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posted by  mj00614 on 11/5/2009 12:06:52 AM  |  status: Closed  |  Earned Karma: 67

seesaw please help!!!!

Course Textbook Chapter Problem Needs by
Calculus Based Physics Physics for Scientists and Engineers A Strategic Approach (2nd) by Knight N/A N/A 11/5/2009 at 8:00:00 AM
Question Details:
A seesaw consists of a board 4m long, which pivots about its center. A 50 kg person sits at one end of the seesaw. How far from the pivot should a person of 75 kg sit that they can balance without touching the ground?
A 1000 kg steel bar of length L is suspended at a point on the bar which is 0.5 m away from its center-of-mass. If a 50 kg person standing at the end of the bar which is closer to the point of suspension is in rotational equilibrium, what is the total length of the bar?
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posted by Golden Monkey on 11/5/2009 12:21:50 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
to balance:
.
   m1 * distance from center  =  m2 * distance from center
.
   50 *  2   =     75 *   distance
.
       distance  =   1.333 meters   for the 75 kg person
.
Again,   weight of bar * distance from pivot  =  weight of person * distance from pivot
.
                      1000 *   0.5    =  50 * (  (1/2) L  -  0.5 )
.
                500   =    25 L   -  25
.
                   525  =  25 L
.
           L  =   21   meters
The Israelites angered the lord when they worshipped the golden calf. They should have worshipped the GOLDEN MONKEY instead!
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posted by Mobiusdick on 11/5/2009 12:28:26 AM  |  status: Live
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Response Details:
If the board is 4m long, there are 2m on either side. The torques from both of the people should be equal for the board to be balanced.

Therefore: (50kg)(9.8m/s2)(2m) = (75kg)(9.8m/s2)(d) 
d = 1.33m


For the bar, given that the center of mass has to exist at the center of the bar, we can say that the center of the mass of the bar is at L/2. Again, the torque of the person must be equal the torque of the bar. The distance of the person can be represented as .5m less than half of the length of the bar.

(50kg)((L/2)-.5m)(9.8m/s2) = (1000kg)(9.8m/s2)(.5m)
((L/2)-.5) = 10m
L/2 = 10.5m
L = 21m
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