Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. if the children are separated by a distance of 3.5 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?

Step-by-step explanation:

This is more of a Physics problem than just a straight "math" problem because you need to know about torque and rotational equilibrium to solve it.  The formula for torque is

torque = weight * length of the lever arm in meters

Since we are given the mass of each child, we need to solve for their weights, which is found by multiplying their masses by the pull of gravity, which is 9.8 m/sec/sec.  This gives the weight of the 20 kg child to be 196 Newtons, and the weight of the 30-kg child to be 294 Newtons.  

If the length between the 2 children is 3.5 meters, then let's say that the distance away that the heavier child is from the fulcrum is r.  That makes the distance that the lighter child is away from the fulcrum as 3.5 - r.  Now we can fill in the rotational equilibrium formula that says that the sum of the torques must equal 0 if the seesaw is to remain balanced.  Because one torque is positive and one is negative, we can move the negative one over to the other side of the equals sign making them equal to each other, which is what rotational equilibrium is all about.  Here's our formula thus far:

196(3.5 - r) = 294r and

686 - 196r = 294r and

686 = 490r so

r = 1.4

That's the distance that the heavier child is.  The lighter child, then, is 3.5 - 1.4 so that distance is 2.1 meters

answer: r there any options?

step-by-step explanation:

a

step-by-step explanation: