The drat is small, the log is long and it can’t roll. The coefficient of friction is 0.2. The surface is horizontal.
Update:
Remo, Scythian:
No tools are allowed. The drat can choose the point where to apply the force and the direction of the force. Nothing else. The magnitude of the force is 0.1*P, where P is the weight of the log.
And let the friction coefficient be 0.23.
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The drat needs to apply a horizontal force at the end of the log perpendicular to the log’s axis. An instantaneous center of rotation exists at a distance of 0.2*L from the actual center of the log resulting in a forward acceleration of the log’s center. The drat can keep this up until he/she/it reaches home.
http://i941.photobucket.com/albums/ad260/dcheddie0…
O’ is the instantaneous center of rotation, a is the acceleration of the actual center, O and α is the angular acceleration of the log. The portion above O’ moves to the left hence the direction of the frictional forces. We are assuming here that the distribution of the normal reaction is unaffected by horizontal forces.
Horizontal forces:
mg/10 + (μmg/L)*(L/2 – h) – (μmg/L)*(L/2 + h) = ma
Moments about O:
(mg/10)*L/2 – (μmg/L)*(L/2 – h)*(L/2 + h)/2 – (μmg/L)*(L/2 + h)* (L/2 – h)/2 = I*α
= (1/12)mL² * a/h
Simplifying and solving leads to:
120μ*(h/L)³ + (6-10μ)*(h/L) – 1 = 0
which for μ = 0.2 yields
24(h/L)³ + 4(h/L) – 1 = 0
h/L ≈ 0.20116 < 0.5
a ≈ 0.0195*g > 0
This means that the above scenario is possible as shown. The drat IS ABLE to accelerate the log from rest.
The diagram shows this for zero velocity, but the drat can keep this up little by little until he drags the log home. If we take into account velocity, then we’ll have an a_x and a_y since we must consider the centripetal acceleration of the center O.
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This sounds like one of those “anything goes” questions, in that I don’t know if there are any restrictions on what the drat can do to move the log. Creative use of levers and pulleys can the job done. But I think First Grade Rocks! is on the right path, in that if we assume that the log weighs 1 pounds, and the drat can only push 10 pounds, and it takes 20 pounds to push the log, then the drat can push one end of the log that has 10 pounds of resistance, not 20. Something like that. Of course, some math is needed here to verify that, and maybe I’ll do that if First Grade Rocks! doesn’t beat me to it. I am not so sure if even a block is needed.
Edit: Thanks for the clarification, Zo Maar. Now I understand that the problem is finding the most advantageous spot to push the log.
Edit 2: Always a race, these Y!A questions. Dr D has beat me to this, and has provided a fine answer. Thumbs up for him.
Our drat picks an end she wishes to move towards her house. She then goes to the other end and puts a block on that end to keep it from moving backwards. She then goes back to the end she wishes to move and she can just push the log. The log will rotate around the block she placed against the far end.
When she has pushed the one end far enough, she reverses herself and begins to push the formerly fixed end.
It turns out the the log acts as a lever. Since the drat travels exactly twice as far as the center of the log, the force she needs to overcome friction is k * 1/2mg or 1/10 mg, which is exactly what she can lift. (And where does the other half of the force come from? The block she placed on the far end of the log).
*******Addendum
Our Drat is an extraordinarily smart drat. Not only does she know the principles of leverage, but she also knows that sin(θ) + cos(θ) can be greater than 1.
Her best strategy is to lift and push at the end of the log (probably at ~15 degrees — but I’ll do the math later). This has the added advantage of kicking the axis of rotation down past the center of mass. … (to be continued)
Lifting at θ = asin(1/10) gives you a little advantage. You can move a log with a k = 2cosθ/(10-sinθ) = 0.201 … but that is not enough!
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Dr. D has the right idea. Assuming for some reason that you could not get the instaneous axis of rotation to be offset from the center with a straight horizontal push as he proposes (I need to check his approach), you could with an angled upwards push. (It is this offsetting of the center of rotation which allows you to reduce the effective friction).
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Dr. D appears to be right on spec. It also works with μ = 0.23. Note: the drat would be able to accelerate a little better with a very slight upward push — it reduces friction and pushes the center of rotation a bit closer to optimum — but the benefits are marginal.
Source(s): “With a long enough lever, I can move the earth.” Archimedes Von Drat
I think it can! bus i never have seen….