1. A 400 N child and a 300 N child sit on either end of a 2.00 m long seesaw. where along the seesaw should the pivot support be placed to ensure rotational equilibrium?
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Thread: physics homework help
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01-09-2011, 04:32 PM #1
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01-09-2011, 04:38 PM #2
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01-09-2011, 04:40 PM #3
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01-09-2011, 04:44 PM #4
- Join Date: Mar 2010
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01-09-2011, 04:57 PM #5
thankyou so much, repped. that actually makes a lot of sense. i hate how physics makes as much sense as reading a language you dont know until you have it broken down and explained.
if anyone else is willing to explain another problem id really appreciate it.
a bridge 20m long weighing 4x10^5N is supported by two pillars located 3m from each end. if a 1.98x10^4N car is parked 8m from one end of the bridge how much force does each pillar exert?270 bench
390 squat
435 deadlift
210 power clean
sophomore. Taylor Gang or die.
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01-09-2011, 05:08 PM #6
- Join Date: Mar 2010
- Location: Statesboro, Georgia, United States
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...............8m
..............o--o
---------------------------------
/.../...../....../...../....../......../ |
-------------------------------
3m..|....20m........................|3m
.....F1................................F2
force down of 400,000 N
Car weighing 19,800 N(force down) is 8m from one end of the bridge.
sum force down = sum force up
400,000N+19,800N= F1+F2(the two pillars)
419800N=F1+F2
sum torque cw=sum torque ccw
torque=fd
ccw forces: bridge,car
(400,000N)(7m)+(19,800N)(5m)=F1(0)+F2(14m)
14F2=2,800,000N/M+ 99000N/M
F2=207,071.43N
Plug into the equation from the force up/force down part,
419,800N=F1+207071.43
F1=212,728.57N***SCC***
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"We are what we repeatedly do. Excellence, then, is not an act, but a habit." Aristotle
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01-09-2011, 05:50 PM #7
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01-09-2011, 06:21 PM #8
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01-09-2011, 09:39 PM #9
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