Physics exercises_solution: Chapter 11

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 11 | Take the origin to be at the center of the small ball then x 0 0387m cm kg . from the center of the small ball. The calculation of Exercise becomes kg 0 kg m kg m x ------------------------------------------------------ kg This result is smaller than the one obtained in Exercise . In the notation of Example take the origin to be the point S and let the child s distance from this point be x. Then M - D 2 mx n MD 1 scm ---------------- 0 x ---- m cm M m 2 m which is L 2 - D 2 2 halfway between the point S and the end of the plank. a The force is applied at the center of mass so the applied force must have the same magnitude as the weight of the door or 300 N. In this case the hinge exerts no force. b With respect to the hinge the moment arm of the applied force is twice the distance to the center of mass so the force has half the magnitude of the weight or 150 N . The hinge supplies an upward force of 300 N -150 N 150 N. F sin 40 2800N so F kN keeping an extra figure. The other person lifts with a force of 160 N - 60 N 100 N. Taking torques about the point where the 60 - N force is applied 100N x 160N or x m 6 m. 1100 N I If the board is taken to be massless the weight of the motor is the sum of the m 600 N applied forces 1000 N. The motor is a distance 1000N m from the end where the 400-N force is applied. The weight of the motor is 400 N 600 N - 200 N 800 N. Of the myriad ways to do this problem a sneaky way is to say that the lifters each exert 100 N to the lift the board leaving 500 N and 300 N to the lift the motor. Then the distance of the motor frnm ttiA AnH ax Piapa ttiA iSDD X Pnt PA ic annlippl ic 2-00 m 3QQ X n __ TPia ppnfpr np from the end where the 600_fn force is applied is 800 n 111 .The center of omvitv ic InpatArl at 200 N m 800 N 0-75 m _ n on frn pnr u liorp thn AiTTlM .