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Q2(a) A box sliding at point B as shown in Figure Q2(a) with acceleration of 3.8 m/s². Assume the coefficient of kinetic friction is same for each section. Detemine the acceleration of the box at point A. Sketch the free body diagram at both points to help you answer. 35 20 Figure Q2(a) (b) A ball with mass of 3kg is acted upon by four forces of magnitude with the direction as shown in the Figure Q2(b). Given the forces as: F1 = 50N, F2 = 25N, F3 = 8N, F4 = 5N, (i1) Calculate the magnitude of the ball's acceleration and its direction. (11) Relate the unbalanced forces in the system shown with the equation of motion. 40° F1 Figure Q2(b)

Q2(a) A box sliding at point B as shown in Figure Q2(a) with acceleration of 3.8 m/s². Assume the coefficient of kinetic friction is same for each section. Detemine the acceleration of the box at point A. Sketch the free body diagram at both points to help you answer. 35 20 Figure Q2(a) (b) A ball with mass of 3kg is acted upon by four forces of magnitude with the direction as shown in the Figure Q2(b). Given the forces as: F1 = 50N, F2 = 25N, F3 = 8N, F4 = 5N, (i1) Calculate the magnitude of the ball's acceleration and its direction. (11) Relate the unbalanced forces in the system shown with the equation of motion. 40° F1 Figure Q2(b)

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Q2(a) A box sliding at point B as shown in Figure Q2(a) with acceleration of 3.8 m/s². Assume the coefficient of kinetic friction is same for each section. Detemine the acceleration of the box at point A. Sketch the free body diagram at both points to help you answer. 35 20 Figure Q2(a) (b) A ball with mass of 3kg is acted upon by four forces of magnitude with the direction as shown in the Figure Q2(b). Given the forces as: F1 = 50N, F2 = 25N, F3 = 8N, F4 = 5N, (i) Calculate the magnitude of the ball's acceleration and its direction. (ii) Relate the unbalanced forces in the system shown with the equation of motion. 40° Figure Q2(b)
Transcribed Image Text:Q2(a) A box sliding at point B as shown in Figure Q2(a) with acceleration of 3.8 m/s². Assume the coefficient of kinetic friction is same for each section. Detemine the acceleration of the box at point A. Sketch the free body diagram at both points to help you answer. 35 20 Figure Q2(a) (b) A ball with mass of 3kg is acted upon by four forces of magnitude with the direction as shown in the Figure Q2(b). Given the forces as: F1 = 50N, F2 = 25N, F3 = 8N, F4 = 5N, (i) Calculate the magnitude of the ball's acceleration and its direction. (ii) Relate the unbalanced forces in the system shown with the equation of motion. 40° Figure Q2(b)
Fundamental Equation of Dynamics KINAMATICS Equations of Motion: Particle Rectilinear motion: Constant a = de EF = ma EF, = m(ag), EF, = m(ag), Variable a Particle dv ds v = vo + at 1 Rigid Body (Plane Motion) dt s = , + vạt +at 12 = vỷ + 2a.(s - 50) ads = vdv EMG = Iça a EMp = E(M)p Particle Curvilinear Motion: Principle of Work and Energy: T, + U,-2 = T, Cartesian Coordinates (1,y,z) V = * a, = * Vy = ý ay = jỹ Kinetic Energy Particle T = : T=mvå + lgw?