Blocks 1 and 2 of masses m1 and m2, respectively are connected by a light string, as show above. These blocks are further connec
2 answers:
Answer:
1) μ₁ = (1 + m₂/m₁ )/3 tan θ - m3/3m₁ scs θ , 2) W₃ = g (m₁ + m₂) sin θ + 3f , 3) a₁ = g sin θ - f / m₁
Explanation:
For this problem we must use Newton's second law, as the speed of the blocks is constant zero acceleration
Let's look at the attachment for the names of the forces. Let's write the equations for each body
Body 3
T- W₃ = 0
T = W₃
Body 2
Y Axis
N- Wy₂ = 0
N = ₂
X axis
Wₓ₂- T₃ + T₁ -fr₂ = 0
Let's use trigonometry to find the weight components
sin θ = Wₓ₂ / W₂
cos θ = ₂ / W₂
Wₓ₂ = W₂ sin θ
₂ = W₂ cos θ
They indicate that the friction force of block 2 is
fr₂ = 2f
We replace
W₂ sin θ - T₃ + T₁ - 2f = 0
Body 1
Y Axis
N₁ - ₁ = 0
X axis
Wₓ₁ -T₁- fr₁ = 0
If we use the same trigonometry equation as for body 2
Wₓ₁ = W₁ sin θ
₁ = W₁ cos θ
Give the rubbing force of block 1
fr₁ = f
We replace
W₁ sin θ - T₁ - f = 0
Let's write our system of equations
T₃ - W₃ = 0
W₂ sintea - T₃ + T₁ - 2f = 0
W₁ sintea - T₁ - f = 0
Add the three equations
W₂ sin θ + W₁ sin θ -W₃ - 3f = 0
g sin θ (m₂ + m₁) - m₃ g = 3f
f = g [(m₁ + m₂) sin θ - m₃] / 3
The formula for friction force is
fr₁ = μ N₁
fr₁ = f = μ W₁ cos θ
μ₁ = f / m₁ g cos θ
μ₁ = g [(m₁ + m₂) sin θ - m₃] / 3 m₁ g cos θ
μ₁ = (1 + m₂/m₁ )/3 tan θ - m3/3m₁ scs θ
2) add 2 and 3 equation
W₂ sin θ + W₁ sin θ - T₃ - 3f = 0
T₃ = g (m₁ + m₂) sin θ + 3f
W₃ = g (m₁ + m₂) sin θ + 3f
3) when cutting the rope T1 = 0
W₁ sin θ - f = m₁ a₁
a₁ = g sin θ - f / m₁
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