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Chapter 09, Problem 36 A 8.72 -m ladder with a mass of 23.8 kg lies flat on the ground. A painter grabs the top end of the ladder and pulls straight upward with a force of 269 N. At the instant the top of the ladder leaves the ground, the ladder experiences an angular acceleration of 1.76 rad/s about an axis passing through the bottom end of the ladder. The ladder's center of gravity lies halfway between the top and bottom ends. (a) What is the net torque acting on the ladder? (b) What is the ladder's moment of inertia? (a) Number Units (b) Number Units Click if you would like to Show Work for this question: Open Show. Work

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Answers r =Fysino 9.- Torque due to pulling force is provide anticlockwise moment I on ladder. Te FL =(269N)(8.72 m) Tp =+2345.68 Nm Torque due to weight of the ladder is provide clockwise moments is Tu = mg) (1) --[(23.8kg) 6-8 m/s2)](31.7.2m) Tv=-1016.9264 Nm. Net torque acting on the ladder is. Thet = Tft Tv = 2345.68 NM-1016.9264 Nm = 1328.75 Tnet E 1329 Nm. 10: Tnet - I< I = Thet I 1329Nm - 1176 rad/s2 = 755.1 kg m? 1:755k9m2