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Two identical blocks are tied together with a string and placed along the same radius of a turntable that is spinning about its center. The inner block is 2 cm from the center and the outer block is 5 cm from the center. The coefficient of static friction between the turntable and the blocks is µs = 0.79, and the string is taut.

2007-02-09 08:02:29 · 2 answers · asked by Pritesh P 2 in Science & Mathematics Physics

Also, suppose that the blocks each have a mass m = 24 g. For the value of w you just found, what is the tension in the string?

2007-02-09 08:03:25 · update #1

2 answers

These should answer your question....

http://www.physics247.com/physics-homework-help/pulley.php

http://www.physicsforums.com/showthread.php?p=903240

2007-02-09 08:18:19 · answer #1 · answered by B*Family 4 · 0 0

Oh, that's a good one! the only factor you already understand is that, for an merchandise to maintain the comparable orientation, the sum of the torques from all forces could be equivalent to 0.  what isn't so evident is a thank you to establish this problem so which you're observing torques. Neither end of the ladder can transmit any torques quickly.  The wall end can purely transmit typical forces (horizontal), on an identical time as the floor end can transmit the two typical (vertical) and frictional (horizontal) forces.  The frictional rigidity and the traditional rigidity from the wall end could be equivalent and opposite by using definition.  The question is, what are the torques from the different forces, and what's a good place to degree them from? enable's degree from the floor end of the ladder.  that's (6.0m * sin 22° = 2.25 m) from the wall.  the burden of the ladder is 17.0 kg * g, and is a million.a hundred twenty five m from the portion of length; the torque is for that reason 19.a hundred twenty five g N-m.  the guy on the ladder is a million.688 m from the portion of length, and has a weight of seventy six g N; the torque is for that reason 128.25 g N-m.  The sum of the torques is 147.375 g N-m. This torque could be resisted by using the traditional rigidity from the wall.  the portion of touch with the wall is (6.0 m * cos 22°=5.56m) from the portion of reference, so the traditional rigidity could be 147.375 g N-m / 5.fifty six m = 25.8 g N. because of the fact the touch with the wall is frictionless, each and all the burden is exerted on the touch with the floor.  This sums to ninety 3 kg * g.  The minimum coefficient of friction for the ladder to stay placed is: 25.8 g N / ninety 3 g N = 0.277.

2016-10-01 21:14:41 · answer #2 · answered by ? 4 · 0 0

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