Mechanics 8.5 - Addition and resultant forces in dynamics

Mechanics 8.5 - Addition and resultant forces in dynamics

Combine forces by adding their vectors, then use the resultant to find acceleration in a plane. You will turn magnitude-and-direction descriptions into components, recover directions without losing the quadrant, and use a given acceleration direction to determine unknown forces. The same two component equations connect force diagrams to motion.

Resultant force in a plane

Can a 3 N3\,\mathrm N force and a 4 N4\,\mathrm N force combine to make just 5 N5\,\mathrm N? Yes: let one point east and the other north. Their component triangle has hypotenuse 32+42=5\sqrt{3^2+4^2}=5, so adding their magnitudes would give the wrong physical effect.

You already know

Vectors add component by component. Newton's second law uses the resultant of all forces on the chosen particle: ∑F=ma\sum\mathbf F=m\mathbf a.

The useful surprise is broader: as the angle between those same two forces changes, their resultant magnitude can take any value from 11 to 7 N7\,\mathrm N. Opposite directions give 4−3=14-3=1; the same direction gives 4+3=74+3=7. Direction is part of the data.

Resultant force

The resultant force is the vector sum of the forces acting on a particle. It is the single force with the same effect on the particle's acceleration as those forces acting together.

Use bold R\mathbf R for the resultant here. It is a vector, unlike a normal reaction magnitude also sometimes denoted by RR.

Add force components

If Fk=Xki+Ykj\mathbf F_k=X_k\mathbf i+Y_k\mathbf j, then

R=∑Fk=(∑Xk)i+(∑Yk)j\mathbf R=\sum\mathbf F_k =\left(\sum X_k\right)\mathbf i+\left(\sum Y_k\right)\mathbf j

With constant positive mass mm, a=R/m\mathbf a=\mathbf R/m.

Not in the formula booklet: learn it.

The head-to-tail vector diagram and the component sum describe the same addition. Never place a third-law partner from another body's diagram in this sum.

[DIAGRAM: asset_slug: p3_8_5_understand_and_use_addition_of_forces_and_resultant_forces_in_dynamics__diagram_01; asset_name: Add force vectors; description: Perpendicular 3 N east and 4 N north forces on one particle, followed by a head-to-tail component triangle with 5 N resultant. Acceleration is parallel to the resultant for positive mass.]
Diagram

Add the signed components before dividing by the mass.

Worked example 1: three forces and a mass

Question. Forces (3i+7j) N(3\mathbf i+7\mathbf j)\,\mathrm N, (−5i+2j) N(-5\mathbf i+2\mathbf j)\,\mathrm N and (8i−4j) N(8\mathbf i-4\mathbf j)\,\mathrm N act on a 2 kg2\,\mathrm{kg} particle. Find resultant and acceleration.

Plan. Sum each component, then apply Newton's second law.

R=(3−5+8)i+(7+2−4)jsigned sums=(6i+5j) Na=R2divide by mass=(3i+52j) m s−2\begin{aligned} \mathbf R&=(3-5+8)\mathbf i+(7+2-4)\mathbf j &&\text{signed sums}\\ &=(6\mathbf i+5\mathbf j)\,\mathrm N\\ \mathbf a&=\frac{\mathbf R}{2} &&\text{divide by mass}\\ &=(3\mathbf i+\tfrac52\mathbf j)\,\mathrm{m\,s^{-2}} \end{aligned}

Why add first? Newton's second law concerns the combined effect, not any one force in isolation.

Answer. R=(6i+5j) N\mathbf R=(6\mathbf i+5\mathbf j)\,\mathrm N and a=(3i+52j) m s−2\mathbf a=(3\mathbf i+\frac52\mathbf j)\,\mathrm{m\,s^{-2}}.

Check. Multiplying each acceleration component by 2 kg2\,\mathrm{kg} reproduces the resultant components.

Change just the direction of the third force by reversing it. Its magnitude stays unchanged, but both component signs reverse.

