(3) 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. (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: Fbottom-on-top = Ftop-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
(3) 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. (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: Fbottom-on-top = Ftop-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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:(3) 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.
(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: Fbottom-on-top = Ftop-on-bott
-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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