Math Problem Statement

What method would you use to solve this problem?: A 400 N and a 300 N child sit on either end of 2 m long see-saw. Where along the see-saw should the support be placed so the children are in rotational equilibrium?

Solution

To solve this problem, we need to use the principle of rotational equilibrium. For an object to be in rotational equilibrium, the sum of all torques (moments) acting on it must be zero. Torque is the rotational equivalent of force and is calculated as:

τ=F×d\tau = F \times d

Where:

  • τ\tau is the torque,
  • FF is the force (in this case, the weight of the children),
  • dd is the perpendicular distance from the pivot point (support) to the point of application of the force.

Steps to solve:

  1. Set up a coordinate system: Choose a point along the seesaw as the pivot point. This is where the support will be placed. Let's denote this point as xx (the distance from the left end of the seesaw).

  2. Set up the forces:

    • The first child has a weight of 400 N, and sits at the left end of the seesaw (at x=0x = 0 and a distance of 2 m from the pivot).
    • The second child has a weight of 300 N, and sits at the right end of the seesaw (at x=2x = 2 m and a distance of 2x2 - x m from the pivot).
  3. Balance the torques: Since the seesaw is in rotational equilibrium, the clockwise torque must equal the counterclockwise torque.

The torque exerted by the first child is: τ1=400N×x\tau_1 = 400 \, \text{N} \times x

The torque exerted by the second child is: τ2=300N×(2x)\tau_2 = 300 \, \text{N} \times (2 - x)

For equilibrium, the sum of the torques should be zero: 400x=300(2x)400x = 300(2 - x)

  1. Solve for xx: 400x=600300x400x = 600 - 300x 400x+300x=600400x + 300x = 600 700x=600700x = 600 x=600700=67m0.857mx = \frac{600}{700} = \frac{6}{7} \, \text{m} \approx 0.857 \, \text{m}

So, the support should be placed 0.857 m from the left end of the seesaw.


Would you like me to explain any part of the solution in more detail?

Related questions:

  1. How would the position of the support change if the weights of the children were reversed?
  2. How do you calculate torque when forces are not perpendicular to the axis of rotation?
  3. What other factors could affect the balance of the seesaw in real life, aside from the weights of the children?
  4. How would the length of the seesaw affect the location of the support point?
  5. How would the answer change if there were additional children on the seesaw?

Tip:

When working with rotational equilibrium problems, always ensure that you account for the direction of torques (clockwise or counterclockwise), and remember that torques at the pivot will cancel each other out.

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Math Problem Analysis

Mathematical Concepts

Rotational Equilibrium
Torque
Algebra

Formulas

\tau = F \times d
400x = 300(2 - x)

Theorems

Principle of Rotational Equilibrium

Suitable Grade Level

Grades 9-12