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CHAPTER 13 Kinetics of a Particle: Force and acceleration ( Newton’s method) KINEMATICS Rectilinear motion Curvilinear motion s = so + vot + v2 a t ds vdv a t v v = vo + act 1 2 a t 2 c 2 = vo + 2ac(s-so) constant acceleration an v2 a an at 2 2 Normal & Tangential Components KINETICS Newton’s Method CHAPTER 13 Work and Energy Method Impulse and momentum STATICS Newton’s first law A particle originally at rest, will remain in rest SF=0 Newton’s second law A particle acted upon by an unbalanced force, SF=ma experiences acceleration that has the same direction as the force and a magnitude that is directly proportional to the force DYNAMICS Free body diagram Kinetics diagram Example 1 The crate has a mass of 50 kg. If the crate is subjected to a 400[N] towing force as shown, determine the velocity of the crate in 3[s] starting from rest. ms= 0.5, mk= 0.3, Equations of Motion : Kinematics : The acceleration is constant, P is constant Example 2 The crate has a mass of 80 kg. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration. ms= 0.5, mk= 0.3, [solution] ms= 0.5 mk= 0.3 80(9.81) = 784.8[N] T 20o Ff = 0.5N : verge of slipping : impending motion Equations of equilibrium : SFx=0 ; SFy=0 ; N Tcos20o – 0.5N = 0 N + Psin20o – 784.8 = 0 T = 353.29 [N] , ……..(i) ……...(ii) N = 663.79 [N] [solution] 784.8 [N] 353.29 [N] a 20o Ff = 0.3N N Equations of Motion : SFy=may ; N – 784.8 + 353.29sin20o = 80(0) N = 663.97 [N] SFx=max ; 353.29cos20o – 0.3(663.97) = 80a a = 1.66 [m/s2] Equations of Motion: Normal and Tangential Coordinates o t When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates. b SFbub o SFnun t SFtut the force components acting on the particle SFt = mat SFn = man SFb = 0 Ft : tangential force Fn : centripertal force Fb : binormal force Example 1 The 3 kg disk is D is attached to the end of a cord as shown. The other end of the cord is attached to a ball-and-socket joint located at the center of the platform. If the platform is rotating rapidly, and the disk is placed on it and released from rest as shown, determine the time it takes for the disk to reach a speed great enough to break the cord. The maximum tension the cord can sustain is 100 N. mk = 0.1 Frictional force : Direction : F = mkND oppose the relative motion of the disk with respect to the platform Equations of Motion : SFn=man ; T = m(v2/) = 3v2 SFt=mat ; 0.1ND = 3at SFb=0; ND – 29.43 = 0 Setting T = 100 N to get critical speed vcr ND = 29.43 N at = 0.981 m/s2 vcr = 5.77 m/s Kinematics : Since at is constant, vcr = vo + att 5.77 = 0 + (0.981)t T = 5.89 s Example 2 At the instant q = 60o, the boys center of mass is momentarily at rest. Determine the speed and the tension in each of the two supporting cords of the swing when q = 90o. The boy has a weight of 300 N ( ≈ 30 kg). Neglect his size and the mass of the seat and cords At q = 60o, At q = 90o, v=0 v=? Free-body diagram n 2T q T=? Kinetic diagram n man mat W t t 2T n n q man θ W mat t t Equations of Motion : SFn=man ; SFt=mat ; 2T – Wsinq = man 2T – 300sinq = 30(v2/3) Wcosq = mat 300cosq = 30at 10cosq = at an = v2/ = v2/3 ……(i) ……(ii) 2T – 300sinq = 30(v2/3) 10cosq = at ……(i) ……(ii) Kinematics : (to relate at and v) vdv = at ds vdv = 10cosq dq v 90 0 60 s = q ds = dq vdv ( 10 cos q )dq v = 2.68 [m/s] Solving, we get The speed of the boy at q = 90o T = 186 [N] Quiz 1 The 400 [kg] mine car is hoisted up the incline using the cable and motor M. For a short time, the force in the cable is F = 3200t 2 [N], where t is in seconds. If the car has an initial velocity v1=2 [m/s] at s=0 and t=0, determine the distance it moves up the plane when t=2 [s]. Ans = 5.43 m 8 8 15 17 15 17