² or T= 1,w? Polar Coordinates (r,8,z) Rigid Body a, = # - re? ag = rê + 2rẻ (Plane Motion) V = ré Work Uş = SF cos e de U, = (F cos 8)As Uw = -WAy Variable Force : Normal-Tangential Coordinates (n,t,b) Constant Force: a = i = v, Weight U, = -(kei -kei) Spring [1+ (dy/dx)*]a/2 |d²y/dx*| Couple of Moment: UM = M AO Where p= Power and Efficiency du Uout Pout P = = Fv, E= Pin Conservation of Energy Theorem !! dt Relative Motion VB = VA + VB/A T +V = T2 + V2 Potential Energy Rigid Body Motion About a Fixed Axis V = V, + V. Where: V = ± Wy , V, = +ks? Иariable a Constant a = de Principle of linear Impulse and Momentum: dw w = w, +at a = dt 1 e = 0, + wat + w? = w3 + 2a.(0 - 0.) mv, +ES Fåt = mvz Rigid Body: m(vc)ı +ES Fdt = m(va)a Particle de dt wdw = ade For Point P Conservation of Linear Momentum: s= er, v= wr. a = ar, an = w'r Σ(mv), Σ (mv), Relative General Plane Motion-Translating Coefficient of Restitution (v3), – (v)2 Axes Vg = VA + vn/Acpin) B = Ga+ aB/Apin) Principle of Angular Impulse and Momentun: : (H,), +ES M,dt = (H,)2 Where H, = (d)(mv) Relative General Plane Motion-Tran. And Rot. Axis Particle Vg = VA + w X TB/a + (VB/A)vz ag = an + i x P/a +w x (w x rayA) + 2w x (V8/A)v Rigid Body : (H.h +ES M,dt = (H)2 (Plane Motion) + Where He = Igw Kineties (H.)1 + M,dt = (H,)2 Mass Moment of Inertia Parallel-Axis Theory 1= [r*dm 1 = lg + m d? Where H, = 1,w Conservation of Angular Momentum Radius of Gyration k = Σ(Η), Σ (Η),
Transcribed Image Text:Fundamental Equation of Dynamics KINAMATICS Equations of Motion: Particle Rectilinear motion: Constant a = de EF = ma EF, = m(ag), EF, = m(ag), Variable a Particle dv ds v = vo + at 1 Rigid Body (Plane Motion) dt s = , + vạt +at 12 = vỷ + 2a.(s - 50) ads = vdv EMG = Iça a EMp = E(M)p Particle Curvilinear Motion: Principle of Work and Energy: T, + U,-2 = T, Cartesian Coordinates (1,y,z) V = * a, = * Vy = ý ay = jỹ Kinetic Energy Particle T = : T=mvå + lgw?² or T= 1,w? Polar Coordinates (r,8,z) Rigid Body a, = # - re? ag = rê + 2rẻ (Plane Motion) V = ré Work Uş = SF cos e de U, = (F cos 8)As Uw = -WAy Variable Force : Normal-Tangential Coordinates (n,t,b) Constant Force: a = i = v, Weight U, = -(kei -kei) Spring [1+ (dy/dx)*]a/2 |d²y/dx*| Couple of Moment: UM = M AO Where p= Power and Efficiency du Uout Pout P = = Fv, E= Pin Conservation of Energy Theorem !! dt Relative Motion VB = VA + VB/A T +V = T2 + V2 Potential Energy Rigid Body Motion About a Fixed Axis V = V, + V. Where: V = ± Wy , V, = +ks? Иariable a Constant a = de Principle of linear Impulse and Momentum: dw w = w, +at a = dt 1 e = 0, + wat + w? = w3 + 2a.(0 - 0.) mv, +ES Fåt = mvz Rigid Body: m(vc)ı +ES Fdt = m(va)a Particle de dt wdw = ade For Point P Conservation of Linear Momentum: s= er, v= wr. a = ar, an = w'r Σ(mv), Σ (mv), Relative General Plane Motion-Translating Coefficient of Restitution (v3), – (v)2 Axes Vg = VA + vn/Acpin) B = Ga+ aB/Apin) Principle of Angular Impulse and Momentun: : (H,), +ES M,dt = (H,)2 Where H, = (d)(mv) Relative General Plane Motion-Tran. And Rot. Axis Particle Vg = VA + w X TB/a + (VB/A)vz ag = an + i x P/a +w x (w x rayA) + 2w x (V8/A)v Rigid Body : (H.h +ES M,dt = (H)2 (Plane Motion) + Where He = Igw Kineties (H.)1 + M,dt = (H,)2 Mass Moment of Inertia Parallel-Axis Theory 1= [r*dm 1 = lg + m d? Where H, = 1,w Conservation of Angular Momentum Radius of Gyration k = Σ(Η), Σ (Η),
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