Page 79 - E-modul Mekanika Teknik 2 Fase E/Kelas X SMK

X = 180 cm = 5 cm
3
36 cm
2

Y = F1.Y1 + F2.Y2 + F3.Y3
∑F
2
2
2
Y = 12 cm . 9 cm + 12 cm . 5 cm + 12 cm . 1 cm
2
36 cm
3
3
3
Y = 108 cm + 60 cm + 12 cm
36 m 2
3
Y = 180 m = 5 m
2
36 m
Didapat koordinat titik berat Zo (5:5) m.
Y

2 cm

b1

X b2 X 6 cm

b3
Y

2 cm

a1 a2 a3
X Y
4 cm 2 cm 4 cm

Gambar 52. Penentuan Letak Momen Kelembaman Latihan Soal Nomor 2 Momen
Inersia

Jarak Z1 & Z2 akan sumbu x dan y:
b1 = 5 cm – 1 cm = 4 cm
b2 = 0 cm
b3 = 5 cm – 1 cm = 4 cm
a1 = 5 cm – 3 cm = 2 cm
a2 = 0 cm
a3 = 5 cm – 3 cm = 2 cm

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