Diketahui:
- Massa mobil (m) = 1.250 kg
- Radius₁ = 30 cm
- Radius₂ = 50 cm
- g = 10 m/s²
Ditanyakan:
- Gaya minimum pada penampang kecil (F₁) = ......?
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Jawab:
[tex]\small{\boxed{\begin{aligned} \sf \frac{F_1}{A_1} &= \sf \frac{F_2}{A_2} \\ \sf \frac{F_1}{\cancel{\pi} \times {r_1}^{2}} &= \sf \frac{m \times g}{\cancel{\pi} \times {r_2}^{2}} \\ \sf \frac{F_1}{{r_1}^{2}} &= \sf \frac{m \times g}{{r_2}^{2}} \\ \sf \frac{F_1}{{(30 \: cm)}^{2}} &= \sf \frac{1.250 \: kg \times {10 \: m/s}^{2}}{{(50 \: cm)}^{2}} \\ \sf \frac{F_1}{{900 \: cm}^{2}} &= \sf \frac{\cancel{12.500} \: N}{{\cancel{2.500} \: cm}^{2}} \\ \sf \frac{F_1}{{900 \: cm}^{2}} &= \sf \frac{5 \: N}{{1 \: cm}^{2}} \\ \sf F_1 &= \sf \frac{5 \: N \times {900 \: \cancel{cm}^{2}}}{{\cancel{1 \: cm}^{2}}} \\ \sf F_1 &= \sf 5 \: N \times 900 \\ \sf F_1 &= \sf \boxed{\pink{\bf 4.500 \: N}} \end{aligned}}}[/tex]
∴ Jadi, gaya minimum yang harus dikerjakan pada penampang kecil agar dapat mengangkat mobil adalah [tex] \pink{\bf 4.500 \: N} [/tex]
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