3.03b–3.03d, 3.03h - Newton's laws and equations of motion

3.03b–3.03d, 3.03h - Newton's laws and equations of motion

Use all three Newton laws to connect a body's forces with its motion. You will complete force diagrams, form directed equations for vehicles and lifts, work with force vectors, and decide when several bodies can be treated as one particle.

The first law and zero resultant force

Does a moving trolley need a forward resultant force to keep moving? Imagine an ideal trolley already rolling at a steady velocity: a forward pull of 6 N6\,\mathrm N exactly balances a backward resistance of 6 N6\,\mathrm N. Its velocity stays steady, although forces act on it.

You already know

Velocity has magnitude and direction. Zero acceleration means constant velocity, including the possible velocity zero. Add force vectors by adding corresponding components.

The distinction is between a force and the resultant of all forces on your chosen body. We use an inertial reference frame, such as the usual stationary-ground model: a frame in which Newton's laws apply.

Newton's first law

A particle remains at rest or moves with constant velocity unless acted on by a non-zero resultant external force. Equilibrium means zero resultant force; it permits uniform straight-line motion as well as rest.

Not in the formula sheet: learn it.

A force diagram shows actual external forces on that body. There is no additional “force of motion”. To complete an equilibrium diagram, add the force that cancels the sum of the known arrows.

[DIAGRAM: asset_name: Completing an equilibrium force diagram; asset_slug: equilibrium_missing_force; description: On a particle, forces (6,0) and (-2,3) newtons have sum (4,3), so the added force (-4,-3) newtons balances them; equal force-component scale and labelled directions; caption: Completing an equilibrium force diagram]
Completing an equilibrium force diagram

Worked example 1: completing a moving particle's diagram

Question. A particle moves with constant velocity. Its only forces are 6i N6\mathbf i\,\mathrm N, (−2i+3j) N(-2\mathbf i+3\mathbf j)\,\mathrm N and F\mathbf F. Find F\mathbf F and its magnitude.

Plan. Add the known forces, then reverse their resultant.

S=6i−2i+3j=4i+3jadd componentsF=−S=−4i−3jmake the sum zero∣F∣=(−4)2+(−3)2=5find the length\begin{aligned} \mathbf S&=6\mathbf i-2\mathbf i+3\mathbf j=4\mathbf i+3\mathbf j &&\text{add components}\\ \mathbf F&=-\mathbf S=-4\mathbf i-3\mathbf j &&\text{make the sum zero}\\ |\mathbf F|&=\sqrt{(-4)^2+(-3)^2}=5 &&\text{find the length} \end{aligned}

Why reverse the sum? Constant velocity requires both component resultants to vanish.

Answer. F=(−4i−3j) N\mathbf F=(-4\mathbf i-3\mathbf j)\,\mathrm N, magnitude 5 N5\,\mathrm N.

Check. 6−2−4=06-2-4=0 and 3−3=03-3=0; both directions balance.

Change only the motion statement from uniform motion to rest. The force calculation should stay the same: equilibrium describes forces, not a unique velocity.

Worked example 2: completing a stationary particle's diagram

Question. A particle remains at rest. Its only forces are 6i N6\mathbf i\,\mathrm N, (−2i+3j) N(-2\mathbf i+3\mathbf j)\,\mathrm N and F\mathbf F. Find F\mathbf F and its magnitude.

Plan. Use zero acceleration to cancel the known resultant.

S=(6−2)i+3j=4i+3jknown resultantF=−4i−3jequilibrium∣F∣=16+9=5Pythagoras\begin{aligned} \mathbf S&=(6-2)\mathbf i+3\mathbf j=4\mathbf i+3\mathbf j &&\text{known resultant}\\ \mathbf F&=-4\mathbf i-3\mathbf j &&\text{equilibrium}\\ |\mathbf F|&=\sqrt{16+9}=5 &&\text{Pythagoras} \end{aligned}

Why no new force for being at rest? Rest is one of the motions allowed by zero resultant.

Answer. F=(−4i−3j) N\mathbf F=(-4\mathbf i-3\mathbf j)\,\mathrm N, magnitude 5 N5\,\mathrm N.

Check. The three vectors sum to 0\mathbf0, exactly as in example 1.

If the velocity is non-zero, its direction still comes from the velocity vector, not from a vanished resultant.

Watch out

“Moving right, therefore resultant right” is wrong. Moving right at constant velocity gives zero resultant; a rightward resultant gives a rightward change in velocity.

Now complete a diagram with a different known resultant.

The second law and directed equations

When the forces stop balancing, use their signed resultant to find how velocity changes.

Equation of motion

An equation of motion applies Newton's second law to a named body in a stated direction. The force in the law is the resultant external force, not just the largest force or the driving force.

For constant mass, force and acceleration are parallel. One newton is the force that gives a mass of 1 kg1\,\mathrm{kg} acceleration 1 m s−21\,\mathrm{m\,s^{-2}}.

