dua balok A=3kg dan B=5kg dihubungkan dengan tali dan diteletakkan di atas lantai yang mempunyai koefisien gerak statis dan kinetis 0,2 dan 0,1. Jika balok B di

• Hk. II Newton
[tex] \mathrm{m_A = 3 \ kg} \\
\mathrm{m_B = 5 \ kg} \\
\mathrm{\mu_s = 0,2} \\
\mathrm{\mu_k = 0,1} \\
\mathrm{F = 40 \ N} \\
\mathrm{g = 10 \ m/s^2 \ ... \ anggap} \\
\mathrm{\alpha = 37^o \ \Rightarrow \sin \alpha = \frac35 \ \Rightarrow \cos \alpha = \frac45} \\
\\
\mathrm{T = ??}
\\ \\
\heartsuit \ \mathrm{Gaya \ Normal \ B \ :} \\
\mathrm{N_B = w_B - F \sin \alpha} \\
\mathrm{N_B = (5 \times 10) - (40 \times \frac35)} \\
\mathrm{N_B = 26 \ N} \\
\\
\heartsuit \ \mathrm{Gaya \ Normal \ A \ :} \\
\mathrm{N_A = w_A} \\
\mathrm{N_A = 3 \times 10 = 30 \ N} \\ \\
\diamondsuit \ \mathrm{Gaya \ Gesek \ A \ :} \\ \\
\mathrm{f_A = \mu_kN_A} \\
\mathrm{f_A =0,1 \times 30 = 3 \ N} \\
\\
\diamondsuit \ \mathrm{Gaya \ Gesek \ B \ :} \\
\mathrm{f_B = \mu_kN_B} \\
\mathrm{f_B = 0,1 \times 26= 2,6 \ N} \\
\\
\mathrm{Maka \ :} \\
\sum{F} = (\sum{m})a \\
\mathrm{F \cos \alpha - f_A-f_B} = (m_A+m_B)a [/tex]
(40 × ⅘) - 3 - 2,6 = (3 + 5)a
a = 3,3 m/s²

Sekarang tinjau benda A :
∑ F = mₐ a
T - fₐ = mₐ a
T - 3 = 3 × 3,3
T = 9,9 + 3
T = 12,9 N