Force Problem and Solution

Force Problem and Solution - Space Shu)le Li-off • ...

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Unformatted text preview: Space Shu)le Li-off •  At li-off, the space shu)le orbiter (m=109,000 kg) is a)ached to a fuel tank (m=760,000 kg). The fuel tank is a)ached to two rocket boosters (m=590,000 kg each). The shu)le orbiter is not directly connected to the rocket boosters. Each rocket booster provides 12,500,000 N. This causes the shu)le to accelerate straight upwards. A) What is the value of the upwards acceleraNon? B) Draw a force diagram for the shu)le orbiter. Include proper labeling and numerical values. C) Draw a force diagram for the fuel tank. Include proper labeling and numerical values. D) Draw a force diagram for each rocket booster. Include proper labeling and numerical values. •  •  •  •  Space Shu)le Li-off SoluNon Page 1 •  At li-off, the space shu)le orbiter (m=109,000 kg) is a)ached to a fuel tank (m=760,000 kg). The fuel tank is a)ached to two rocket boosters (m=590,000 kg each). The shu)le orbiter is not directly connected to the rocket boosters. Each rocket booster provides 12,500,000 N upwards. This causes the shu)le to accelerate straight upwards. A) What is the value of the upwards acceleraNon? The total mass is 2,049,000 kg. The downwards force due to gravity is 20,4900,000 N. The upwards thrust is 25,000,000 N. So, the net force is 4,510,000 upwards. This makes the acceleraNon: ΣF=4,510,000 = m a = 2,049,000 * a a = 2.2 m / s^2 FYI: The acceleraNon actually increases as fuel is used up, since the shu)le becomes lighter. The shu)le reaches a height of about 46km a-er 2 minutes. •  •  •  •  •  •  •  •  •  B) Draw a force diagram for the shu)le orbiter. Include proper labeling and numerical values. Each piece of the shu)le is a)ached, so they are all acceleraNng at 2.2 m / s^2. The orbiter’s net force is then: ΣF = m a = 109,000 * 2.2 = 239,800 N The downwards force of gravity is mg = 1,090,000, so the upwards force of the fuel tank on the orbiter must be 239,800+1,090,000 = 1,329,800 N Ftank on orbiter = 1,329,800 N Fearth on orbiter = 1,090,000 N Space Shu)le Li-off SoluNon Page 2 •  •  C) Draw a force diagram for the fuel tank. Include proper labeling and numerical values. Gravity is pulling down on the tank with a force of 7,600,000 N. From Newton’s 3rd law, we know that the orbiter is pushing downwards on the tank with a force of 1,329,800 N. The net force on the tank must be: ΣF = m a = 760,000 * 2.2 = 1,672,000 N The upwards force must therefore be 1,672,000+7,600,000+1,329,800=10,601,800 N Each rocket provides half that force. Frocket2 on tank = 5,300,900 N •  •  •  Frocket1 on tank = 5,300,900 N Forbiter on tank = 1,329,000 N Fearth on tank = 7,600,000 N Space Shu)le Li-off SoluNon Page 3 •  •  D) Draw a force diagram for each rocket booster. Include proper labeling and numerical values. Gravity is pulling down on the rocket with a force of 5,900,000 N. From Newton’s 3rd law, we know that the tank is pushing downwards on the rocket with a force of 5,300,900 N. The net force on the rocket must be: ΣF = m a = 590,000 * 2.2 = 1,298,000 N The upwards force must therefore be 1,298,000+5,900,000+5,300,900=12,498,900 N This is basically the same (minus some rounding error) as the force we were given for each rocket at the beginning. This is a good check to make sure everything was done correctly. The upwards force is provided by the gas that is being expelled out of the rocket Each rocket has the same force diagram. •  •  •  Fgas on rocket1/2 = 12,500,000 N Ftank on rocket1/2 = 5,300,900 N •  •  Fearth on rocket1/2 = 5,900,000 N ...
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