the true answer to this problem must lie between the 800 and 1,200 ft-lb values calculated above. It so happens that Eq. 3 will also give the work that must be supplied by the monkey to climb the first 10 ft. of the chain, because this section of chain is, in fact, lifted from the floor and none of that portion of the chain's weight is supported by the ceiling. In this case the integration limits on dx are between 0 and 10 ft. and the integration step is straightforward. The work for this portion of the climb should come out to 225 ft-lb. After the 10 ft. height is exceeded, the total length of the chain is no longer the same as the height climbed by the monkey because the chain begins to pull upward and away from the floor after the 10 ft. limit is reached. So now it would be helpful to have another expression for the chain mass as a function of the height climbed at all heights between 10 and 40 ft. and to then integrate (i.e. add up) the work during that portion of the climb. See the following table: Table 1. Relation Between Monkey's Height and Length of Supported Chain Columns Below Are: (1) Monkey's Height, (2) Length of Chain Supported by Monkey, (3) Length of Chain Supported by Ceiling, (4) Total Length of Chain - all in feet. (1) (2) (3) (4) --- --- --- --- 0 0 40 50 10 10 40 50 20 15 35 50 30 20 30 50 40 25 25 50 From this table it can be seen that the length of chain supported by the monkey, only at or after the 10 ft. height limit is reached, is equal to (x/2 + 5), where x is either greater than or equal to 10 ft. So the work that the monkey needs to do, between 10 and 40 ft, may be found from: 6) dw = Fdx = {M + (0.5/32)(x/2 + 5)}gdx, where the integration limits here, on dx, are between 10 and 40 ft. If the results using Eq. 3 (with integration limits on dx between 0 and 10 ft.) and Eq. 6 (with integration limits on dx between 10 and 40 ft.) are added together, the minimum work that must be exerted by the monkey to climb the chain will be found. Note: the final sum here must be between 800 and 1,200 ft-lb. It is believed that the numerical result of integrating Eq. 6 between the limits of 10 and 40 ft. will produce a result of 862.5 ft-lb. Adding this result to the 225 ft-lb result above will produce a final answer of 1087.5 ft-lb for the total amount of "mechanical" work that must be performed by the monkey. I hope that seeing this solution helps you to understand this problem. - Doctor Fwg, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/