Saturday

September 20, 2014

September 20, 2014

Posted by **amanda and leah** on Wednesday, February 8, 2012 at 11:45pm.

How far from the 450.8 N child should the pivot be placed to ensure rotational equilibrium? Disregard the mass of the seesaw. Answer in units of m

part 2 of 2

Suppose a 367 N child sits 0.193 m from the 450.8 N child. How far from the pivot must a 261 N child sit to maintain rotational equilibrium? Answer in units of m

- Physics -
**Elena**, Thursday, February 9, 2012 at 6:02amThe torques have be equaled

P1x=P2(2.2-x)

x=(P2•2.2)/( P1+ P2)=0.4 m

Part 2

P1•0.4+P3(0.4-0.193) = P2•1.8+P4•x, P4

where P1=450.8 N, P2=101.0 N, P3= 367 N, and P4=261 N.

Calculate x

**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 437.7 N child and a 237.1 N child sit on either end of a 2.5 meter ...

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 ...