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nomor 1
Gaya yang bekerja F membentuk sudut θ yang berjarak d dari poros
maka momen gayanya:
τ = F sin θ
ada digambar
nomor 2
F1 = 10 N
F2 = 4 N
F3 = 5 N
F4 = 4 N
d1 = 0 m (poros)
d2 = 2 m
d3 = 2 + 1 = 3 m
d4 = 2 + 1 + 3 = 6 m
τ?
penyelesaian:
τ = τ2 + τ4 - τ1 - τ3
τ = (F2 d2) + (F4 d4) - (F1 d1) - (F3 d3)
τ = (4 . 2) + (4 . 6) - (10 . 0) - (5 . 3)
τ = 8 + 24 - 0 - 15
τ = 17 Nm searah jarum jam