Cliff and Will are carrying a uniform 2.0m board of mass 69kg. Will is supporting the board at the end while cliff is 0.6m from the other end as shown in the following figure. Cliff has attached his lunch to his end of the board, and the tension in the string supporting the lunch is 203N. Find the normal forces exerted by Cliff and Will.

Sum the vertical forces:

C+W-69*9.8-203=0
C+W= ? you do it.

Then sum moments about either end. I will Sum moments about Will.

1*69*9.8+2*203-C(1.4)=0
solve for Cliff here, then put it in the first equation, solve for Will

To solve this problem, we need to consider the forces acting on the board and use the principle of torque equilibrium.

First, let's identify the forces acting on the board:
1. The weight of the board itself, acting downward at its center of mass.
2. The normal forces exerted by Cliff and Will, acting upward at their respective supports.
3. The tension in the string supporting the lunch, acting upward at Cliff's end of the board.

Since the board is in equilibrium, the net torque acting on the board must be zero. Torque is calculated by multiplying the force vector by the perpendicular distance from the axis of rotation. In this case, the axis of rotation is at Cliff's end of the board.

Let's denote the distance from Cliff's end to the center of mass of the board as "d1" and the distance from Will's end to the center of mass as "d2".

Considering the torques acting clockwise as positive, we can write the torque equation:

(Tension × d1) - (Weight of the board × (d1 + d2)) = 0

To find the normal force exerted by Cliff and Will, we need to resolve the forces vertically.

The normal force exerted by Cliff can be calculated as the sum of the weight of the board and the tension from the string:

Normal force by Cliff = Weight of the board + Tension

The normal force exerted by Will can be calculated as the difference between the weight of the board and the tension:

Normal force by Will = Weight of the board - Tension

Now, let's plug the given values into the equations:

Given:
Mass of the board (m) = 69 kg
Length of the board (L) = 2.0 m
Tension in the string supporting the lunch = 203 N

Weight of the board = mass × gravity
= 69 kg × 9.8 m/s^2
= 676.2 N

Substituting these values into our equations:

For the torque equation:
(203 N × 0.6 m) - (676.2 N × (0.6 m + d2)) = 0

Solving this equation will give us the value of d2, the distance from Will's end to the center of mass.

For the normal forces:
Normal force by Cliff = 676.2 N + 203 N
Normal force by Will = 676.2 N - 203 N

Solving these equations will give us the normal forces exerted by Cliff and Will.

Note: If the value for d2 is negative, it means the center of mass is located to the left of Will's end, which would be physically unrealistic in this scenario.