Worked example 2: reverse one force

Question. Keep the first two forces and mass above, but replace the third force by (−8i+4j) N(-8\mathbf i+4\mathbf j)\,\mathrm N. Find resultant and acceleration.

Plan. Insert the reversed vector in the sum and divide by the same mass.

R=(3−5−8)i+(7+2+4)jnew signed sums=(−10i+13j) Na=(−5i+132j) m s−2divide by 2\begin{aligned} \mathbf R&=(3-5-8)\mathbf i+(7+2+4)\mathbf j &&\text{new signed sums}\\ &=(-10\mathbf i+13\mathbf j)\,\mathrm N\\ \mathbf a&=(-5\mathbf i+\tfrac{13}{2}\mathbf j)\,\mathrm{m\,s^{-2}} &&\text{divide by }2 \end{aligned}

Why reverse both signs? Multiplying a vector by −1-1 reverses its entire direction.

Answer. R=(−10i+13j) N\mathbf R=(-10\mathbf i+13\mathbf j)\,\mathrm N, a=(−5i+132j) m s−2\mathbf a=(-5\mathbf i+\frac{13}{2}\mathbf j)\,\mathrm{m\,s^{-2}}.

Check. The resultant changed by (−16i+8j) N(-16\mathbf i+8\mathbf j)\,\mathrm N, exactly twice the negative of the former third force.

The components may cancel even when none of the individual forces is zero.

Watch out

The line ∣R∣=∣F1∣+∣F2∣\lvert\mathbf R\rvert=\lvert\mathbf F_1\rvert+\lvert\mathbf F_2\rvert is valid only for forces in the same direction. Add the vectors before finding a magnitude.

Now combine three new component forces.

Magnitude and direction of the resultant

A component answer tells you how far a vector points along each axis. A magnitude-and-direction answer describes the same vector by its length and orientation. Sketch its quadrant before pressing the inverse tangent key.

Magnitude and direction from components

For R=Xi+Yj\mathbf R=X\mathbf i+Y\mathbf j,

∣R∣=X2+Y2\lvert\mathbf R\rvert=\sqrt{X^2+Y^2}

For non-zero components the acute reference angle satisfies tan⁡β=∣Y/X∣\tan\beta=\lvert Y/X\rvert; choose the required direction from the component signs. A zero vector has no direction.

Not in the formula booklet: learn it.

If X=0X=0, the vector is vertical, so no division is needed. If Y=0Y=0, it is horizontal. When both are non-zero, your sketch tells you whether to add or subtract the acute angle from a straight or full turn.

Worked example 3: a first-quadrant resultant

Question. Find the magnitude of (6i+2j) N(6\mathbf i+2\mathbf j)\,\mathrm N and its direction above the positive i\mathbf i-direction, to 11 decimal place.

Plan. Use Pythagoras, then inverse tangent within the known first quadrant.

∣R∣=62+22=210 Nperpendicular componentsθ=tan⁡−1(2/6)=18.4349…∘first quadrant\begin{aligned} \lvert\mathbf R\rvert&=\sqrt{6^2+2^2}=2\sqrt{10}\,\mathrm N &&\text{perpendicular components}\\ \theta&=\tan^{-1}(2/6)=18.4349\ldots{}^\circ &&\text{first quadrant} \end{aligned}

Why is the acute calculator answer sufficient here? Both components are positive, so the direction lies between east and north.

Answer. Magnitude 210 N2\sqrt{10}\,\mathrm N; direction 18.4∘18.4^\circ above the positive i\mathbf i-axis.

Check. 210 cos⁡θ=62\sqrt{10}\,\cos\theta=6 and 210 sin⁡θ=22\sqrt{10}\,\sin\theta=2, using unrounded θ\theta.

Change only the horizontal sign. The magnitude and acute reference angle stay the same, but the quadrant changes.

