A rigid bar pendulum is attached to a cart, which moves along the horizontal plane. The rigid bar has a center of mass at L/2. T
he pendulum is pinned to the cart, and a torsional spring is located between the cart and the pendulum. Assume the torsional spring does not cause a dynamic effect on the cart (but does affect the rigid bar). Recall for a pinned attachment we can assume reaction forces in the horizontal and vertical directions. There is a horizontal force F(t) that acts on the cart, which is also attached to a wall on the opposite side through a linear spring. Ignore friction.
a. Identify the number of degrees of freedom, and define a choice of generalized coordinates.
b. Find the equations of motion using Newton's second law.
a. Identify the number of degrees of freedom, and define a choice of generalized coordinates.
b. Find the equations of motion using Newton's second law.
1 answer:
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Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.