ch12-p031 - force has the same magnitude (though opposite...

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31. The problem states that each hinge supports half the door’s weight, so each vertical hinge force component is F y = mg /2 = 1.3 × 10 2 N. Computing torques about the top hinge, we find the horizontal hinge force component (at the bottom hinge) is () 2 (27kg)(9.8m/s ) 0.91 m/2 80 N. 2.1m 2(0.30m) h F == Equilibrium of horizontal forces demands that the horizontal component of the top hinge force has the same magnitude (though opposite direction).
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Unformatted text preview: force has the same magnitude (though opposite direction). (a) In unit-vector notation, the force on the door at the top hinge is 2 top ( 80 N)i (1.3 10 N)j F = + . (b) Similarly, the force on the door at the bottom hinge is 2 bottom ( 80 N)i (1.3 10 N)j F = + +...
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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