Explanation Given data: θ 1 = 60.0 ∘ θ 2 = 25.0 ∘ Mass of the block on the left incline, M 1 = 6.0 kg Formula Used: Newton's second law: F net = m × a Calculation: Let the block on the right has a mass ( m = M 2 ). We'll use two different but related coordinate systems for the two blocks. For both blocks the axes will be parallel and perpendicular to the inclined plane. For block M 1 up the ramp and out of the ramp are positive, while down the ramp and out of the ramp are positive for M 2 . We can draw free-body diagrams and apply Newton's second law in component form for both blocks. The tension acting on each block will be identical in magnitude. We are only interested in the forces and acceleration acting parallel to the planes because the blocks do not leave the plane; the acceleration in the perpendicular direction is always zero. Both blocks are at rest, which means the acceleration of each block is zero and we can calculate M 2 . Free-body diagram of M 1 : Newton's second law for M 1 : ∑ F ext ,parallel = T − w M 1 , parallel = T − M 1 g sin ( 60.0 ∘ ) = M 1 a parallel = 0 = > T = M 1 g sin ( 60

2nd Edition

ISBN: 9781464196393

Author: Freedman

Publisher: MAC HIGHER

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Chapter 4, Problem 75QAP

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The mass of the block on the right incline.

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Chapter 4, Problem 74QAP

Chapter 4, Problem 76QAP

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The mass of the block on the right incline.

Solution Summary: The author calculates the mass of the block on the right incline, using Newton's second law. The tension acting on each block will be identical in magnitude.

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