- 1 Rf Rf2 Wte Wth D Et 0 Rf2 1 W D Wto 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Board 1 (34.52 KiB) Viewed 60 times
1 RF RF2 Wte Wth d ET, = 0 = (RF2) (1) - (W)(d) - (Wto)('/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long board
-
answerhappygod
- Site Admin
- Posts: 899604
- Joined: Mon Aug 02, 2021 8:13 am
1 RF RF2 Wte Wth d ET, = 0 = (RF2) (1) - (W)(d) - (Wto)('/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long board
1 RF RF2 Wte Wth d ET, = 0 = (RF2) (1) - (W)(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. At what percent of body height is the person's center of mass located? 57% 63% 60% 54%
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!