Tuesday

March 3, 2015

March 3, 2015

Posted by **Kendra** on Wednesday, April 2, 2008 at 9:28pm.

How far from the 437.7 N child should the pivot be placed to ensure rotational equilibrium? disregard the mass of the seesaw. answer in units of m.

- Physics -
**bobpursley**, Wednesday, April 2, 2008 at 10:35pmI will be happy to critique your thinking.

The sum of the moments about the piviot has to be zero.

- Physics -
**drwls**, Wednesday, April 2, 2008 at 10:36pmLet the heavier child be x meters from the fulcrum. The lighter child will be 2.5 - x meters from the fulcrum.

For equilibrium,

437.7 x = 237.1 (2.5 -x)

Solve for x.

674.8 x = 592.75

x = 0.88 m

**Answer this Question**

**Related Questions**

Physics - A 450.8 N child and a 101.0 N child sit on either end of a 2.2 m long ...

Physics - A 450.8 N child and a 101.0 N child sit on either end of a 2.2 m long ...

PHYSics - a child with a mass of 29 kg sits at a distance of 4m from the pivot ...

Physics - The seesaw in the figure below is 4.5 m long. Its mass of 20 kg is ...

Physics - A 20-kg child sits on a see saw on the playground, 3.0 m from the ...

physics - A fulcrum is placed under the center of a uniform 4.00m long, 3.00kg ...

Physics - A seesaw is 4 meters long and is pivoted in the middle. There is a ...

physics - two children sit at opposite ends of a 60 lb. 10 ft. seesaw which is ...

Math - Two children sit on a seesaw that is 14 ft long. One child weighs 95 lb, ...

physics - An adult and a child are on a seesaw 14.5 ft long. The adult weighs ...