yang ngerti tolong ajarin gue soal fisika begini dong:') makasih

Diket :
Dari gambar diperoleh ada 4 kawat dng panjang masing-masing adalah
Kawat I, L1 = 4
Kawat II, L2 = 5/sin α = 5/0,6 = 25/3
Kawat III, L3 = 4
Kawat IV, L4 = 2

Titik pusat massa masing-masing kawat adalah,
Kawat I, (2,0) maka :
X1 = 2 dan y1 = 0
Kawat II, (2, 5/2) maka :
X2 = 2 dan y2 = 5/2
Kawat III, (4,3) maka :
X3 = 4 dan y3 = 3
Kawat IV, (4,5) maka :
X4 = 4 dan y4 = 5

Tanya :
Xpm = __? Dan Ypm = __?

Jawab :
Xpm = (X1.L1 + x2.L2 + x3.L3 + x4.L4) / (L1 + L2 + L3 + L4)

Xpm = (2•4 + 2•25/3 + 4•4 + 4•2) / (4 + 25/3 + 4 + 2)

Xpm = (8 + 50/3 + 16 + 8) / (10 + 25/3)

Xpm = (32 + 50/3) / (30/3 + 25/3)

Xpm = (96/3 + 50/3) / (55/3)

Xpm = (146/3) / (55/3)

Xpm = 146/55

Xpm = 2,65

Dan

Ypm = (y1.L1 + y2.L2 + y3.L3 + y4.L4) / (L1 + L2 + L3 + L4)

Ypm = (0•4 + 5/2•25/3 + 3•4 + 5•2) / (4 + 25/3 + 4 + 2)

Ypm = (0 + 125/6 + 12 + 10 / (10 + 25/3)

Ypm = (32 + 125/6) / (30/3 + 25/3)

Ypm = (192/6 + 125/6) / (55/3)

Ypm = (317/6) / (55/3)

Ypm = 317/6 • 3/55

Ypm = 317/2• 1/55

Ypm = 317/110

Ypm = 2,88

Jadi pusat massa sistem nya adalah
Xpm = 2,65
Ypm = 2,88

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FZA