Question

2. A child weighing 120 N is sitting on oneside of a see-saw at a distance of 1.5 m from
the pivot. Where should a man weighing400 N sit on the other side of the see-saw for
rotational equilibrium?

3. A mechanic tightens a nut by applying a
force of 10 N on a spanner with a handle of
1.2 m. How long should the handle be if he
wants to tighten the nut by applying a force
of 15 N?​

Answer 1

Answer:

2) solution.

let man weighing 400 N sit at a distance of x from the pivot.

By moment of equilibrium,

sum of anticlockwise moment = sum of clockwise moment

120× 1.5 = 400 × x

180 = 400x

x = 180/400

x = 0.45 m

man weighing 400 N should sit at a distance of 0.45 m from the pivot point.

3) solution.

let the length of the handle be x m.

moment = 10×1.2 = 12 Nm

this moment will be equal to the new moment

so, 12 = 15×x

so, x = 12/15 = 0.8 m

the length of the handle of the spanner should be 0.8 m.

Answer 2

Step-by-step explanation:

let man weighing 400N sit at a distance of x from the pivot by moment of equilibrium