Explanation Given Information: The initial velocity ( v 0 ) at an angle of ( θ ) of 6 ° . The diameter ( d ) of the grinding wheel is 350 mm . The acceleration due to gravity ( g ) is 9.81 m/s 2 . The angle ( α ) is 10 ° . The angle ( β ) is 30 ° . Calculation: Write the equation for x component of velocity ( v x ) 0 : ( v x ) 0 = v 0 cos θ Substitute 6 ° for θ . ( v x ) 0 = v 0 cos 6 ° Calculate the x component acceleration ( a x ) using the relation: a x = 0 Write the equation for y component of velocity ( v y ) 0 : ( v y ) 0 = v 0 sin θ Substitute 6 ° for θ . ( v y ) 0 = v 0 sin 6 ° Calculate the x component acceleration ( a y ) using the relation: a y = − g Calculate the value of x at point B using the relation: x = d 2 sin α Substitute 350 mm for d and 10 ° for α . x = ( 350 mm × 1 m 1000 mm ) 2 sin ( 10 ° ) = 0.175 sin ( 10 ° ) Calculate the value of y at point B using the relation: y = d 2 cos α Substitute 350 mm for d and 10 ° for α . y = ( 350 mm × 1 m 1000 mm ) 2 cos ( 10 ° ) = 0.175 cos ( 10 ° ) Write the position equation for horizontal motion. Calculate the time ( t c ) at point c using position equation for horizontal motion: x = x 0 + ( v x ) 0 t + 1 2 a x t 2 Here, x 0 is the initial horizontal position and t is time coordinate. Substitute 0 for a x . x = x 0 + ( v x ) 0 t + 1 2 ( 0 ) t 2 x = x 0 + ( v x ) 0 t (1) Substitute − 20 mm for x 0 , 0.175 sin ( 10 ° ) for x , and v 0 cos 6 ° for ( v x ) 0 . 0.175 sin ( 10 ° ) = − ( 20 mm × 1 m 1000 mm ) + v 0 cos 6 ° t 0.175 sin ( 10 ° ) = − 0.02 + v 0 cos 6 ° t 0.030388 + 0.02 = v 0 cos 6 ° t 0.050388 = v 0 cos 6 ° t Solve the above equation for time ( t ) at point B : t B = 0.050388 v 0 cos 6 ° t (2) Write the position ( y ) equation for vertical motion: y = y 0 + ( v y ) 0 t + 1 2 a y t 2 Here, y 0 is the initial vertical position. Substitute – g for a y . y = y 0 + ( v y ) 0 t + 1 2 ( − g ) t 2 y = y 0 + ( v y ) 0 t − 1 2 g t 2 (3) Substitute 205 mm for y 0 , 32.2 ft/s 2 for g , 0.175 cos ( 10 ° ) for y and v 0 sin 6 ° for ( v y ) 0 . 0.175 cos ( 10 ° ) = ( 205 mm × 1 m 1000 mm ) + v 0 sin 6 ° t − 1 2 ( 9.81 ) t 2 0.17234 = 0.205 + v 0 sin 6 ° t B − 1 2 ( 9.81 ) t B 2 (4) Calculate the initial velocity ( v 0 ) at point B : Substitute equation (2) in equation (4). 0.17234 = 0.205 + v 0 sin 6 ° ( 0

Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977251

Author: BEER

Publisher: MCG

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Chapter 11.4, Problem 11.109P

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The range of values of the initial velocity (v0) for which the water will land on the grinding wheel between point B and C.

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Chapter 11.4, Problem 11.108P

Chapter 11.4, Problem 11.110P

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Vector Mechanics for Engineers: Statics and Dynamics

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The range of values of the initial velocity ( v 0 ) for which the water will land on the grinding wheel between point B and C .

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