Explanation Given: The mass of the block A is 60 lb . The mass of the block B is 40 lb . The displacement of the block B is 2 ft . The friction between the blocks and incline is μ k = 0.1 . Draw the free body diagram of blocks as shown in Figure (1). Refer Figure (1). The forces N A , N B , R do not create any displacement of the blocks. Hence the work done by these forces is zero. From Figure (1). The total length of the cord is equal to the addition of two times of length of the cord of block A and the length of the cord of block B . 2 s A + s B = l Calculate the change in lengths of the cord. 2 s A − s B = 0 2 s A = s B (I) Similarly, calculate the velocities of the cords. 2 v A − v B = 0 2 v B = v B (II) Refer Figure (1). Consider free body diagram of blocks A and B only. Resolve the forces along y − axis to calculate the force N A . ∑ F y = m a y Here, the acceleration along y − axis is zero. a y = 0 Substitute 0 for a y . ∑ F y = m ( 0 ) ∑ F y = 0 (III) Write the formula for friction force F . F = μ k N (IV) Here, μ k is the friction coefficient and N is the normal reaction force. Write the formula for work done by the friction ( F f ) . U F = F s (V) Here, F f is the friction force and s is the displacement due to friction force ( F f ) . Write the formula for work done by their weights. U W = W s (VI) Here, W is the weight and s is the displacement due to force ( W ) . Write the formula for kinetic energy ( T ) T = 1 2 m v 2 (VII) Here, m is the mass of the object and v is the velocity of the object. Write the formula for Principle of Work and Energy. [ ( T 1 ) A + ( T 1 ) B ] + ∑ U = ( T 2 ) A + ( T 2 ) B (VIII) Here, ( T 1 ) A , ( T 1 ) B is initial kinetic energy, ( T 2 ) A , ( T 2 ) B is final kinetic energy of the block A and B respectively and ∑ U is total work done by all the forces on the block. Conclusion: Calculate the change in length of the cord which is connected to block A . Substitute 2 ft for s B in Equation (I). 2 s A = 2 ft s A = 2 ft 2 s A = 1 ft Refer Figure (1), consider free body diagram of block A only. Calculate the normal reaction force ( N A ) . Resolve the forces along y − axis and substitute in Equation (III). N A − 60 cos 60 ° = 0 N A = 30 lb Calculate the friction force ( F A ) . Substitute 0.1 for μ k and 30 lb for N in Equation (IV). F A = 0.1 ( 30 lb ) = 3 lb Calculate the work done by the friction force ( F A ) . Substitute 3 lb for F and 1 ft for s in Equation (V). U F A = 3 lb ( 1 ft ) = 3 ft ⋅ lb Calculate the work done by the weight of the block A . The weight of the block 60 lb is acting at an angle of 60 ° while resolving along the x − axis is 60 sin 60 ° . Substitute 60 sin 60 ° for W and 1 ft for s in Equation (VI). U W A = 60 sin 60 ° lb ( 1 ft ) = 51.9615 ft .lb Refer Figure (1), consider free body diagram of block B only. Calculate the normal reaction force ( N B ) . Resolve the forces along y − axis and substitute in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 14.3, Problem 21P

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(b) A steel 'hot rolled structural hollow section' column of length 5.75 m, has the cross-section shown in Figure Q.5(b) and supports a load of 750 kN. During service, it is subjected to axial compression loading where one end of the column is effectively restrained in position and direction (fixed) and the other is effectively held in position but not in direction (pinned). i) Given that the steel has a design strength of 275 MN/m², determine the load factor for the structural member based upon the BS5950 design approach using Datasheet Q.5(b). [11] ii) Determine the axial load that can be supported by the column using the Rankine-Gordon formula, given that the yield strength of the material is 280 MN/m² and the constant *a* is 1/30000. [6] 300 600 2-300 mm wide x 5 mm thick plates. Figure Q.5(b) L=5.75m Pinned Fixed

Chapter 14 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 14.3 - Determine the work of the force when it displaces...Ch. 14.3 - Determine the kinetic energy of the 10-kg block.Ch. 14.3 - Prob. 1FP Ch. 14.3 - If the motor exerts a constant force of 300 N on...Ch. 14.3 - If the motor exerts a force of F = (600 + 2s2) N...Ch. 14.3 - The 1.8-Mg dragster is traveling at 125 m/s when...Ch. 14.3 - When s = 0.5 m, the spring is unstretched and the...Ch. 14.3 - The 5-lb collar is pulled by a cord that passes...Ch. 14.3 - The 20-kg crate is subjected to a force having a...Ch. 14.3 - Prob. 2P

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The velocity of the block A .

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