Explanation Given: The mass of the block A , m A = 100 lb . The mass of the block B , m B = 50 lb . The velocity of the block A while being attached to the block B , v O = 5 ft/s . The co-efficient of kinetic friction between the blocks, μ k = 0.2 . Show the free diagram of the blocks A as in figure (1), blocks A as in figure (2), pulley and cables as in figure (3), and the movement of block B downward along and inclined plane movement in block A as in figure (4). Refer figure (1) Block A . Write the formula for equation of motion in x -axis direction. ∑ F x = m a x (I) Here, summation of total force acting on radial direction along x -axis direction is ∑ F x mass is m and acceleration radial direction along x -axis direction is a x . Refer Figure (1) calculate the components of force. ∑ F x = − T A − μ k N A + m A sin θ (II) Here, the mass of block A m A , the angle is θ , the normal force at block A is N A , and coefficient of kinetic friction is μ k . Write the formula for equation of motion in y -axis direction. ∑ F y = m a y (III) Here, summation of total force acting on radial direction along y -axis direction is ∑ F y mass is m and acceleration radial direction along y -axis direction is a y . Refer Figure (1) calculate the components of force. ∑ F x = N A + m A cos θ (IV) Here, the mass of block A m A , the angle is θ , the normal force at block A is N A , and coefficient of kinetic friction is μ k . Refer figure (2) Block B . Write the formula for equation of motion in y -axis direction. ∑ F y = m a y (V) Here, summation of total force acting on vertical direction along y -axis direction is ∑ F y mass is m and acceleration vertical direction along y -axis direction is a y . Refer Figure (2) calculate the components of force. ∑ F y = T B + m B (VI) Here, the mass of block B m B . Refer figure (3) Pulleys at C and D . Write the formula for equation of motion in y -axis direction. ∑ F y = 0 (VII) Here, summation of total force acting on vertical direction is ∑ F y . Refer Figure (3) calculate the components of force. ∑ F y = 2 T A − 2 T B Determine the position of the block A . v 2 = v O 2 + 2 a A ( s − s O ) (VIII) Conclusion: Refer figure (1) Substitute 100 for m A , and 4 5 for cos θ in Equation (IV) ∑ F y = N A − ( 100 ) ( 4 5 ) Substitute 0 for a y , ( 100 32.2 ) for m A , and N A − ( 100 ) ( 4 5 ) for ∑ F y in Equation (III) N A − ( 100 ) ( 4 5 ) = ( 100 32.2 ) × 0 N A = 80 lb Substitute 80 lb for N A , 0.2 for μ k , 100 for m A , and 3 5 for sin θ in Equation (II) ∑ F x = − T A − ( 0.2 ) ( 80 ) + ( 100 ) ( 3 5 ) Substitute 100 32.2 for m A , and − T A − ( 0.2 ) ( 80 ) + ( 100 ) ( 3 5 ) for ∑ F x in Equation (I) − T A − ( 0.2 ) ( 80 ) + ( 100 ) ( 3 5 ) = ( 100 32.2 ) a A T A − 44 = − 3.1056 a A T A = 44 − 3.105 a A (IX) Refer figure (2) Substitute 50 lb for m B in Equation (VI) ∑ F y = T B − 50 Substitute 50 32.2 for m B , and T B − 50 for ∑ F y in Equation (V) T B − 50 = 50 32.2 a B T B − 50 = 1.553 a B T B = 50 + 1.553 a B (X) Refer figure (3) Substitute 2 T A − 2 T B for ∑ F y in Equation (VII) 2 T A − 2 T B = 0 T A = T B (XI) Substitute 44 − 3

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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Chapter 13, Problem 28P

To determine

The acceleration of block A.

The distance block A slides before it stops.

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Chapter 13, Problem 27P

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