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Seesaw: simple statics problem?

  • #1
Seesaw: simple statics problem?????

Q:
A uniform seesaw is of length 3.90 m and weighs 235 N. The seesaw is pivoted about a point on its bottom surface and the location of this pivot along the length of the seesaw can be adjusted as far as a distance 0.170 m from the center of the seesaw. Little Karen, who weighs 440 N, has her center of gravity over the right end, with the seesaw set so she exerts the most torque about the pivot. The seesaw balances horizontally when Elwood has his center of gravity is a distance 0.160 m from the left end. Both have their feet off the ground.

I have no idea what's up with this problem. I figured that we could just put the pivot in the exact center of the seesaw and then use that the net torque must be zero. So basically it'd be 1.95(Karens radius) times 440 (karens weight) is equal to 1.79 (Elwood's radius = 3.90/2 - 0.16) times Elwood's weight. So 1.95(440)=1.79(W). Since the pivot is in the center of the seesaw, the weight of the seesaw would just cancel itself out. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
siddharth
Homework Helper
Gold Member
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What's the question? Is it to find Elwood's weight?

If so, first find the point where the seesaw is pivoted.It's given that Karen exerts the most torque about the pivot. You also know that Torque=rxF. So where should the seesaw be pivoted for Torque(and hence 'r') to be maximum?

Once you find that, balance the clockwise and counter-clockwise torques, and you will be able to get Elwood's weight.
 
  • #3
OH, thanks a lot. I missed that little detail.
 

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