Penjelasan dengan langkah-langkah :
Jawaban: (B)
Penjelasan:
tan θ = 3/4 --> θ = 37°
F cos 37 - f₁ - f₂ = (m₁+ m₂). a
20. 4/5 - μ.m₁.g - μ.(m₂.g+F.sin 37) = (2+1). a
16 - 1/3. 2. 10 - 1/3. (1. 10+20. 3/5) = 3.a
16 - 20/3 - 22/3 = 3.a
16 - 14 = 3.a
a = 2/3 m/s²
T - f₁ = m₁.a
T - μ.m₁.g = m₁.a
T - 1/3. 2. 10 = 2. 2/3
T = 20/3 + 4/3 = 24/3
T = 8 N
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Penjelasan dengan langkah-langkah :
Jawaban: (B)
Penjelasan:
tan θ = 3/4 --> θ = 37°
F cos 37 - f₁ - f₂ = (m₁+ m₂). a
20. 4/5 - μ.m₁.g - μ.(m₂.g+F.sin 37) = (2+1). a
16 - 1/3. 2. 10 - 1/3. (1. 10+20. 3/5) = 3.a
16 - 20/3 - 22/3 = 3.a
16 - 14 = 3.a
a = 2/3 m/s²
T - f₁ = m₁.a
T - μ.m₁.g = m₁.a
T - 1/3. 2. 10 = 2. 2/3
T = 20/3 + 4/3 = 24/3
T = 8 N