Forces in action

EasyMediumHard
123456789101112131415161718192021222324252627282930313233343536373839404142434445464748
Question 17
Medium

A schematic diagram showing a horizontal seesaw with a central triangular pivot. Child A sits on the left at a distance of 1.0 m from the pivot, with an arrow pointing down representing a weight force of 360 N. Child B sits on the right at a distance of 1.0 m from the pivot, with an arrow pointing down representing a weight force of 450 N.

The diagram shows two children sitting on a seesaw.

a.

Explain what happens to the seesaw when both children lift their feet off the ground.

[2]
b.

Calculate the distance from the pivot that Child B must sit to balance the seesaw when Child A sits 0.8 m from the pivot.

Use the equation: moment of a force=force×distance\text{moment of a force} = \text{force} \times \text{distance}moment of a force=force×distance

[3]
c.

Child A has a weight of 360 N.

Calculate the mass of Child A.

Use the equation: gravitational force=mass×gravitational field strength\text{gravitational force} = \text{mass} \times \text{gravitational field strength}gravitational force=mass×gravitational field strength (Take gravitational field strength, g=10 N/kgg = 10\text{ N/kg}g=10 N/kg)

[2]
d.

Child A stands on one foot when they leave the seesaw. The area of their foot in contact with the ground is 1.8 × 10-2 m2.

Calculate the pressure Child A exerts on the ground.

[2]

Forces in action Questions

  1. GCSE
  2. /Physics
  3. /Forces in action