a 10 foot long teeter-totter pivots in the middle. a 50 lb boy sits on one end. Where must a 100 lb boy sit to balance the beam?

Half the distance from fulcrum (center) to the end, which is 5 m from the end.

At that point, the moments are balanced.

Thanks a lot you are a genius

To balance the teeter-totter, the weight on one side must equal the weight on the other side. In this case, we have a 10-foot-long teeter-totter and a 50 lb boy on one end.

To find out where the 100 lb boy must sit to balance the beam, we need to consider the concept of torque. Torque is the force applied to an object about an axis of rotation. In this case, the pivot point of the teeter-totter acts as the axis of rotation.

The torque exerted by an object is calculated by multiplying the force applied by the distance at which the force is applied from the axis of rotation. In equation form, torque (T) is equal to force (F) multiplied by distance (D): T = F × D.

Since we already know the force applied by the 50 lb boy on one end of the teeter-totter, we can use this information along with the distance from the pivot point to determine where the 100 lb boy must sit to balance the beam.

Let's assume that the 50 lb boy is sitting x feet away from the pivot point. Since the total length of the teeter-totter is 10 feet, the 100 lb boy must sit (10 - x) feet away from the pivot point.

Now, we can create an equation:

Torque exerted by the 50 lb boy = Torque exerted by the 100 lb boy.

50 lb * x feet = 100 lb * (10 - x) feet.

Simplifying the equation:

50x = 100(10 - x).

50x = 1000 - 100x.

Adding 100x to both sides of the equation:

50x + 100x = 1000.

150x = 1000.

Dividing both sides of the equation by 150:

x = 1000 / 150.

x = 6.67 feet.

Therefore, the 100 lb boy must sit approximately 6.67 feet away from the pivot point to balance the beam.