I'm not sure what this is, really important, please help!

lillybeth

lillybeth

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This is for a a friend of mine who doesn't have an account right now, I haven't done this yet in Math, so here is his question:
A cable weighing 0.8 pounds per foot is attached to a small 80-pound robot, and then the robot is lowered into a 60-foot deep well to retrieve a 7 pound wrench. The robot gets out of the well (carrying the wrench) by climbing up the cable with one end of the cable still attached to the robot. How much work does the robot do in climbing to the top of the well?
Thanks guys!

(sorry about the title markFL, i didnt know what to put, i dont know the name of this problem)
 
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This is an application of work done by a varying force. The weight of the robot-wrench-cable system varies because the amount of cable hanging from and supported by the robot increases as it climbs up. We need the formula:

\(\displaystyle \displaystyle W=\int_{x_1}^{x_2}F_x\,dx\)

We may let:

\(\displaystyle x_1=0,\,x_2=60\)

and

\(\displaystyle F_x=87+0.8\cdot0.5x=87+0.4x\)

You see, we must consider the weight of the robot and wrench which remains constant, and the cable the length of which that is supported by the robot is half the distance the robot has climbed, which is \(\displaystyle x\). And so length times weight per length gives total weight of the cable.

Thus we have:

\(\displaystyle \displaystyle W=\int_{0}^{60}87+0.4x\,dx=\left[87x+0.2x^2 \right]_0^{60}=5940 \text{ ft}\cdot\text{lb}\).

edit: If Perl hasn't had calculus yet, let me know as there is a geometrical way to get the work by finding the area of a trapezoid.
 
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lillybeth said:
This is for a a friend of mine who doesn't have an account right now, I haven't done this yet in Math, so here is his question:
A cable weighing 0.8 pounds per foot is attached to a small 80-pound robot, and then the robot is lowered into a 60-foot deep well to retrieve a 7 pound wrench. The robot gets out of the well (carrying the wrench) by climbing up the cable with one end of the cable still attached to the robot. How much work does the robot do in climbing to the top of the well?
Thanks guys!
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This is as much a physics problem as it is calculus, and since I am a physicist I will take that approach. The calculus approach is work = force × distance
[FONT=MathJax_Math]W[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math]F[/FONT][FONT=MathJax_Main]d[/FONT][FONT=MathJax_Math]y[/FONT]
but physics tells us work is change of energy,

[FONT=MathJax_Math]W[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]△[/FONT][FONT=MathJax_Math]E[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Size2]∑[/FONT][FONT=MathJax_Math]m[/FONT][FONT=MathJax_Math]g[/FONT][FONT=MathJax_Main]△[/FONT][FONT=MathJax_Math]y[/FONT]

The robot and the wrench with a combined weight 87 pounds are raised by 60 feet.

The center of gravity of the cable is initially at 30 feet and finally at 45 feet from the bottom, and the total weight of the cable is 48 pounds. [This is much easier than integrating the Force!]

Edit: Then the total work is (87 pounds)*(60 ft) + (48 pounds)*(15 ft) = 5940 ft-pound

Concerning units: the unit of "pound" is a Force, mg. Thus you multiply pounds times distance to find Work in units of ft-lb. [In the metric system, the weight "1 pound" is replaced by the mass 2.2 kg, which must be multiplied by g = 9.8 m/s^2 to find the Force. Force×distance --> energy in Joules.]
 
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(Replying to MarkFL and Dr. Phil), Perl figured it out on openstudy. Thanks anyway! He appreciates the effort! :)
 
Denis said:
You earlier said "He", so Perl is a guy?! I pity him :confused:
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I spelled it wrong, and its just his handle on OS. Watch out because he plans to post here. I think.

I corrected spelling above, Dennis. Sorry 4 the error.
 
I am not sure the analysis of this problem is correct. Since the robot is climbing up the cable the weight of the cable is supported by the top of well. Using the integral of 87 + 0.8x is correct if the robot is climbing up the wall and the cable is hanging below the robot in ever increaing length.

I got 87 lbs times 60 ft equals 5220 ft lbs.
 
guitarguy said:
I am not sure the analysis of this problem is correct. Since the robot is climbing up the cable the weight of the cable is supported by the top of well. Using the integral of 87 + 0.8x is correct if the robot is climbing up the wall and the cable is hanging below the robot in ever increaing length.

I got 87 lbs times 60 ft equals 5220 ft lbs.
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The total weight supported at the top of the well is robot+wrench+60 feet of rope. Hence the Tension is 135# where the rope is fastened. Since the rope has weight, Tension decreases as you go down, till you reach the point where the robot is hanging on. Just above that point,
T(x+) = 135# - (60ft - x)(0.8 #/ft) = 87# + (x)(0.8 #/ft)

Below that point hangs a loop of rope. If there were a pulley at the bottom supporting some mass you would recognize that there is a mechanical advantage of 2. The weight of rope, (0.8 #/ft)*x, is supported equally by the two strands. Thus at a point immediately below x, the tension in the rope is
T(x-) = (1/2)(x)(0.8 #/ft)

The force exerted by the robot on the rope is the difference of those two tensions:
F(x) = T(x+) - T(x-) = 87# + (0.4 #/ft)(x)
The increment of Work is
F(x)dx = [87# + (0.4 #/ft)(x)]dx
This integrates to 5940 ft-pounds, in agreement with the response of MarkFL on Feb. 6. [I also edited my response of Feb 6 because I had said the total weight of cable was 4.8 pounds instead of 48 pounds!]
 
Denis said:
Throwing your weight around, Phil? :rolleyes:
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Hmmm. If I am throwing an object, I am dealing with mass, NOT weight. So I can truthfully answer, "No." [Maintaining the distinction between mass and weight is one of my pet crusades.]

I guess I should have used SI units - how about 8054 J?

This problem illustrates why the majority of physics questions use massless ropes. How many different answers did people come up with?
 
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