Question

1/1 pts Question 4 A weightlifter hoists a barbell above his head. The total length of the bar is 2 meters. Eighty kg of welght are added to each side, with the center of the welghts lying 23 m from the outside tip of the bar on both sides If the barbell was hoisted from the very center of the bar, what would be the torque applied on each side? (lust one side, not the sum of both together) (Disregard the weight of the bar). 621.1 Nm 592.3 Nm 604.3 Nm None of the possibilities are correct 0/1 pts IncoctQuestion 5 If the same barbellpength ” 2 meters, was now loaded with ony 20 kg on one side and 30 kg on the other, how far from the 30 kg side should the bar be grasped (one handed) in order to carry the bar balanced? Assume the centers of the weights are stil 23 m from each side. Give the final answer in distance from the weight on the 30 kg side. (Again disregard the weight of the bar). 041 85 2 72

Answer 1

The torque is given by,

The torque is given by, tau = r times F Here the perpendicular distance from the center is 1-0.23 = 0.77 m. The force acting on the end is 80 kgf. Hence the torque is, rightwardsdoublearrow tau = (0.77) (80 times 9.81) = 604.3 Nm

Here the perpendicular distance from the center is 1-0.23 = 0.77 m. The force acting on the end is 80 kgf. Hence the torque is,

$\Rightarrow \tau = (0.77)(80 \times 9.81)=604.3 \, Nm$