(2) A stack of two crates is being lifted upwards by two cables, as shown in the figure below. The tension in each cable is 2270 N. You may assume that the masses of the cables are negligible compared to the masses of the crates. Do not use Newton's 3rd Law; use only Newton's 2nd Law. (a) Using Newton's 2nd Law, calculate the force that the top crate exerts on the bottom crate and the force that the bottom crate exerts on the top crate. (answer: Fotom on-top = Fiop-on-bottom= 3062 N) (b) The maximum tension that each cable can handle without breaking is 3230 N. Calculate the maximum upward acceleration that the load can have without breaking the cables. (answer: 8.97 m/s²) 232 kg 112 kg

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Chapter1: Units, Trigonometry. And Vectors
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(2) A stack of two crates is being lifted upwards by two cables, as shown in the figure below. The tension
in each cable is 2270 N. You may assume that the masses of the cables are negligible compared to the
masses of the crates. Do not use Newton's 3rd Law; use only Newton's 2nd Law.
(a) Using Newton's 2nd Law, calculate the force that the top crate exerts on the bottom crate and the force
that the bottom crate exerts on the top crate.
(answer: Fotom on-top = Fop on-bottom= 3062 N)
(b) The maximum tension that each cable can handle without breaking is 3230 N. Calculate the maximum
abottom-on-top
%3D
mtop-on-
upward acceleration that the load can have without breaking the cables. (answer: 8.97 m/s²)
232 kg
112 kg
Transcribed Image Text:(2) A stack of two crates is being lifted upwards by two cables, as shown in the figure below. The tension in each cable is 2270 N. You may assume that the masses of the cables are negligible compared to the masses of the crates. Do not use Newton's 3rd Law; use only Newton's 2nd Law. (a) Using Newton's 2nd Law, calculate the force that the top crate exerts on the bottom crate and the force that the bottom crate exerts on the top crate. (answer: Fotom on-top = Fop on-bottom= 3062 N) (b) The maximum tension that each cable can handle without breaking is 3230 N. Calculate the maximum abottom-on-top %3D mtop-on- upward acceleration that the load can have without breaking the cables. (answer: 8.97 m/s²) 232 kg 112 kg
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