Chapter 11, Problem 28A

If you know your own weight and have seesaw and a meterstick available, how can you determine the approximate weight of a friend?

To determine

The approximate weight of a friend using a person’s weight, a seesaw, and a meter stick.

Answer to Problem 28A

If seesaw bends down the end where first person is sitting, his friend would weigh more. The approximate weight of his friend is $W\times x/(L-x)$ .

Explanation of Solution

Given: A person and his friend are supposed to use a seesaw to measure his weight using his weight and a meter stick. Let’s say the first person’s weight is $W\text{'}=W\times x/(L-x)$ kg and his friend’s weight is $W\text{'}$ kg. Length of the seesaw is $L$ cm.

Formula used:

Using Torque equation,

Force applied x Distance from center of symmetry = Constant

Calculation:

In this case, the weight would be the Force applied at the end of the seesaw. Now, to balance the seesaw, if weight is not equal then the product of weight and distance from the center of seesaw must be symmetrical. First person is sitting at $x$ cm distance from the center of seesaw. His friend would be sitting at $(L-x)$ cm from the center of seesaw on the other side.

$W\times x=W\text{'}\times (L-x)$

$W\text{'}=W\times x/(L-x)$

Conclusion:

Thus, the approximate weight of his friend is $W\times x/(L-x)$ . Or, in the absence of a meter stick, one who goes down after sitting on the seesaw would weigh more.