Explanation Given information: The flow rate of sand ( Δ m Δ t ) in the conveyor belt is 100 kg/s . The velocity of the sand at A and B are v A = 4.5 m/s and v B = 4.5 m/s . The combined weight of the component and the sand is W = 4 kN . Calculation: Apply the principle of impulse momentum to the conveyor system as shown in Figure 1. Refer Figure 1. Consider C x , C y and D y are the reactions at point C and D . Take Moment about the point C . ( Δ m ) v A × 0.9 + D ( Δ t ) × 3 − W ( Δ t ) × 1.8 = − ( Δ m ) v B × 1.65 3 D ( Δ t ) = 0.9 ( Δ m ) v A − 1.65 ( Δ m ) v B + 1.8 W ( Δ t ) Divide both side of the Equation by Δ t . 3 D = 0.9 ( Δ m Δ t ) v A − 1.65 ( Δ m Δ t ) v B + 1.8 W (1) Substitute 100 kg/s for ( Δ m Δ t ) , 4.5 m/s for v A , 4.5 m/s for v B , and 4 kN for W in Equation (1). 3 D = ( 0.90 × 100 × 4.5 ) − ( 1.65 × 100 × 4.5 ) + ( 1.8 × 4 kN × 1000 N 1 kN ) D = 6862

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ISBN: 9781260265453

Author: BEER

Publisher: MCG

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Chapter 14, Problem 14.114RP

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Find the reactions at the support C and D.

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Chapter 14 Solutions

VECTOR MECH...,STAT.+DYN.(LL)-W/ACCESS

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Find the reactions at the support C and D .

Solution Summary: The author explains the principle of impulse momentum to the conveyor system as shown in Figure 1.

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Find the reactions at the support C and D.

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