Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

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1. Draw a free-body diagram of the wooden board. Represent the board as a straight line with forces acting at the ends and possibly at the center. Label the forces: 60 N upward force at one end, and unknown force F at the other end. The weight of the board (160 N) acts downward at the center of the board, which is at 1.50 m from either end.
2. Set up the equation for the rotational equilibrium (torque balance) about one end of the board. Assume the pivot point is at the end where the 60 N force is applied. The torque due to the weight of the board will try to rotate the board clockwise, and the torque due to the force F at the other end will try to rotate it counterclockwise.
3. Calculate the torque due to the weight of the board. Torque (\(\tau\)) is given by the product of the force and the perpendicular distance from the pivot point to the line of action of the force. Here, \(\tau_{weight} = 160 \, \text{N} \times 1.50 \, \text{m}\).
4. Set up the torque due to the unknown force F at the other end. Since this force also acts upward, its torque will oppose that of the weight. The distance from the pivot to this force is the full length of the board, 3.00 m. Thus, \(\tau_F = F \times 3.00 \, \text{m}\).
5. Use the principle of torque equilibrium, which states that the sum of clockwise torques must equal the sum of counterclockwise torques for the system to be in equilibrium. Set up the equation: \(160 \, \text{N} \times 1.50 \, \text{m} = F \times 3.00 \, \text{m}\). Solve this equation to find the magnitude of the force F that the other person must apply.