helicopter is lifting two crates simultaneously. One crate with a mass of 232 kg is attached to the helicopter by a cable. The second crate with a mass of 140 kg is hanging below the first crate and attached to the first crate by a cable. As the helicopter accelerates upward at a rate of 1.42 m/s2, what is the tension in each of the two cables. Enter T1, the upper crate first. Then enter T2.
Question
helicopter is lifting two crates simultaneously. One crate with a mass of 232 kg is attached to the helicopter by a cable. The second crate with a mass of 140 kg is hanging below the first crate and attached to the first crate by a cable. As the helicopter accelerates upward at a rate of 1.42 m/s2, what is the tension in each of the two cables. Enter T1, the upper crate first. Then enter T2.
Solution
To solve this problem, we need to use the formula for force, which is mass times acceleration (F = ma).
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First, we calculate the force exerted by the helicopter to lift both crates. The total mass of the crates is 232 kg + 140 kg = 372 kg. The acceleration is 1.42 m/s^2. So, the total force (F_total) is 372 kg * 1.42 m/s^2 = 528.24 N.
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The tension in the cable attached to the helicopter (T1) is equal to the total force exerted by the helicopter, so T1 = 528.24 N.
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Next, we calculate the force needed to lift the second crate. The mass of the second crate is 140 kg and the acceleration is 1.42 m/s^2. So, the force (F2) is 140 kg * 1.42 m/s^2 = 198.8 N.
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The tension in the cable attached to the first crate (T2) is equal to the force needed to lift the second crate, so T2 = 198.8 N.
So, the tension in the cable attached to the helicopter (T1) is 528.24 N and the tension in the cable attached to the first crate (T2) is 198.8 N.
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