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1. Use equations (3) and (6) to obtain and m+m' = 2M sin a f = ₁₂ (m-m')g. (9) 2. Draw the free body diagram for the block when it is sliding downward on the incline plane with a constant speed. 3. Draw the free body diagram for the block when it is accelerating upward on the incline plane. 4. The block never moves in the y direction, the acceleration in the y direction is 0. From ΣFy = may, we have ΣΕ Since the kinetic friction f = μN, show that N-Mg cos α = 0. f με = Mg cos a 15 == Newton's second law is ΣF = mã. In the component form, it becomes ΣFx = max for motion in the x direction and Σ Fy = may for motion in the y direction. For a block moving up an inclined plane under the influence of friction f and tension in the string T, the x equation is T- f-Mg sina = max. For a block moving down an inclined plane under the influence of friction, the x equation is T + f - Mg sina = max, simply changing the sign of friction f. == Now suppose the block is moving up the plane at a constant speed. In this case, ax = 0, and ΣFx = max becomes T-f-Mg sina = 0. (1) We next consider the hanging mass, with the coordinates and the free body diagram shown to the right. For the hanging mass, ΣFx = max gives T-mg = 0. (2) Substituting mg from equation (2) for T in equation (1), we get mg-f-Mg sin a (3) +y mg tr If the block is moving down the plane with a constant speed, the kinetic friction f is up the plane, and equation (1) becomes - T'+f Mg sin a = 0 (4) and equation (2) is now T' - m'g = 0. (5) Equation (3) now becomes m'g+f Mg sin a = 0. (6)

1. Use equations (3) and (6) to obtain and m+m' = 2M sin a f = ₁₂ (m-m')g. (9) 2. Draw the free body diagram for the block when it is sliding downward on the incline plane with a constant speed. 3. Draw the free body diagram for the block when it is accelerating upward on the incline plane. 4. The block never moves in the y direction, the acceleration in the y direction is 0. From ΣFy = may, we have ΣΕ Since the kinetic friction f = μN, show that N-Mg cos α = 0. f με = Mg cos a 15 == Newton's second law is ΣF = mã. In the component form, it becomes ΣFx = max for motion in the x direction and Σ Fy = may for motion in the y direction. For a block moving up an inclined plane under the influence of friction f and tension in the string T, the x equation is T- f-Mg sina = max. For a block moving down an inclined plane under the influence of friction, the x equation is T + f - Mg sina = max, simply changing the sign of friction f. == Now suppose the block is moving up the plane at a constant speed. In this case, ax = 0, and ΣFx = max becomes T-f-Mg sina = 0. (1) We next consider the hanging mass, with the coordinates and the free body diagram shown to the right. For the hanging mass, ΣFx = max gives T-mg = 0. (2) Substituting mg from equation (2) for T in equation (1), we get mg-f-Mg sin a (3) +y mg tr If the block is moving down the plane with a constant speed, the kinetic friction f is up the plane, and equation (1) becomes - T'+f Mg sin a = 0 (4) and equation (2) is now T' - m'g = 0. (5) Equation (3) now becomes m'g+f Mg sin a = 0. (6)

Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter6: Circular Motion And Other Applications Of Newton’s Laws
Section: Chapter Questions
Problem 6.4CQ: Describe the path of a moving body in the event that (a) its acceleration is constant in magnitude...
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1. Use equations (3) and (6) to obtain and m+m' = 2M sin a f = ₁₂ (m-m')g. (9) 2. Draw the free body diagram for the block when it is sliding downward on the incline plane with a constant speed. 3. Draw the free body diagram for the block when it is accelerating upward on the incline plane. 4. The block never moves in the y direction, the acceleration in the y direction is 0. From ΣFy = may, we have ΣΕ Since the kinetic friction f = μN, show that N-Mg cos α = 0. f με = Mg cos a 15
Transcribed Image Text:1. Use equations (3) and (6) to obtain and m+m' = 2M sin a f = ₁₂ (m-m')g. (9) 2. Draw the free body diagram for the block when it is sliding downward on the incline plane with a constant speed. 3. Draw the free body diagram for the block when it is accelerating upward on the incline plane. 4. The block never moves in the y direction, the acceleration in the y direction is 0. From ΣFy = may, we have ΣΕ Since the kinetic friction f = μN, show that N-Mg cos α = 0. f με = Mg cos a 15
== Newton's second law is ΣF = mã. In the component form, it becomes ΣFx = max for motion in the x direction and Σ Fy = may for motion in the y direction. For a block moving up an inclined plane under the influence of friction f and tension in the string T, the x equation is T- f-Mg sina = max. For a block moving down an inclined plane under the influence of friction, the x equation is T + f - Mg sina = max, simply changing the sign of friction f. == Now suppose the block is moving up the plane at a constant speed. In this case, ax = 0, and ΣFx = max becomes T-f-Mg sina = 0. (1) We next consider the hanging mass, with the coordinates and the free body diagram shown to the right. For the hanging mass, ΣFx = max gives T-mg = 0. (2) Substituting mg from equation (2) for T in equation (1), we get mg-f-Mg sin a (3) +y mg tr If the block is moving down the plane with a constant speed, the kinetic friction f is up the plane, and equation (1) becomes - T'+f Mg sin a = 0 (4) and equation (2) is now T' - m'g = 0. (5) Equation (3) now becomes m'g+f Mg sin a = 0. (6)
Transcribed Image Text:== Newton's second law is ΣF = mã. In the component form, it becomes ΣFx = max for motion in the x direction and Σ Fy = may for motion in the y direction. For a block moving up an inclined plane under the influence of friction f and tension in the string T, the x equation is T- f-Mg sina = max. For a block moving down an inclined plane under the influence of friction, the x equation is T + f - Mg sina = max, simply changing the sign of friction f. == Now suppose the block is moving up the plane at a constant speed. In this case, ax = 0, and ΣFx = max becomes T-f-Mg sina = 0. (1) We next consider the hanging mass, with the coordinates and the free body diagram shown to the right. For the hanging mass, ΣFx = max gives T-mg = 0. (2) Substituting mg from equation (2) for T in equation (1), we get mg-f-Mg sin a (3) +y mg tr If the block is moving down the plane with a constant speed, the kinetic friction f is up the plane, and equation (1) becomes - T'+f Mg sin a = 0 (4) and equation (2) is now T' - m'g = 0. (5) Equation (3) now becomes m'g+f Mg sin a = 0. (6)
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