besar momen gaya yang dialami benda tersebut?

diket: f1=6N, f2=4N, f3=10n, r1=6, r2=0, r3=4

dt = r.F
[tex]dt = 6(6 \times {10 }^{ - 2} ) + 4(0) + 4(10 \times {10}^{ - 2} )[/tex]
[tex]dt = 36 \times {10}^{ - 2} + 40 \times {10}^{ - 2} [/tex]
[tex]dt = 76 \times {10}^{ - 2} [/tex]

[tex]dt = 0.76[/tex]

(t = torsi)

Momen Gaya - Gerak Rotasi

∑σ = F.R

σ = Momen Gaya
F = Gaya
R = Jarak ke poros

σ = 10 . 4 - 4 . 0 - 6. 6
σ = 40 - 36
σ = 4 N