Explanation Given: The mass flow of sand is d m d t = 50 kg/s . The initial vertical velocity of the sand ( v A ) y = 10 m/s . The velocity of the conveyor belt is v B = 1.5 m/s . The total mass of the sand and conveyor is m = 750 kg at the center of mass point G . Draw the free body diagram of the chute as shown in Figure 1. Refer Figure (1). Write the steady flow equation of moments about the point B ∑ M B = d m d t [ d B B ( v B ) y − d A B v A ] (I). Here, d m d t is the rate of mass flow, d is the perpendicular distance of forces acting about the point B , v A is the velocity of the sand at A , v B is the velocity of the sand at B and ∑ M B the net moment about the point B by all the forces. Write the steady flow equation of forces along x and y − a x e s . ∑ F x = d m d t [ ( v B ) x − ( v A ) x ] (II) ∑ F y = d m d t [ ( v B ) y − ( v A ) y ] (II) Here, ∑ F x is the net forces along x − a x i s , ∑ F y is the net forces along y − a x i s . Conclusion: Refer Figure (1). The net moment about the point B by the forces is ∑ M B = W ( 4 m ) − A y ( 8 m ) Substitute 750 ( 9.81 ) N for W . ∑ M B = 750 ( 9.81 ) N ( 4 m ) − A y ( 8 m ) = 29430 − 8 A y From Figure (1). d B B = 0 d A B = 8 m Here, v A = 10 m/s ( v B ) y = 1.5 m/s ( sin 30 ° ) = 0.75 m/s Substitute 29430 − 8 A y for ∑ M B , 50 kg/s for d m d t , 10 m/s for v A , 0.75 m/s for ( v B ) y , 0 for d B B and 8 m for d A B in Equation (I). 29430 − 8 A y = 50 kg/s [ 0.75 m/s ( 0 ) − 10 m/s ( 8 m ) ] 29430 − 8 A y = − 4000 A y = 29430 + 4000 8 A y = 4178

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ISBN: 9781292088785

Author: HIBBELER

Publisher: INTER PEAR

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Chapter 15.9, Problem 131P

To determine

The horizontal and vertical force components at the pin support B and the roller

support A .

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Chapter 15.9, Problem 130P

Chapter 15.9, Problem 132P

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The horizontal and vertical force components at the pin support B and the roller support A .

Solution Summary: The author determines the horizontal and vertical force components at the pin and roller support A.