PHY 101: Lecture 4C 4.1 The Concepts of Force and Mass 4.2 Newton’s First Law of Motion 4.3 Newton’s Second Law od Motion 4.4 The Vector Nature of Newton’s Second Law of Motion 4.5 Newton’ Third Law of Motion 4.6 Types of Forces: An Overview 4.7 The Gravitational Force 4.8 The Normal Force 4.9 Static and Kinetic Frictional Forces 4.10 The Tension Force 4.11 Equilibrium Applications of Newton’s Laws of Motion 4.12 Nonequilibrium Applications of Newton’s Laws of Motion

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4.11 Equilibrium Applications of Newton’s Laws of Motion

Types of Problems Equilibrium Problems Nonequilibrium Problems Sum of forces = 0 No acceleration Static Problems - Velocity = 0 Dynamic Problems – Velocity not 0 Nonequilibrium Problems Sum of forces not 0 Acceleration

Static Equilibrium – Example 1A You and two friends find three ropes tied together with a single knot and decide to have a three-way tug-of-war Alice pulls to the west with 100 N of force Bob pulls to the south with 200 N How hard, and in which direction, should you pull to keep the knot from moving?

Static Equilibrium – Example 1B T2 / T1 = tanq = tan-1(T2/T1) = tan-1(200/100)=63.40 T3 = T1/cosq T3 = 100/cos(63.4) T3 = 223 N SFx = T1x + T2x + T3x SFy = T1y + T2y + T3y T1x = - T1 T2x = 0 T3x = T3cosq T1y = 0 T2y = - T2 T3y = T3sinq -T1 + T3cosq = 0 -T2 + T3sinq = 0 T1 = T3cosq T2 = T3sinq

Static Equilibrium - Example 2A What are tensions on each rope? Mass Upper: SFy = Wy + Ny + fy + Ty + Py=0 Wy=-(mU)g=-(10)(9.8)=-98 Ny = 0 fy = 0 Ty = TU-TL Py = 0 -98 + 0 + 0 + TU – TL + 0 = 0 TU – TL = 98 N Mass Lower: SFy = Wy + Ny + fy + Ty+Py=0 Wy=-mLg = -(10)(9.8) = -98 Ny = 0 fy = 0 Ty = TL Py = 0 -98+0+0+TL+0 =0 TL = 98 N

Static Equilibrium - Example 2B What are tensions on each rope? TU – TL = 98 N TL = 98 N Tu = 98 + 98 = 196 N

Static Equilibrium - Alternative Example 2 What are tensions on each rope? Mass Upper and Mass Lower: SFy = Wy + Ny + fy + Ty + Py=0 Wy=-(mU+mL)g=-(20)(9.8)=-196 Ny = 0 fy = 0 Ty = TU Py = 0 -196 + 0 + 0 + TU + 0 = 0 TU = 196 N Mass Lower: SFy = Wy + Ny + fy + Ty+Py=0 Wy=-mLg = -(10)(9.8) = -98 Ny = 0 fy = 0 Ty = TL Py = 0 -98+0+0+TL+0 =0 TL = 98 N

Dynamic Equilibrium – Example 1A A car with a weight of 15,000 N is being towed up a 200 slope at constant velocity Friction is negligible The tow rope is rated at 6000 N maximum tension Will it break?

Dynamic Equilibrium – Example 1B SFx = Wx + Nx + fx + Tx + Px SFy = Wy + Ny + fy + Ty + Py WX = -wsinq Wy = -wcosq Nx = 0 Ny= n Tx = T Ty = 0 T – Wsinq = 0 N – Wcosq = 0 T = (15000)sin(20) = 5130 N Rope will not break

PHY 101: Lecture 4 Force and Newton’s Laws of Motion 4.12 Non-Equilibrium Applications of Newton’s Laws of Motion

NonEquilibrium – Example 1 What is the mass of an object that accelerates at 3.0 m/s2 under the influence of a 5.0-N net force? Assume direction is horizontal F = ma m = F/a m = 5.0 / 3.0 = 1.7 kg

NonEquilibrium – Example 2 Two forces act on an object of mass 4.00 kg One force is 40.0 N in the +x-direction The other force is 60.0 N in the + y-direction Find magnitude and direction of the acceleration of the object x-direction SFx = max 40 = 4ax 40/4 = ax = 10 m/s2 y-direction SFy = may 60 = 4ay 60/4 = ay = 15 m/s2 magnitude of a = sqrt(102 + 152) = 18 m/s2 q = tan-1(15/10) = 560

NonEquilibrium – Example 3 Horizontal force of 300 N is applied to a 75.0 kg box Box slides on a level floor, opposed by a force of kinetic friction of 120 N What is magnitude of the acceleration of box? SFx = Wx + Nx + (fk)x + Tx + Px = max Wx = 0 Nx = 0 (fk)x = -120 Tx = 0 Px = 300 0 + 0 -120 + 0 + 300 = 75ax ax = 180/75 = 2.4 m/s2

