5. A massless string passes over a frictionless pulley and carries
1 answer:
Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
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Answer:
Explanation:
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So we have:
Answer:
26b) 66.7%
27) 500 N
Explanation:
26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.
26.b) Mechanical advantage is the ratio between the forces:
MA = load force / effort force
Efficiency is the ratio between the work:
e = work done on load / work done by effort
Work is force times distance.
e = (F load × d load) / (F effort × d effort)
Rearranging:
e = (F load / F effort) × (d load / d effort)
e = MA × (d load / d effort)
In a two pulley system, the load moves half the distance of the effort. So the efficiency is:
e = (4/3) × (1/2)
e = 2/3
e = 66.7%
27) In a three pulley system, the load moves a third of the distance of the effort.
e = (F load / F effort) × (d load / d effort)
0.40 = (600 N / F) × (1/3)
F = 500 N
