# A 56 kg uniform board 2.36 m long is supported by a pivot 79 cm from the left end and by a scale at the right end?

A 56 kg uniform board 2.36 m long is supported by a pivot 79 cm from the left end and by a scale at the right end (see image).

1. Where should the 43 kg child sit if the scale is to read 100 N? in cm FROM THE LEFT END OF THE BOARD.

2. Where should the 43 kg child sit if the scale is to read 300 N? in cm FROM THE LEFT END OF THE BOARD.

### 2 Answers

- Andrew SmithLv 74 weeks ago
This also is the same question that I answered both two days ago and also today https://answers.yahoo.com/question/index?qid=20191...

- billrussell42Lv 74 weeks ago
that means there is a force of 100 N downwards on the right end.

CW torques:

1. board itself, on the right is 2.36 – 0.79 = 1.57 m and the center of that section is half way from the pivot. Half is 0.785 m from the pivot. Weight of that section is (1.57/2.36)56 = 37.25 kg. so torque is 37.25 kg x 0.785 m = 29.24 kg-m

2. force from scale = 100 N x 1.57 m = 157 Nmconverting to kg-m that is 157/9.8 = 16.02 kg-m(technically, I should convert all the torques to Nm but it is less work using kg-m)

CCW torques

1. board itself. On the left is 2.36-1.57 = 0.79 m and the center of that section is half way from the pivot. Half is 0.395 m from the pivot. Weight of that section is (0.79/2.36)56 = 18.75 kg. so torque is 18.75 kg x 0.79 m = 14.81 kg-m

2. boy, 43 kg x D m = 43D

set torques equal

29.24 + 16.02 = 14.81 + 43D

43D = 30.45

D = 0.71 m

that is from pivot.

from left end it is 0.79 – 0.71 = 0.08 m

(2) is the same, just change a number....

I gave you the technique, that should be all you need.