course PHY231 I've had some trouble with 4.44 from the text. I think I finally got the upward and downward acceleration reasoned out, but I'm not very confident that it's correct. A Gymnast of mass m climbs a vertical rope attached to the ceiling. You can ignore the weight of the rope. Draw a free-body diagram for the gymnast. Calculate the tension in the rope if the gymnast : a)Climbs at a constant rate b)hangs motionless c)accelerates up rope with an acceleration of magnitude a d)slides down rope with accleration of magnitude a For each free-body diagram, the gymnast is represented as a particle, and the only forces exerted on the gymnast are vertical. The tension in the rope exerts a force opposite to gravity on the gymnast. Upwards is considered the positive direction. When the gymnast climbs the rope at a constant rate, there is no acceleration and therefore no net force. The gymnast must exert a force on the rope that is sufficient to oppose gravity, and the rope exerts a force, tension (T) of equal magnitude and opposite direction. T = (m*9.8m/sec^2) When the gymnast is motionless, there is no acceleration and therefore no net force. The tension of the rope (T) and the gymnast's weight (m*g) sum to zero. T = (m*9.8m/sec^2) When the gymnast is accelerating up the rope, there must be a net force. The force causing acceleration is expressed (m*a). The forces on the gymnast are T, m*a, m*g. The tension in the rope and m*a act in the upward direction, and m*g acts in the downward direction: T = m(g - a). The rope exerts a force greater than the weight of the gymnast on the gymnast. When the gymnast is accelerating down the rope, there must be a net downward force. Since the gymnast is not falling freely, the rope exerts some force on the gymnast. This force is less than the gymnast's weight. The net force is the difference between the weight of the gymnast and the tension in the rope. The tension can be expressed T = m*g - Fnet"