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

Engineering Mechanics: Dynamics; Modified Mastering Engineering with Pearson eText -- Standalone Access Card -- for Engineering Mechanics: Dynamics (14th Edition)

14th Edition

ISBN: 9780134229294

Author: HIBBELER

Publisher: PEARSON

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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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= MMB 241- Tutorial 1.pdf 1/3 80% + + MMB 241 - Dynamics of Particles Tutorial 1 Topic: Kinematics of Particles:- Displacement, velocity and acceleration of particles. QUESTIONS 1. If v = (4t+5) m/s, where t is in seconds, determine a when t = 2 s. 2. If s = (2t³) m, where t is in seconds, determine v when t = 2 s. 3. If a = 2 m/s², determine v at s = 4 m if v = 3 m/s at s = 0. 4. If a = 4 m/s², determines when t = 3 s if v = 2 m/s and s = 2 m when t=0. 5. A particle moves along a straight line such that its position is defined by s = (t² - 6t + 5) m. Determine the average velocity, the average speed, and the acceleration of the particle when t = 6 s. 6. A particle is moving with a velocity of vo when s=0 and t=0. If it is subjected to a deceleration of a = -kv³ where k is a constant, determine its velocity and position as functions of time. 7. A particle travels along a straight-line path such that in 4s it moves from an initial position SA = -8 m to a position SB = +3 m. Then in…

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Engineering Mechanics: Dynamics; Modified Mastering Engineering with Pearson eText -- Standalone Access Card -- for Engineering Mechanics: Dynamics (14th Edition)

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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.