tolong dijawab soal fisika

Diketahui:
l = 4m.
m Batang = 8 kg.

Ditanya = momen gaya (τ)? poros dianggap di dinding.
Στ = τ batang + τ N lantai + τ Gaya gesek.
Στ = τ batang + τ N lantai + 0 (karena licin)
Στ = F Batang.l Batang + F Normal.l Normal
Στ = -(m Batang.g.cos37°).2m + (m batang.g.cos37°).4m
Στ = -8kg.(10m/s^2).(4/5).2m + 8kg.(10m/s^2).(4/5).4m
Στ = -128Nm+256Nm.
Στ = 128 Nm.
Momen gaya sebesar 128Nm berlawanan arah jarum jam.

[tex]\displaystyle l=4\,\text{m}\\m=8\,\text{kg}\\\theta=37^\circ\\\\\Sigma\tau=?\\\\\text{ambil titik poros di dinding}\\\Sigma\tau=w_{\text{batang}}\cdot r_w-N_{\text{lantai}}\cdot r_\text{N}-f_{\text{lantai}}\cdot r_f\\\Sigma\tau=mg\cdot\frac12l\cos\theta-mgl\cos\theta-0\cdot l\sin\theta\\\Sigma\tau=-\frac12mgl\cos\theta\\\Sigma\tau=-\frac12\cdot8\cdot10\cdot4\cos37^\circ\\\Sigma\tau=-40\cdot4\cdot0,8\\\boxed{\boxed{\Sigma\tau=-128\,\text{Nm}}}\\\\\text{Tanda}-\text{berarti searah jarum jam}[/tex]