Explanation Given: The weight is 40 lb . The distance s x is R . The distance s y is 4 ft . The distance from A to B is 20 ft . The angle is 30 ° . The distance and angle is shown in Figure (1). Write the expression for final velocity. υ B 2 = υ 0 2 + 2 a c ( s − s 0 ) (I). Here, final velocity is υ B , initial velocity is υ 0 , acceleration due to gravity is a c , distance is s , and initial distance is s 0 . Write the expression for final distance in x direction. s x = ( s x ) 0 + ( υ x ) A t B C (II). Here, final distance in the horizontal direction is s x , initial distance in the horizontal direction is ( s x ) 0 , initial velocity in the horizontal direction is ( υ x ) A and time is t . Write the expression for final distance in y direction. s y = ( s y ) 0 + ( υ y ) A t B C + 1 2 a c t B C 2 (III). Here, final distance in the vertical direction is s y , initial distance in the vertical direction is ( s y ) 0 , initial velocity in the vertical direction is ( υ y ) A , acceleration due to gravity is a y and time is t . Write the expression for final velocity in y direction. υ B = ( υ 0 ) + a c t A B (IV). Here, final velocity is υ B , initial velocity is ( υ 0 ) , acceleration due to gravity is a c and time is t . Write the expression to calculate total time from A to C . t = t A B + t B C (V). Here total distance is t , time traveled from A to B is t A B and time traveled from B to C is t B C . From Figure (1) write the expression for ( υ x ) 0 . ( υ x ) 0 = υ B cos θ (VI). From Figure (1) write the expression for ( υ y ) 0 . ( υ y ) 0 = υ B sin θ (VII). Draw the free body diagram as shown in Figure (2). Refer Figure (2) and write the formula to calculate horizontal force components ∑ F x = m a (VIII). Here, summation of total force acting on horizontal direction is ∑ F x , the mass is m and acceleration is a . Refer Figure (2) and resolve the forces along x − axis . ∑ F x = 40 sin 30 ° (IX). Write the expression for mass. m = W 32.2 ft/s 2 (X). Conclusion: Substitute 40 lb for W in Equation (X). m = W 32.2 ft/s 2 = 40 lb 32.2 ft/s 2 Substitute 40 lb 32.2 ft/s 2 for m in Equation (VIII). ∑ F x = m a x = 40 32.2 a Substitute 40 32.2 a for ∑ F x in Equation (IX). ∑ F x = 40 sin 30 ° 40 32.2 a = 40 sin 30 ° a = 16.1 ft/s 2 From Figure (1) substitute 0 for υ 0 , 0 for s 0 , 20 ft for s and 16

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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The point where suitcase strikes the ground at C time required for suitcase to travel from A to C .

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