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this is weird.
an initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1100N. The coefficient of static friction between the box and the floor is 0.35. (a) what should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?
(a) is easy, about 19 degrees and the book agrees with my answer. But (b) makes little sense. Clearly we determined the angle which maximizes the pulling power independent of weight, because the mass of the object was not given. What do they mean by weight in that situation? I thought perhaps they meant the normal force on the box from the floor, but also seems impossible to determine.
My book claims the answer to (b) is about 3.3 kilonewtons. No idea where they got this.
A logarithm is just a misspelled algorithm.
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Are you considering that the tension can't exceed 1100N?
Why did the vector cross the road?
It wanted to be normal.
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Let the weight of the sand+box be W, the tension in the cable be T, and the normal reaction be R. Suppose the cable is angled at θ to the horizontal.
Resolving forces perpendicularly:
Resolving forces horizontally:
This gives the value of θ which you have found. And
so that does give a maximum value for x. And for θ in the first quadrant
which is approximately 3330 N.
Last edited by JaneFairfax (2008-02-16 01:54:01)
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thanks, Jane! though all i really needed was the last line. I just wasn't sure what they wanted.
A logarithm is just a misspelled algorithm.
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thanks, Jane! I learned a lot from that!
Love the forces set-up, and love the
negative 2nd deriv for maximum,
and love the way you used trig to
get from tangent to cosine and sine.
Totally awesome!!
igloo myrtilles fourmis
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