Chapter 3 Force and Motion 4th Edition

What You Will Learn Newton’s 3 Laws of Motion Kinematics In SI and Engineering English units gc usage to clarify units Free body diagrams Kinematics Speed vs. time diagrams Exploring Engineering

Newton’s Laws of Motion Newton’s First Law. The speed v of a body remains constant unless the body is acted upon by an external force F. Newton’s Second Law. The acceleration a of a body is parallel to and directly proportional to the net force F applied to it and inversely proportional to its mass m, or a  F/m. Newton’s Third Law. The mutual forces of action and reaction between two bodies are equal, opposite, and collinear. Exploring Engineering 3

Newton’s 1st Law First law implies that speed v = constant An object at rest will stay at rest unless an unbalanced force acts upon it. An object in motion will not change its speed unless an unbalanced force acts upon it. Throw up a very heavy ball in an open sport’s car and a stationary observer sees the ball execute a parabola In the car, an observer sees it go straight up and back down Exploring Engineering 4

Newton’s 2nd Law An object will accelerate if the unbalanced force on the object is not zero The direction of the acceleration is the same as the direction of the unbalanced force. The magnitude of the acceleration is directly proportional to the unbalanced force, and inversely proportional to the mass of the object and written F  ma. Exploring Engineering 5

Newton’s 3rd Law For every action, there is an equal and opposite reaction. Free Body Diagrams, FBD, are used for systems in static equilibrium such as bridges and buildings and thus occupy much of the activities of civil engineers Exploring Engineering 6

Back to Newton’s 2nd Law Recall the definition of force as proportional to mass × acceleration or F  ma What is the missing proportionality constant? Suppose we knew a force F1 corresponding to a mass m1 being accelerated by a1. Then F1  m1a1 Therefore Exploring Engineering 7

More Newton’s 2nd Law If we say the length scale is the meter, the mass scale is the kilogram and the time scale is the second, and F is defined to be in newtons (N). To complete the definition let’s say F1 is 1 N when m1 is 1 kg and a1 is 1 m/s2. Then m1a1/F1 = 1 [dimensionless] and F = ma, which is the way you have previously seen Newton’s 2nd Law. Now we know the units of the force and its magnitude. This is called the SI or MKS (meter-kilogram-second) system. Exploring Engineering 8

Still More Newton’s 2nd Law If we say the length scale is in feet (ft), the mass is in pounds mass (lbm) and the time scale in seconds (s), and F is defined to be in pounds force (lbf). To complete the definition let’s say F1 is 1 ft when m1 is 1 lbm and a1 is 32.174 m/s2. Then gc = m1a1/F1 = 1 × 32.174/1 [lbm][ft/s2]/[lbf] and F = ma/gc. This is called the Engineering English unit system Now we know the units of the force and its magnitude. Note gc = 32.174 [lbm][ft]/[lbf s2] and not to be confused with g = acceleration due to gravity, 32.174 ft/s2 Exploring Engineering 9

Ok, More Newton’s 2nd Law What force in N is needed to accelerate a frictionless motorcycle of 250. kg by 7.50 m/s2? Need: F on a mass of 250. kg accelerated by 7.50 m/s2 Know: Newton's 2nd Law How: In SI units F = ma Solve: F = ma = 250. × 7.50 = 1.875 × 103 [kg][m/s2] = 1.88 × 103 N Note: 1 N = is about the weight of a small apple on earth! Exploring Engineering 10

Even More of Newton’s 2nd Law What’s the force on a frictionless wheel-barrow of 50.0 lbm being accelerated at 2.10 ft/s2? What’s the weight of the barrow? Need: a) F on a mass of 50. lbm accelerated by 2.10 ft/s2; b) wt of wheel-barrow Know: Newton's 2nd Law, acceln due to gravity g = 32.2 ft/s2; gc = 32.2 [lbm][ft]/[lbf s2] How: Engineering English units F = ma/gc Solve: a) F = ma/gc = 50. × 2.10/32.2 = 3.26 [lbm ft/s2][lbf s2/lbm ft] = 3.3 lbf; b) Wt = mg/gc = 50. × 32.2/32.2 = 50. lbf Exploring Engineering 11

Newton’s 3rd Law If you push on something it pushes back with the same force (but not the same acceleration since the pushed masses may be different) If you fire a 15.0 gram bullet from a 5.0 kg rifle. If the initial acceleration of the bullet from the 60.0 cm barrel is 1000. m/s, what is the recall force you experience from firing the gun at your shoulder? Exploring Engineering 12

More Newton’s 3rd Law Need: Recoil force from the rifle. Know: Rifle weighs 5.0 kg, the bullet 0.015 kg. The barrel (down which the bullet accelerates) is 0.60 m and the bullet leaves the muzzle at 1000. m/s. How: Assume the acceleration in the barrel is uniform and use Newton’s 3rd law to calculate the recoil force. Exploring Engineering 13

Newton’s 3rd Law - Example Accept (for the moment) that the acceleration of the bullet = (½v2)/barrel length, or a = ½ × 1000.2/0.60 [m/s]2/[m] = 8.3 × 105 m/s2. Therefore the Force on the bullet is F = ma = 0.015 × 8.3 × 105 [kg][m/s2] = 1.3 × 104 N The reaction on the rifle is in the opposite direction to that on the bullet = - 1.3 × 104 N. (If you also want the initial acceleration of the rifle it is - 1.3 × 104/5.0 = -2.6 × 103 [kg m/s2][1/kg] = -2.6 × 103 m/s2) Exploring Engineering 14

Free Body Diagram (FBD) Example FBDs are used to keep track of forces A stationary crate weighing 100. N sits on a 30.0 degree ramp Resolve the forces on the crate and the slope parallel to the hill and normal to it. Need: FBD for crate on a 30 degree slope Know: Forces on the crate and the slope are equal and opposite How: Resolve forces parallel and normal to the slope 100. N Fr 30º 87.N 50.N Ramp Crate Exploring Engineering 15

Free Body Diagram Example Solve: The weight of the block resolves to 100. N cos30º = 86.6 N normal to the slope and 100.sin30º = 50.0 N parallel to the slope. The friction force Fr on the block must be equal and opposite to the force on the block. The frictional force is thus 50.0 N. Exploring Engineering 16

Kinematics The study of motion without consideration of the associated forces. In this elementary treatment we will only use 1-dimensional methods. In > 1-D, speed is not equivalent to velocity because velocity is a vector with components in each direction. In 1-D speed is the magnitude of velocity and is a scalar. We will graphical solutions only. Basically no equations are used and the method emphasizes understanding not formula plugging. Exploring Engineering 17

Cartesian Geometry Exploring Engineering 18

Kinematics and Cartesian Geometry Exploring Engineering 19

Kinematics, v-t and Constant a Max speed v, Speed Start accn Slope is acceleration Area under curve is distance travelled. Time Exploring Engineering 20

Kinematics Example You accelerate from rest at a constant 5.0 m/s2 for 13 s. What’s 1) Your final speed and 2) how far will you have travelled? Slope = v/t = 5.0 m/s2 v = v = 5.0 × 13 [m/s2][s] = 65 m/s Distance travelled = ½ v t = ½ × 65 × 13 [ m/s][s] = 4.2 × 102 m Speed, m/s 5.0 m/s2 v t 0 s Time, s 13 s Exploring Engineering 21

Summary Newton’s 3 Laws: Kinematics: 1) Bodies that remain undisturbed continue until external forces act on them 2) Force proportional to mass × acceleration and how to manipulate units for gc formulation 3) Action and reaction are equal and opposite Kinematics: Central position of speed vs. t diagrams. Exploring Engineering 22