Question

need help asap!! thanks in advance

A diving board has length of 1.8 m and a total mass of 65 kg. However, the mass is not uniformly distributed, so its center of mass is off center as shown in the diagram below. However, we know that the moment of inertia around the center of mass is 20 kg m². The board is fixed to rotate around its far left end. A support pushes up with a force of 750 N at the midpoint of the board at the angle shown. A boy steps on the far right end of the board, exerting a downward force of 500 N. What is the magnitude of the board's angular acceleration at the moment the boy first steps onto it? cm 0.75m 404 750N 0 -2 rad/s2 0 -3.15 rad/s? 0 -15.2 rad/s2 0 -20 rad/s2 -25.1 rad/s2

Answer 1

som - Thet I xa T=8X m 65kg, I= 1.8m 12 ( Given) - 2okg-m' Io = I com Q = 40 1/2 500N Fsince P d 0.75m Fcosa mg f Let the angular acceleration of dod 2 by perpendicular axis theorem" moment of inertia of rod about point up! Ip I com + m (0.75,2 (T-1 I = I com +md? 20+ 65101752 Ip = 56.56 Kg-m2 1 Inet)about = Ip Xd = r F about point P Ip x a= rf r = perpendicular distance from force

It is clear the that the torque about point 'p' of borizontal component of force f that is 'Frings is → clockwise Now u → michac louise. zero. 500 x I + mg x 0.75 - fcosa x 22 - Ip x 2 500 X 1.8) + (65*9.8 X0.75) (750 cosyox7.52 56.56 x 2 517.08) 56.56 a = 1900 + 477.75 25 15.2 radis? since angular acceleration toward upward ga 25 15.2 rodis (C) - 15.2 gadis? Ang