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A rocket of.mass 1000kg containing propellant gas of 3000kh is to be launche vertically. If the fuel is consumed at a steady rate of 60kg/s, calculate the least velocity of the exhaust gases if the rocket and the content will lift off the launching pad immediately after firing?

A 15kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 200N and breaks as the elevator accelerates. What was the elevatos's minimum acceleration

1 answer

To solve both problems, we'll go step by step.

Problem 1: Rocket Launch

For the rocket to lift off, we need to consider the forces involved and the principle of thrust. The rocket must exert a thrust greater than its weight to start ascending.

  1. Weight of the Rocket: \[ \text{Weight} = m g \] where \( m = 1000 , \text{kg} + 3000 , \text{kg} = 4000 , \text{kg} \) (total mass of the rocket and propellant) and \( g \approx 9.81 , \text{m/s}^2 \).

    \[ \text{Weight} = 4000 , \text{kg} \times 9.81 , \text{m/s}^2 = 39240 , \text{N} \]

  2. Thrust Required: For the rocket to lift off, the thrust \( T \) provided by the exhaust gases must be greater than the weight: \[ T > 39240 , \text{N} \]

  3. Thrust Equation: The thrust produced by the rocket can be expressed as: \[ T = \dot{m} v_e \] where \( \dot{m} = 60 , \text{kg/s} \) (mass flow rate of the exhaust) and \( v_e \) is the velocity of the exhaust gases.

  4. Finding the Least Exhaust Velocity (\(v_e\)): We set the thrust equal to the weight for the rocket to just lift off: \[ \dot{m} v_e = 39240 , \text{N} \]

    Substituting \( \dot{m} \): \[ 60 , \text{kg/s} \cdot v_e = 39240 , \text{N} \] \[ v_e = \frac{39240 , \text{N}}{60 , \text{kg/s}} = 654 , \text{m/s} \]

Therefore, the least velocity of the exhaust gases required for the rocket to lift off is 654 m/s.

Problem 2: Elevator Acceleration and Monkey's Cord Tension

  1. Weight of the Monkey: The weight \( W \) of the monkey can be calculated using: \[ W = m g \] where \( m = 15 , \text{kg} \) and \( g \approx 9.81 , \text{m/s}^2 \). \[ W = 15 , \text{kg} \times 9.81 , \text{m/s}^2 = 147.15 , \text{N} \]

  2. Condition for the Cord to Break: The tension \( T \) in the cord must satisfy: \[ T = W + F \quad (F = ma \text{ is the force due to elevator's acceleration}) \] Here, \( F = 15a \), where \( a \) is the elevator's acceleration.

    The cord can withstand a maximum tension of \( T_{max} = 200 , \text{N} \): \[ 200 , \text{N} = 147.15 , \text{N} + 15a \]

  3. Solving for Elevator's Minimum Acceleration \( a \): Rearranging the equation: \[ 15a = 200 , \text{N} - 147.15 , \text{N} \] \[ 15a = 52.85 , \text{N} \] \[ a = \frac{52.85 , \text{N}}{15 , \text{kg}} \approx 3.5233 , \text{m/s}^2 \]

Therefore, the elevator's minimum acceleration before the cord breaks is approximately 3.52 m/s².