NonEquilibrium – Example 4 A boy pulls a box of mass 30 kg with a force of 25 N at an angle of 300 above the horizontal (a) Ignoring friction, what is the acceleration of the box? (b) What is the normal force exerted on the box by the ground? SFx = Wx + Nx + (fk)x + Tx + Px = max SFy = Wy + Ny + (fk)y + Ty + Py = may Wx = 0 Wy = mg = 30(-9.8) = -294 Nx = 0 Ny = N Tx = 0 Ty = 0 (fk)x 0 (fk)y = 0 Px = 25 cos30 = 21.7 Py = 25 sin30 = 12.5 (a) x-direction 0 + 0 + 0 + 0 + 21.7 = 30ax ax = 21.7/30 = 0.72 m/s2 (b) y-direction -294 + N +12.5 = -281.5 + N = 30ay = 0 N = 281.5 newtons

NonEquilibrium – Example 5A A 1500 kg car is pulled by a tow truck The tension in the tow rope is 2500 N A 200 N friction force opposes the motion What is the car’s acceleration

NonEquilibrium – Example 5B SFx = Wx + Nx + (fk)x + Tx + Px = max SFy = Wy + Ny + (fk)y + Ty + Py = may Wx = 0 Wy = mg = 1500(-9.8) = -14700 Nx = 0 Ny = N Tx = 2500 Ty = 0 (fk)x = -200 (fk)y = 0 Px = 0 Py = 0 0 + 0 -200 +2500 + 0 = 1500ax ax = 2300/1500 = 1.53 m/s2

NonEquilibrium – Example 6 The Atwood machine consists of two masses suspended from a fixed pulley m1 = 0.55 kg m2 = 0.80 kg (a) What is the acceleration of the system? (b) What is magnitude of tension in string?

NonEquilibrium – Example 6a Mass 1: SFy = Wy + Ny + (fk)y + Ty + Py = may Wy = -m1g Ny = 0 Ty = T (fk)y = 0 Py = 0 -0.55g +0 + 0 + T + 0 = 0.55ay Mass 2: Wy = -m2g Ny = 0 -0.80g +0 + 0 + T + 0 = -0.80ay Subtract equations: 0.25g = 1.35ay ay = 1.8 m/s2

NonEquilibrium – Example 6b -0.55g + T = 0.55ay T = 0.55ay + 0.55g T = 0.55(1.8) + 0.55(9.8) = 6.4 N

NonEquilibrium Alternative Example 6A The Atwood machine consists of two masses suspended from a fixed pulley m1 = 0.55 kg m2 = 0.80 kg (a) What is the acceleration of the system? (b) What is magnitude of tension in string?

NonEquilibrium Alternative Example 6B Pull = m2g Pull = m1g a Lift masses m1 and m2 so that they are horizontal Gravity on m1 becomes a pull to the left Gravity on m2 becomes a pull to the right Treat two masses and rope between as a single system Tension in the rope is an Internal force and is ignored in Newton’s Second Law

NonEquilibrium Alternative Example 6C Mass 1 + Mass2: SFx = Wx + Nx + (fk)x + Tx + Px = max Wx = 0 Nx = 0 Tx = 0 (fk)x = 0 Px = -m1g + m2g -0.55g+0.80g + 0 + 0 + 0 + 0 = (0.55 + 0.8)ax ax = .25(9.8) / 1.35 = 1.8 m/s2 Mass 1: Wx = -m1g Nx = 0 Tx = T (fk)x = 0 Px = 0 -0.55g +0 + T + 0 + 0 = 0.55(1.8) T = 0.99 + 0.55(9.8) = 6.4 N

NonEquilibrium – Example 7A Three blocks are pulled along a frictionless surface by a horizontal force of F = 18.0 N (a) What is the acceleration of the system? (b) What are tension forces in the strings?

NonEquilibrium – Example 7B Mass 1: SFx = Wx+Nx+(fk)x+Tx+Px =m1ax Wx = 0 Nx = 0 Tx = T1 (fk)x = 0 Px = 0 0 + 0 + 0 + T1 + 0 = m1ax T1 = m1ax T1 = 1.0ax (Equation 1) Mass 2: SFx=Wx+Nx+(fk)x+Tx+Px=m2ax Wx = 0 Nx = 0 Tx = -T1+T2 (fk)x = 0 Px = 0 0 + 0 + 0 - T1 + T2 + 0 = m2ax -T1 + T2 = m2ax -T1 + T2 = 2.0ax (Equation 2)

NonEquilibrium – Example 7C Mass 3: SFx = Wx Nx+(fk)x+Tx+Px =m3ax Wx = 0 Nx = 0 Tx = -T2 (fk)x = 0 Px = P 0 + 0 + 0 - T2 + P = m3ax -T2 + P = m3ax -T2 + 18.0 = 3.0ax (Equation 3) Add Equations 1 and 2 T2 = 3.0ax (Equation 4) Add Equations 3 and 4 18.0 = 6.0ax ax = 18.0/6.0 = 3.0 m/s2 Substitute into Equation 1 T1 = 1.0(3.0) = 3.0 N Substitute into Equation 3 T2 = 18.0 – 3.0(3.0) = 9.0 N

