Explanation Given information: The angle between the two straight section of the rotating arm as θ = 120 ° . The rate of discharge ( Q ) of water through the sprinkler is 20 L/min . The velocity ( v ) of the rotating arm is 18 m/s . The magnitude of the friction couple ( M F ) is 0.375 N ⋅ m . Calculation: Calculate the mass flow rate ( Δ m Δ t ) of the water using the relation: d m d t = ρ Q (1) Here, ρ is the density of the water. Consider the density of the water is 1 , 000 kg/m 3 . Substitute 1 , 000 kg/m 3 for ρ and 20 L/min for Q in Equation (1). Δ m Δ t = 1 , 000 kg/m 3 × ( 20 L/min × 1 m 3 1 , 000 L × 1 min 60 s ) = 1 , 000 kg/m 3 × ( 20 L/min × 1 m 3 1 , 000 L × 1 min 60 s ) = 0.33333 kg/s Show the geometry of the arm of the sprinkler as shown in Figure 1. Refer Figure 1. The centre of the four rotating arm of the sprinkler is denoted by O . Refer the triangle Δ A B O . Apply the cosine law follows: ( O A ) 2 = ( O B ) 2 + ( B A ) 2 − 2 ( O B ) ( B A ) cos θ = 150 2 + 100 2 − 2 × 150 × 100 cos ( 120 ° ) = 150 2 + 100 2 − 2 × 150 × 100 × ( − 0.5 ) = 32 , 500 + 15 , 000 O A = 47 , 500 = 217.95 mm Apply the sine rule as follows: sin β A B = sin θ O A sin β 100 = sin ( 120 ) 217.95 sin β = ( 0.866 217.95 ) × 100 β = sin − 1 ( 0.39733 ) β = sin − 1 ( 0.39733 ) ≈ 23.411 ° Calculate the value of the angle β as follows: α = 180 ° − 120 ° − β = 60 ° − 23

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 14, Problem 14.115RP

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Find the rate of rotation of the garden sprinkler.

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Find the rate of rotation of the garden sprinkler.

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