Applying Newton's laws: springs, drag, curves and accelerating frames

Lesson 3 set out Newton's three laws and the free-body method, and applied them to blocks, inclines, friction, connected bodies and circular motion. This lesson extends the same method to forces and situations that engineers meet daily. Springs push back in proportion to how far they are stretched; air and water resist motion with a force that grows with the square of speed, which sets the terminal velocity of a falling object and most of the fuel use of a highway vehicle; tires roll with a resistance of their own; banked roads and loops let gravity and the normal force supply centripetal force; and inside an accelerating vehicle or lift, objects seem to feel extra forces. The method never changes: isolate the body, draw every force, choose axes, apply \(\sum\mathbf{F} = m\mathbf{a}\) and check the answer.

Springs and Hooke's law

Most solid objects deform slightly under load and push back. For a spring, and for many structural members within their elastic range, the restoring force is proportional to the deformation:

where x is the extension or compression from the natural length and k is the spring constant (stiffness) in N/m. The minus sign says the force opposes the deformation. Springs combine like this: in parallel (side by side, sharing a load) their stiffnesses add, \(k = k_1 + k_2\); in series (end to end, carrying the same force) their flexibilities add, \(1/k = 1/k_1 + 1/k_2\).

Fluid drag and terminal velocity

An object moving through air or water at ordinary engineering speeds feels a drag force that grows with the square of its speed:

where ρ is the fluid density, A the frontal area and \(C_D\) a dimensionless drag coefficient that depends on shape: about 1.0 to 1.3 for a person or a flat plate broadside to the flow, 0.47 for a sphere, 0.25 to 0.35 for a modern car and under 0.1 for a streamlined body. (At very low speeds, for dust or droplets, drag is proportional to v instead, the case of Lesson 2.) A falling object speeds up until drag balances its weight, reaching the terminal velocity:

Drag and the energy cost of speed

Because drag force grows as v², the power needed to overcome it, \(P = F_Dv\), grows as v³. This is why fuel consumption rises steeply at highway speeds.

Rolling resistance

A tire flexes as it rolls, losing energy in the rubber, which appears as a resisting force roughly proportional to the load: \(F_R = C_{rr}N\), with a rolling resistance coefficient \(C_{rr}\) of about 0.008 to 0.015 for car tires on asphalt, 0.001 to 0.002 for steel wheels on rail, and much higher on gravel or sand. For a 1500 kg car with \(C_{rr} = 0.010\), \(F_R = 0.010 \times 1500 \times 9.81 = 147\) N, independent of speed to a first approximation. So at low speeds rolling resistance dominates, while above about 60 to 80 km/h aerodynamic drag takes over. The very low rolling resistance of steel on rail is the main reason trains are so energy-efficient for freight.

Static and kinetic friction in braking

Lesson 3 distinguished static friction, which can rise to \(\mu_sN\) before slipping begins, from kinetic friction \(\mu_kN\) once surfaces slide, with \(\mu_k\) usually smaller. Braking shows why the difference matters. A tire that is still rolling grips the road with static friction; a locked, skidding tire slides with kinetic friction. On dry asphalt, typical values might be \(\mu_s \approx 0.9\) and \(\mu_k \approx 0.7\). Since the maximum deceleration is \(\mu g\), a car braking at the limit of grip stops at up to 8.8 m/s², but a skidding car only at 6.9 m/s². From 100 km/h, the stopping distances \(v^2/(2\mu g)\) are about 44 m and 56 m, a difference of a car length or two. Worse, a skidding wheel cannot steer. Anti-lock braking systems sense when a wheel is about to lock and release and reapply the brake many times a second, keeping the tire near its peak static grip while preserving steering. On ice, both coefficients may fall below 0.1, multiplying stopping distances by nine or more, which is why winter tires, sanding and lower speed limits matter so much on Canadian and northern US roads.

Curves: banking, friction and loops

A vehicle on a curve needs a net force of \(mv^2/r\) toward the center (Lesson 3). On a flat road, only friction between tires and road can supply it. Banking the road (superelevation) tilts the normal force so that part of it points toward the center. With no friction at all, a curve banked at angle θ is perfect for one speed, the design speed:

With side friction coefficient f available as well, the maximum safe speed becomes \(v = \sqrt{rg\,(\tan\theta + f)/(1 - f\tan\theta)}\).

In a vertical loop, as on a roller coaster, the forces change around the loop. At the top, gravity and the track's push both point down toward the center, and the minimum speed to stay on the track is the one at which gravity alone provides the centripetal force, \(v_{min} = \sqrt{gr}\). Designers keep speeds well above this, and use clothoid-shaped loops (MATH-209) to limit the forces at the bottom, where riders feel heaviest.

Accelerating frames and apparent weight

Newton's laws hold in inertial frames, those not accelerating. Inside an accelerating lift, vehicle or aircraft, it often helps to analyze the motion as seen from the inside. The rule is simple: in a frame accelerating at \(\mathbf{a}_0\), add a fictitious (inertial) force \(-m\mathbf{a}_0\) to every body, then apply the equilibrium or motion equations as if the frame were at rest. The fictitious force is not a real interaction, but it correctly accounts for the frame's acceleration.

A person in a lift accelerating upward at a feels the floor push with \(N = m(g + a)\): they feel heavier. Accelerating downward, \(N = m(g - a)\); in free fall, N = 0 and they feel weightless, as astronauts in orbit do (Lesson 10). What a bathroom scale reads, and what structures and bodies actually feel, is this apparent weight.

Checking your answers

Applying Newton's laws to new situations is where most errors occur, and a few checks catch nearly all of them. Are the units right? Does the answer behave sensibly in limiting cases: does a banking formula give zero speed for a flat road with no friction, does terminal velocity grow with mass? Is the sign of each force consistent with the free-body diagram? And is the size plausible: a car's deceleration of 50 m/s² or a skydiver's terminal velocity of 400 m/s should prompt a second look.