NonEquilibrium Alternative Example 7A Mass 1: SFx=Wx+Nx+(fk)x+Tx+Px=m1ax Wx = 0 Nx = 0 Tx = T1 (fk)x = 0 Px = 0 0 + 0 + 0 + T1 + 0 = m1ax T1 = (1.0)(3.0) = 3.0 N Mass 1, Mass 2, and Mass 3: SFx = Wx+Nx+(fk)x+Tx+Px =max Wx = 0 Nx = 0 Tx = 0 (fk)x = 0 Px = P 0+0+0+0+P = (m1+m2+m3)ax 18.0 = 6.0ax ax=18.0 / 6.0=3.0 (Equation 1)

NonEquilibrium Alternative Example 7B Mass 3: SFx = Wx + Nx + (fk)x + Tx + Px =m3ax Wx = 0 Nx = 0 Tx = -T2 (fk)x = 0 Px = P 0 + 0 + 0 – T2 + P = m3ax T2 = P – m3ax T2 = 18 – (3.0)(3.0) = 9.0 N

NonEquilibrium – Example 8A Pulleys are frictionless There is no friction between mass m3 and the table What is the acceleration of the system? m1=0.25 kg, m2 = 0.50 kg, m3 = 0.25 kg

NonEquilibrium – Example 8B Mass 1: SF =Wy+Ny+(fk)y+Ty+Py=m1a Wy = -m1g Ny = 0 Ty = T1 (fk)y = 0 Py = 0 -m1g + 0 + 0 + T1 + 0 = m1a -m1g+T1=m1a (Equation 1) Mass 2: SF =Wy+Ny+(fk)y+Ty+Py=m2a Wy = -m2g Ny = 0 Ty = T2 (fk)y = 0 Py = 0 -m2g + 0 + 0 + T2 + 0 = m2(-a) -m2g+ T2=-m2a (Equation 2)

NonEquilibrium – Example 8C Mass 3: SFx=Wx+Nx+(fk)x+Tx+Px=ma Wx = 0 Nx = 0 Tx = -T1 + T2 (fk)x = 0 Px = 0 0 + 0 + 0 – T1 + T2 + 0 = m3a – T1+T2=m3a (Equation 3) Subtract equation 2 from equation 1 (m2-m1)g-T2+T1=(m1+m2)a (Eq. 4) Add equations 3 and 4 (m2-m1)g=(m1+m2+m3)a a = (m2-m1)g/(m1+m2+m3) a = (0.50-0.25)(9.8)/(0.50+0.25+0.5) a = 2.45 m/s2

NonEquilibrium Alternative Example 8A Mass 1: SF =Wy+Ny+(fk)y+Ty+Py=m2a Wy = 0 Ny = 0 Ty = T1 (fk)y = 0 Py = -m1g 0 + 0 + 0 + T1 – m1g = m1a -m1g+ T1=m1a T1=m1a+m1g = (0.25)(9.8+2.45) T1 = 3.1 N Mass 1, Mass2, and Mass3 SFx=Wx+Nx+(fk)x+Tx+Px=ma Wx = 0 Nx = 0 Tx = 0 (fk)x = 0 Px = -m1g+m2g 0+0+0+0-m1g+m2g = ma -m1g+m2g=ma a = g(m2-m1)/(m1+m2+m3) a=(9.8)(0.50-0.25)/1=2.45 m/s2

NonEquilibrium Alternative Example 8B Mass 3: SF =Wy+Ny+(fk)y+Ty+Py=m2a Wy = 0 Ny = 0 Ty = T2 (fk)y = 0 Py = -m3g 0 + 0 + 0 + T2 – m3g = -m3a -m3g+ T2=-m3a T2=-m3a+m3g = (0.25)(9.8-2.45) T2 = 1.8 N

NonEquilibrium Alternative Example 9A Weight of the block on table is 422 N Weigh of the hanging block is 185 N Ignoring all friction and assume the pulley is massless (a) Find the acceleration of the two blocks (b) Find the tension in the cord Mass of table block is m1 = 422/9.8 =43.06 kg Mass of hanging block is m2 = 185/9.8 = 18.88 kg

NonEquilibrium Alternative Example 9B Mass 1: SFx=Wx+Nx+(fk)x+Tx+Px=m1a Wx = 0 Nx = 0 Tx = T (fk)x = 0 Px = 0 0+0+0+T+0 = m1a T = m1a T = 43.06(2.99) = 128.8 N Mass 1 and Mass2: SFx=Wx+Nx+(fk)x+Tx+Px=ma Wx = 0 Nx = 0 Tx = 0 (fk)x = 0 Px = m2g 0+0+0+0+m2g = (m1 + m2)a m2g = (m1 + m2)a a = -g(m2/(m1+m2) a=(9.8)(18.88)/(61.94)=2.99 m/s2