For the torque and center of mass:
- The unknown weight (WT) in the first scenario is 1000 N.
- The unknown distance in the second scenario is 4 m.
- The unknown weight of the board in the third scenario is approximately 514.29 N.
What occurs in every scenario?
1. First Scenario (500 N and Unknown WT):
Moment balance:
[tex]\( 500 \, \text{N} \times 4 \, \text{m} = WT \times 2 \, \text{m} \)[/tex]
Solving for WT:
[tex]\( WT = \frac{500 \, \text{N} \times 4 \, \text{m}}{2 \, \text{m}} = 1000 \, \text{N} \)[/tex]
2. Second Scenario (300 N and Unknown Distance):
The same weight is on both sides for balance, hence the distance must also be the same.
Unknown distance:
4 m
3. Third Scenario (600 N and Unknown WT of the Board):
Moment balance:
[tex]\( 600 \, \text{N} \times 3 \, \text{m} = WT_{\text{board}} \times 3.5 \, \text{m} \)[/tex]
(since the board is 7m long, the center is at 3.5m from the pivot)
Solving for WT of the board:
[tex]\( WT_{\text{board}} = \frac{600 \, \text{N} \times 3 \, \text{m}}{3.5 \, \text{m}} \\\approx 514.29 \, \text{N} \)[/tex]