Newton's second law

For constant mass mm,

Fresultant=ma,∑F=maalong a chosen positive direction.\mathbf F_{\mathrm{resultant}}=m\mathbf a, \qquad \sum F=ma\quad\text{along a chosen positive direction}.

Use kg, N and m s−2\mathrm{m\,s^{-2}}. Not in the formula sheet: learn it.

Forces constant in time give constant acceleration. Then the constant-acceleration equations, which are in the formula sheet, connect the force model with subsequent motion.

Worked example 3: drive greater than resistance

Question. A 600 kg600\,\mathrm{kg} vehicle moves right at 4 m s−14\,\mathrm{m\,s^{-1}}. A constant drive of 1500 N1500\,\mathrm N acts right and constant resistance of 300 N300\,\mathrm N acts left. Find its signed acceleration and velocity after 3 s3\,\mathrm s, taking right as positive.

Plan. Subtract resistance from drive, then use v=u+atv=u+at.

1500−300=600avehicle, right positivea=2 m s−2divide by massv=4+2(3)=10 m s−1constant acceleration\begin{aligned} 1500-300&=600a &&\text{vehicle, right positive}\\ a&=2\,\mathrm{m\,s^{-2}} &&\text{divide by mass}\\ v&=4+2(3)=10\,\mathrm{m\,s^{-1}} &&\text{constant acceleration} \end{aligned}

Why subtract 300300? It opposes the chosen positive direction, even though its magnitude is positive.

Answer. a=+2 m s−2a=+2\,\mathrm{m\,s^{-2}}, v=+10 m s−1v=+10\,\mathrm{m\,s^{-1}}.

Check. The positive resultant increases the positive velocity.

Change only the drive to 0 N0\,\mathrm N. Keep the same initial motion, interval, resistance and requested outputs.

Worked example 4: resistance greater than drive

Question. A 600 kg600\,\mathrm{kg} vehicle moves right at 4 m s−14\,\mathrm{m\,s^{-1}}. Its drive is zero and constant resistance is 300 N300\,\mathrm N left. Find its signed acceleration and velocity after 3 s3\,\mathrm s, taking right as positive.

Plan. Use the leftward resultant, then update the velocity.

0−300=600asigned force suma=−0.5 m s−2divide by massv=4−0.5(3)=2.5 m s−1use v=u+at\begin{aligned} 0-300&=600a &&\text{signed force sum}\\ a&=-0.5\,\mathrm{m\,s^{-2}} &&\text{divide by mass}\\ v&=4-0.5(3)=2.5\,\mathrm{m\,s^{-1}} &&\text{use }v=u+at \end{aligned}

Why is negative acceleration compatible with rightward motion? It reduces the rightward velocity; it does not instantly reverse it.

Answer. a=−0.5 m s−2a=-0.5\,\mathrm{m\,s^{-2}}, v=+2.5 m s−1v=+2.5\,\mathrm{m\,s^{-1}}.

Check. Stopping would take 4/0.5=8 s4/0.5=8\,\mathrm s, so the leftward resistance remains appropriate throughout these three seconds.

A resistance that opposes motion must change its description if the motion reverses. A constant-resistance model should not silently continue through a stop into an unphysical reversal. Changing mass also needs a different model; we keep mass fixed.

Watch out

The line 1500=600a1500=600a omits resistance. Write every contributing force before simplifying; mama is not another arrow on the force diagram.

Try a smaller drive, keeping the same style of directed equation.

Weight and support in vertical motion

A lift passenger can move upwards while feeling lighter. The support force responds to acceleration, while weight stays the same in a fixed gravitational field.

Weight and a vertical equation of motion

Weight is mgmg downwards. For a passenger with upward floor reaction RR, taking upwards as positive,

R−mg=ma.R-mg=ma.

Replace RR by tension TT for a cable-supported load with the same force directions. These are applications of Newton's law. Not in the formula sheet: learn it.

Use g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}} unless told otherwise. Mass measures inertia and stays in kg; weight is a force measured in N.

Worked example 5: accelerating upwards

Question. A 60 kg60\,\mathrm{kg} passenger moves upwards with acceleration +1.2 m s−2+1.2\,\mathrm{m\,s^{-2}}. Find the upward reaction and state whether it exceeds the passenger's weight. Take upwards positive and g=9.8 m s−2g=9.8\,\mathrm{m\,s^{-2}}.

Plan. Subtract weight from reaction in the upward equation.

mg=60(9.8)=588 NweightR−588=60(1.2)upward resultantR=660 Nsolve for support\begin{aligned} mg&=60(9.8)=588\,\mathrm N &&\text{weight}\\ R-588&=60(1.2) &&\text{upward resultant}\\ R&=660\,\mathrm N &&\text{solve for support} \end{aligned}

Why include mgmg even inside a lift? Gravity continues to act; the floor supplies a separate force.

Answer. R=660 NR=660\,\mathrm N, exceeding weight by 72 N72\,\mathrm N.

Check. (660−588)/60=1.2(660-588)/60=1.2, the specified acceleration.

Change only the acceleration's direction. The passenger still moves upwards, but now slows down.

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