- A 1 Rf Rf2 Wte Wto D Et 0 Rf20 Wtp D Wt 21 A 75 Kg 1 75 M Tall Person Lays On A 3 Kg 2 M Long Boar 1 (36.47 KiB) Viewed 72 times
a 1 RF RF2 Wte Wto d ET, = 0 = (RF20) – (Wtp)(d) - (Wt.)('/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long boar
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a 1 RF RF2 Wte Wto d ET, = 0 = (RF20) – (Wtp)(d) - (Wt.)('/21) A 75 kg, 1.75 m tall person lays on a 3 kg, 2 m long boar
a 1 RF RF2 Wte Wto d ET, = 0 = (RF20) – (Wtp)(d) - (Wt.)('/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 the torque due to the board? -400 Nm -735.75 Nm -800 Nm -29.43 Nm
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