 Ch 2. Particle Force and Acceleration Multimedia Engineering Dynamics Rect.Coord. Normal/Tang. Coord. PolarCoord. Orbital Mechanics Computational Mechanics
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll DYNAMICS - CASE STUDY SOLUTION Solution of a) - Horizontal Belt Free-body Diagram (FBD) Mass-acceleration Dynamics (MAD) Begin with a free-body diagram (FBD), which shows the forces acting on the crate, and a mass-acceleration diagram (MAD), which shows the inertia vector of the crate. Since there is no force in the z direction, the motion is confined to the x-y plane. Using Newton's Second Law, sum the forces in both directions. Note that there is no acceleration in the y direction.      ΣFy = may      N - mg = 0      N = mg In the x direction, the sum of the force are      ΣFx = -f = max The friction force depends on the static coefficient of friction (no motion) to start moving,      f = μs N = μs mg The equations can be combined to give an expression for ax in terms of μs,      -μs mg = max Horizontal Box Stops Horizontal Box Slides Divide both sides by m,      ax = dv/dt = -μs g Then Integrate ax to find an expression for v, v(t) = dx/dt = -μsgt + vo An expression for x(t) can be found by integrating v(t), x(t) = -1/2 μsgt2 + vot Knowing that vo = 10 ft/s, the expression for v(t) can be used to find the time required for the belt to come to a stop,      v(t) = 0 = -μsgt + 10      0.5(32.2) t = 10      t = 0.6211 s Substituting the time into the x(t) function gives the minimum distance to stop,      x = 10 (0.6211) -1/2 (0.5)(32.2)(0.6211)2        = 3.106 ft Solution of b) - Inclined Belt Free-body Diagram (FBD) Mass-acceleration Dynamics (MAD) As with all dynamic problems, it is best to start with a free-body and a mass-acceleration diagrams. These are shown at the left for this problem. First, rotate the x-y axis to be parallel and perpendicular to the conveyor belt. This will simplify calculations. Using Newton's Second Law, sum the forces in both directions. Noting that there is no acceleration in the y direction, these equations become      ΣFy = may      N - mg cosθ = 0      N = mg cosθ In the x direction,      ΣFx = max       -f + mg sinθ = max The friction force in terms of μs and g is      f = μs N = μs mg cosθ Combining equations, and dividing through by m gives an expression for ax in terms of μs.      ax = dv/dt = g sinθ - μs g cosθ> Inclined Belt Box Stops Inclined Belt Box Slides Integrating ax gives an expression for v, v(t) = gt (sinθ - μs cosθ) + vo Now the next step can be taken by integrating the velocity function to find the position function.      v(t) = dx/dt Integrating v(t) gives x(t) as x(t) = 1/2 gt2 (sinθ - μs cosθ) + vo t Knowing that vo = 10 ft/s, the time required for the belt to come to a stop is      t (32.2) (sin15 - 0.5 cos15) = -10      t = 1.386 s Substitute the time into the expression for x(t) gives the minimum distance to stop,    x = 1/2 (32.2)(1.386)2 (sin15 - 0.5cos15) + 10(1.386)       = 6.928 ft

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