What is the required force to lift a 15.0 kg child on a teeter-totter?

[jhrnndz1]

I have another question,

A 15.0 kg child is sitting on a plyground teeer-totter, 1.50m from the pivot. What force, applied 0.300m on the other side of the pivot, is need to make the cild lift off the ground?

I got the Force to be 735.8N, is this correct?

 

[Doc Al]

Looks good to me. (Too many significant figures though.)

 

[kyle8921]

I cannot confirm the accuracy of your calculation without knowing the specific details of the teeter-totter and its materials. However, I can provide a general explanation of the required force to lift a 15.0 kg child on a teeter-totter.

The required force to lift a 15.0 kg child on a teeter-totter would depend on the weight of the child, the distance of the child from the pivot point, and the length of the teeter-totter. In this scenario, the child weighs 15.0 kg and is sitting 1.50m from the pivot point. The force needed to make the child lift off the ground would be equal to the child's weight multiplied by the distance of the child from the pivot point divided by the distance of the applied force from the pivot point.

Using the formula F = (m x d1)/d2, where F is the required force, m is the mass of the child, d1 is the distance of the child from the pivot point, and d2 is the distance of the applied force from the pivot point, we can calculate the required force.

In this case, the required force would be (15.0 kg x 1.50m)/0.300m = 75N. This means that a force of 75N applied 0.300m on the other side of the pivot point would be needed to lift the child off the ground.

Again, the accuracy of this calculation may vary depending on the specific details of the teeter-totter. It is always important to consider the materials, design, and other factors when calculating the required force in a physical system.