a RF RF2 Wte Wto d ETA = 0 = (RF_)(0) = (Wt) (d) – (Wty)('/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long boar
Posted: Tue May 17, 2022 9:23 pm
- A Rf Rf2 Wte Wto D Eta 0 Rf 0 Wt D Wty 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Boar 1 (35.97 KiB) Viewed 58 times
- A Rf Rf2 Wte Wto D Eta 0 Rf 0 Wt D Wty 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Boar 2 (38.12 KiB) Viewed 58 times
- A Rf Rf2 Wte Wto D Eta 0 Rf 0 Wt D Wty 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Boar 3 (39.21 KiB) Viewed 58 times
- A Rf Rf2 Wte Wto D Eta 0 Rf 0 Wt D Wty 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Boar 4 (35.96 KiB) Viewed 58 times
rd a RF RF2 나 Wtp Wto d TA = 0 = (RF2)() - (Wto)(d) - (W)(1/21) (0 Wty A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long board. The reaction force reading on the scale is 400 N. What is the torque due to the board? 735.75 Nm O-400 Nm -2943 Nm -800 Nm
nd a RF RF2 Wte Wto 1 d ET, = 0 = (RF230) - (Wtp)(d) - (Wto)('/21) = ) A 75 kg 1.75 m tall person lays on a 3 kg, 2 m long board. The reaction force reading on the scale is 400 N. What is d, the distance from the axis of rotation to the person's center of mass? 2 m O 1.75 m O 105 m 125 m
a RFI •RF2 Wte Wto d - ETA = 0 = (RF2)() – (W)(d) - (Wto)(1/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long board. The reaction force reading on the scale is 400 N. At what percent of body height is the person's center of mass located? 63% 60% O 57% 54%