Lecture 8 Applictions of the ewton s Lws Problem-Solving ctics http://web.njit.edu/~sireno/ ewton s Lws I. If no net force ocects on body, then the body s velocity cnnot chnge. II. he net force on body is equl to the product of the body s mss nd ccelertion. III. When two bodies interct, the force on the bodies from ech other re lwys equl in mgnitude nd opposite in direction (F 12 = -F 21 ) Force is vector Force hs direction nd mgnitude Mss connects Force nd ccelertion; Physics 105; Fll 2009 Lecture 6 Andrei Sireno, JI 1 Lecture 7 Andrei Sireno, JI 2 Inertil Frmes: = on-inertil Frme Pseudo Forces Inertil Frme v Lecture 7 Andrei Sireno, JI 3 Lecture 7 Andrei Sireno, JI 4
Inertil Frme; here re no Pseudo Forces: cos = sin = m =g tn on-inertil Frme; here is Pseudo Force: m ewton s Lws do not wor!!! m = sin Lecture 7 Andrei Sireno, JI 5 Lecture 7 Andrei Sireno, JI 6 Problem-Solving ctics : Identify the body / bodies Exmples: bloc, puc, sphere, not, pulley, penguin, etc. Identify the msses of the bodies: m 1 = 5 g, m 2 = 10 g, etc Problem-Solving ctics (cont.): Identify the conditions of the body (moving or t re) t re mens F net = 0 Me setch to visulize the Problem Me choice for the coordinte syem (x-y) Recommended: x horizontl y verticl, or x long the plne of support nd y perpendiculr to the plne of support Lecture 6 Andrei Sireno, JI 7 if moving, then moving with connt velocity F net = 0 ccelerting F net 0 F net = m Lecture 6 Andrei Sireno, JI 8
Problem-Solving ctics (cont.): Problem-Solving ctics (cont.): Identify ll Forces nd their directions: Do the clcultions using FBD down to the ground (lwys) long the ring (if ny) perpendiculr to the support (if ny) f s = µ friction force (if ny) (only for the mx vlue of the force) sin sin Sttic Friction; mximum vlue F fr = µ opposite to the component of other forces prllel to the support cos cos Kinetic Friction; vlue F fr = µ in opposite to the velocity, prllel l to the support Lecture 6 Andrei Sireno, JI 9 For X: F net = sin ; = g sin For X: F net = sin -= 0 For Y: F net = - cos = 0 For Y: F net = - cos = 0 = g sin m = 0 Lecture 6 Andrei Sireno, JI 10 Problem-Solving ctics (cont.): Plug the numbers in the formuls: cos sin For X: F net = sin ; = g sin For Y: F net = - cos = 0 = g sin For = 30 o, = 9.8/2 m/s 2 = 4.9 m/s 2 cos sin For X: F net = sin -= 0 For Y: F net = - cos = 0 m = 0; = cos ; = sin Lecture 6 Andrei Sireno, JI 11 Cse 1 Cse 2 v f f = µ f = sin = 0 f = cos = Cse 3: below the criticl ngle v = 0 sin f = cos µ µ f = sin Lecture 6 Andrei Sireno, JI 12
Cse 4: At the criticl ngle f sin Cse 5 f = 0 f mx = sin = = mx cos µ µ s = tn = cos m = sin mx = g sin Cse 6 = cos = sin Lecture 6 Andrei Sireno, JI 13 Problem #1 two msses re t equilibrium ( no ccelertion, no friction) m 1 = 5 g, m 2 = 7 g; 2 = 30 o 1 =??? pulley m m 1 = 5 g 2 = 7 g 1 2 2 = 30 o Lecture 6 Andrei Sireno, JI 14 Problem #1 Problem #1 two msses re t equilibrium ( no ccelertion, no friction) m 1 = 5 g, m 2 = 7 g; 2 = 30 o 1 =??? 1 1 2 2 m 2 = 7 g For X1: F net = -m 1 g sin 1 + 1 = 0 For Y1: F net = 1 m 1 g cos 1 = 0 m 1 = 0; For X2: F net = m 2 g sin 2-2 = 0 For Y2: F net = 2 -m 2 g cos 2 = 0 m 2 = 0; 1 m 1 = 5 g 2 = 30 o m 1 1g m 2 2g Lecture 6 Andrei Sireno, JI 15 1 = m 1 g cos 1 ; 1 = m 1 g sin 1 2 = m 2 g cos 2 ; 2 = m 2 g sin 2 1 = 2 m 1 1g sin 1 = m 2 2g sin 2 m 1 /m 2 = sin 2 / sin 1 sin 1 =m 2 sin 2 / m 1 = 7 g * sin(30 o )/5 g =0.7; 1 = 44 o Lecture 6 Andrei Sireno, JI 16
Problem #1 L 1 = h / sin 1 L 2 L 2 = h / sin 2 h 2 L 1 m 1 /m 2 = sin 2 / sin 1 = (h / sin 2 )/(h/ sin 1 ) m 1 /m 2 = L 1 / L 2 Is the chin going to move??? Lecture 6 Andrei Sireno, JI 17 Smple Problem A coin of mss m res on boo tht hs been tilted t n ngle with the horizontl. When is incresed to 13, the coin is on the verge of sliding down the boo. Wht is the coefficient of tic friction µ s between the coin nd the boo? Lecture 6 Andrei Sireno, JI 18 Smple Problem For X: For Y: F net = 0 = - sin + f s F net = 0 = - cos = cos ; m = 0 long X direction: 0 = -g (sin - µ s cos ) sin - µ s cos = 0 tn = µ s µ s = tn (13 ) = 0.23 Lecture 6 Andrei Sireno, JI 19 Smple Problem (cont.) is incresed to 20 (the mx ngle is 13 ), the coefficient of tic friction µ s = 0.23 the coefficient of inetic friction µ =015 0.15 Wht is the coin ccelertion? m 0 : For X: F net = m = - sin + f For Y: F net = - cos = 0 = cos ; m = - sin + f = - sin + µ = = - sin + µ cos = = - (sin - µ cos ); = -g (sin - µ cos ); = -9.8*(sin 20-0.15* cos 20 ) m/s 2 = = -9.8*(0.34-0.14) 0.14) m/s 2 = -2 m/s 2 Lecture 6 Andrei Sireno, JI 20
Smple Problem (cont.) is decresed to op the coin. At wht ngle it will move with connt speed? the coefficient of tic friction µ s = 0.23 the coefficient of inetic friction µ = 0.15 his ngle is not 90!!! QZ #8 m 1 /m 2 = sin 2 / sin 1 m = 0 : = -g (sin - µ cos ) sin - µ cos = 0 tn = µ = tn 1 ( µ = tn 1 0.15 = 8.5 ) ( ) ote the difference: 13 nd 8.5 Lecture 6 Andrei Sireno, JI 21 m 2 = 7 g; 2 =30 o nd we cn vry m 1 nd 1 ; eglect friction 1. Wht is the smlle mss m 1 which cn blnce m o 2 = 7 g; 2 = 30 m 1 =??? 2. At wht ngle the smlle mss m 1 cn blnce m 2 = 7 g; 2 =30 o 1 =??? 3. If we cut the ring, which object (#1 with the mss m 1 t the ngle 1 or object #2 with m 2 = 7 g; 2 =30 o ) will hve bigger mgnitude of ccelertion??? (note tht = g sin ), 4. Me setch nd show the direction of 1 nd 2 Lecture 6 Andrei Sireno, JI 22