Final answer:
To maintain static equilibrium on the board, the person's 7 N upward force must be placed at a distance of approximately 1.43 meters from the left edge. This is calculated using the principle of moments, by equating the moments due to the board's weight and the forces applied at the ends of the board.
Step-by-step explanation:
The scenario involves a static equilibrium problem where a 10 N board is in balance due to two forces acting upon it. One force is a 3 N upward force from a string on one end, and the other is an upward force that a person must apply at a specific position to prevent the string from breaking. To find the correct position for the person's 7 N upward force to maintain static equilibrium, we use the principle of moments, which states that clockwise moments equal anticlockwise moments.
Let's designate the distance from the left edge where the 7 N force should be applied as x meters. The board is 5 meters long and has a uniform weight distribution. So, the moment due to the board's own weight (which acts at the center of gravity, at 2.5 m from the left end) is given by:
10 N × 2.5 m = 25 N·m
We can equate this to the sum of the moments due to the two forces keeping the board in equilibrium:
3 N × 5 m + 7 N × x m = 25 N·m
After simplifying, we get:
15 N·m + 7 N × x m = 25 N·m
Moving terms yields:
7 N × x m = 10 N·m
Now we solve for x:
x = 10 N·m / 7 N
x = 1.43 m
Therefore, to keep the board in static equilibrium, the upward force of 7 N must be applied at a distance of approximately 1.43 meters from the left edge of the board.