You lay a board over two wooden pegs. The board has a length of 5m and weighs 25 N. it overhangs the left peg by 1m and the right by 2m. Determine the normal force that each peg exerts on the board.

To determine the normal force that each peg exerts on the board, we need to consider the forces acting on the board in equilibrium. In this case, the weight of the board and the normal forces from the pegs balance each other out.

Let's start by calculating the weight of the board. We are given that the board weighs 25 N.

Since the board overhangs the left peg by 1m and the right peg by 2m, we can divide the weight of the board proportionally between the two pegs.

To determine the weight distribution, we can use the principle of moments. The total weight of the board is acting downward at its midpoint, which is located at 2.5m from either end.

The moment on the left side of the left peg (with overhang) can be calculated as:
Moment1 = weight * distance from left peg
= 25 N * 1 m
= 25 N∙m

The moment on the right side of the right peg (with overhang) can be calculated as:
Moment2 = weight * distance from right peg
= 25 N * 2 m
= 50 N∙m

Since the board is in equilibrium, the moments on both sides must balance each other. Therefore:

Moment1 = Moment2

25 N∙m = 50 N∙m

Now, since the left peg and right peg are supporting the board symmetrically, the normal force exerted by each peg on the board would also be equal.

Let's denote the normal force exerted by each peg as N.

From the equilibrium condition, we know that the sum of the forces in the vertical direction must be zero.

Taking the upward forces as positive, we can write the equation:

N + N - 25 N = 0

Simplifying the equation:

2N - 25 N = 0

N = 25 N / 2

N = 12.5 N

Therefore, each peg exerts a normal force of 12.5 N on the board.