Worked example 4: a leftward component

Question. Find the magnitude of (−6i+2j) N(-6\mathbf i+2\mathbf j)\,\mathrm N and its direction anticlockwise from the positive i\mathbf i-axis, to 11 decimal place.

Plan. Find the acute reference angle, then locate it in quadrant II.

∣R∣=(−6)2+22=210 NPythagorasβ=tan⁡−1(2/6)=18.4349…∘reference triangleθ=180∘−β=161.5650…∘left and up\begin{aligned} \lvert\mathbf R\rvert&=\sqrt{(-6)^2+2^2}=2\sqrt{10}\,\mathrm N &&\text{Pythagoras}\\ \beta&=\tan^{-1}(2/6)=18.4349\ldots{}^\circ &&\text{reference triangle}\\ \theta&=180^\circ-\beta=161.5650\ldots{}^\circ &&\text{left and up} \end{aligned}

Why not report tan⁡−1(2/(−6))\tan^{-1}(2/(-6))? Its principal value is negative and points into the wrong quadrant.

Answer. Magnitude 210 N2\sqrt{10}\,\mathrm N; direction 161.6∘161.6^\circ anticlockwise from the positive i\mathbf i-axis.

Check. Cosine is negative and sine positive at this direction, matching the components.

An angle without a reference direction is ambiguous. Say where you start measuring and whether the turn is clockwise or anticlockwise.

Exam tip

For a requested direction, give the quadrant or bearing as well as a trigonometric calculation. A correct acute angle can still describe the wrong vector.

Use a different second-quadrant vector next.

Resolving forces given by magnitude and direction

A 20 N20\,\mathrm N force at 30∘30^\circ above the horizontal supplies 10 N10\,\mathrm N upwards, with the remaining component to the right. Resolve each angled force before adding it to any other force.

Resolve a magnitude and direction

If a force of magnitude FF has direction θ\theta measured anticlockwise from the positive i\mathbf i-axis, then

F=Fcos⁡θ i+Fsin⁡θ j\mathbf F=F\cos\theta\,\mathbf i+F\sin\theta\,\mathbf j

If an acute angle is given from another reference line, identify adjacent and opposite components and assign their signs from the diagram.

Not in the formula booklet: learn it.

The component adjacent to the stated angle uses cosine. An angle from the vertical therefore puts cosine in the vertical component. Sine and cosine are not permanently attached to axis names.

[DIAGRAM: asset_slug: p3_8_5_understand_and_use_addition_of_forces_and_resultant_forces_in_dynamics__diagram_02; asset_name: Resolve then add; description: Left, a force at angle theta from the horizontal has horizontal component F cos theta and vertical component F sin theta. Right, a head-to-tail triangle shows the summed components and the resultant.]
Diagram

Keep exact trigonometric values when the angle permits them.

Worked example 5: resolve above the horizontal

Question. A 20 N20\,\mathrm N force acts 30∘30^\circ above the positive i\mathbf i-direction. A second force is (5i−6j) N(5\mathbf i-6\mathbf j)\,\mathrm N. Find the resultant exactly.

Plan. Resolve the first force, then add like components.

F1=20cos⁡30∘ i+20sin⁡30∘ jresolve=(103 i+10j) Nexact valuesR=((103+5)i+4j) Nadd second force\begin{aligned} \mathbf F_1&=20\cos30^\circ\,\mathbf i+20\sin30^\circ\,\mathbf j &&\text{resolve}\\ &=(10\sqrt3\,\mathbf i+10\mathbf j)\,\mathrm N &&\text{exact values}\\ \mathbf R&=((10\sqrt3+5)\mathbf i+4\mathbf j)\,\mathrm N &&\text{add second force} \end{aligned}

Why is the vertical component 1010 positive? The force points above the horizontal.

Answer. R=((103+5)i+4j) N\mathbf R=((10\sqrt3+5)\mathbf i+4\mathbf j)\,\mathrm N.

Check. The first force has squared magnitude (103)2+102=400=202(10\sqrt3)^2+10^2=400=20^2.

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