Question

answer each problem please and show work

(20%) Problem 5: You have been hired to design a family-friendly see-saw. Your design will feature a uniform board (mass M-6 kg, length L -2.7 m) that can be moved so that the pivot is a distance d from the center of the board. This will allow riders to achieve static equilibrium even if they are of different mass, as most people are. You have decided that each rider will be positioned so that his/her center of mass will be a distance offer - 12 cm from the end of the board when seated as shown. You have selected a child of mass m - 24 kg (shown on the right), and an adult of mass - 86 kg (shown on the left) to test out your prototype. tatus 1 Otheexpertta.com pleted pleted pleted 50% Part(a) Determine the distance din meters 50% Part (b) Determine the magnitude of the force exerted on the pivot point by the see-saw while in use in newtons. Fy- Grade Summary Deductions 096 Potentis Late Werk 9 T046 Late Potential 709 sin cos t and) colano asin() COS( atand) icotand) Sinh cosh) tanh cotanh() • Degrees Radians Submissions Attempts remaining: ( per attent) detailed stew

Answer 1

Since the adult is on the left side of the pivot point the adults weight will cause the board to rotate counter clockwise. When the pivot. point is moved to the left of the centre of I board, the weight of the board will be on the right coole of pivot point. since the child is on the right side of pivot point, the childs weight will cause the board to rotate clockwise. To balance the Counter clockwise torque must be equal to clockwise torque. Determing the weight, Foo the adult weight 61X9o81 = 843•66M For the board, weight = 6kg x 9.81 = 58.86N. For the child, weight = 24 kg x 9.81 235.440 since the board is 201 m long, each half the booed is 1005 m long since each person is 0.12 m from the board, each person is 0.93 m I from the centre of the board. {1-osm-8-12m = ang! adielt to the prout The distance from the point is 0.93m - d. Torqué = 843.66 NX (0.93 +d). The distance from the centre of the board to the pivot point is de Torque = 58.864 xd The distance from the child to the pivot is 8.93 to Torque = 235.44 MX (0.93+2)

Total clockwise Torque = 58086d + 21809592 +235.44d = 218.9592 + 294-3 d Set the torques equal to each other and sole for do 78406038 - 8430666 = 218.9592 + 294-3 d. - No 343 =VISARE > ONE 1098181896 I check, counter tockouse torque = 484.6038-84366805 4088 2955 mm a 565.6446 = 1137.96 of d = 0.4970689 m Check, Counter clockwise torques = 784.6038 -843-66 of = 365.24859 Nom Clock cause torques : 218.95924294-30 = 365.24659 Nom. Hence, answer is correct. b) The magnitude of the force exerted on the pivot point by the see-saw while in use in equal to the sum of the weigues. F = 843.66N + 58.861 + 235.44 N F = 1137.96N/

Given, 8u = 8.6 m , mn=95 kg re=6-1m, me = 210kg p = 10.1m, mp= bachs 1550kg t - 128, Total moment of Ene Inestia of the Carousel about its axis, in terms of the defined quantities Total moment of Inestia, I = M.I RP hosses +M-If 4 elep + M. I of platform I - In + Ie + Ip nestia of the Carousel about (6) The moment of ite aris; OY I = 8 m n on +4 mere? + / mp op 2 : ** I ={8*98*(8.c)2}+{4x210 -92 +{ {1550X60-12} I = $ 56,209.61 + 61,2564) + (79057.78) I = 166523.75 kg.m2