MF4, MF6, Mn7 - Resolving forces and dynamics in a plane

MF4, MF6, Mn7 - Resolving forces and dynamics in a plane

Find resultants of concurrent forces and choose useful directions for resolution. Then form equations of motion in a line or a plane, including gravity, and check the constraints on a particle moving along a surface.

Resolving a force along chosen axes

Can pulling upwards on a box make it press less hard on the floor without lifting off? A 20 N20\,\mathrm N pull at 30∘30^\circ above horizontal supplies 10 N10\,\mathrm N upwards as well as a forward component. The upward part reduces the floor force; we will calculate both effects.

You already know

For a right-angled triangle, cosine uses the adjacent side and sine the opposite side. Components have signs; magnitudes do not. Choose positive directions before calculating.

Resolving replaces one vector by components whose vector sum equals it. It does not create extra physical forces.

Perpendicular force components

For magnitude FF at angle θ\theta anticlockwise from positive horizontal,

Fx=Fcos⁡θ,Fy=Fsin⁡θ.F_x=F\cos\theta,\qquad F_y=F\sin\theta.

These are signed projections on perpendicular axes. Not in the formula sheet: learn it.

If the angle is measured from vertical, the vertical component is adjacent to it. Redraw the small right-angled triangle rather than learning a second rule.

Worked example 1: angle from horizontal

Question. A 10 N10\,\mathrm N force points up and right at 30∘30^\circ above horizontal. Find its signed horizontal and vertical components exactly, with right and up positive.

Plan. Use the angle's adjacent and opposite projections.

Fx=10cos⁡30∘=53 Nadjacent, rightFy=10sin⁡30∘=5 Nopposite, up\begin{aligned} F_x&=10\cos30^\circ=5\sqrt3\,\mathrm N &&\text{adjacent, right}\\ F_y&=10\sin30^\circ=5\,\mathrm N &&\text{opposite, up} \end{aligned}

Why cosine horizontally? The stated angle touches the horizontal component.

Answer. (53 i+5j) N(5\sqrt3\,\mathbf i+5\mathbf j)\,\mathrm N.

Check. (53)2+52=100=102(5\sqrt3)^2+5^2=100=10^2; the components recover the force's magnitude.

Change only the reference direction: the same numerical angle is now measured from upward vertical. Both component tasks remain exact.

Worked example 2: angle from vertical

Question. A 10 N10\,\mathrm N force points up and right at 30∘30^\circ to upward vertical. Find its signed horizontal and vertical components exactly, with right and up positive.

Plan. Put cosine beside the new reference angle.

Fx=10sin⁡30∘=5 Nopposite, rightFy=10cos⁡30∘=53 Nadjacent, up\begin{aligned} F_x&=10\sin30^\circ=5\,\mathrm N &&\text{opposite, right}\\ F_y&=10\cos30^\circ=5\sqrt3\,\mathrm N &&\text{adjacent, up} \end{aligned}

Why do sine and cosine switch places? The reference line changed, so the adjacent side changed.

Answer. (5i+53 j) N(5\mathbf i+5\sqrt3\,\mathbf j)\,\mathrm N.

Check. The vertical component is now larger, matching a force nearer the vertical; its magnitude is still 10 N10\,\mathrm N.

For up-left, the horizontal component becomes negative. Axis choice can also rotate: on an incline, parallel and perpendicular directions often make the equations shorter.

Watch out

Adding F\mathbf F, FxiF_x\mathbf i and FyjF_y\mathbf j counts one force twice. Use either the force or its two components in an equation.

Keep the vertical reference angle but change the direction and magnitude.

Resultants, magnitude and direction

Start with the directions supplied. Parallel forces need one signed sum: 9 N9\,\mathrm N right and 4 N4\,\mathrm N left give 5 N5\,\mathrm N right. Forces in two perpendicular directions need two sums: 6 N6\,\mathrm N right and 8 N8\,\mathrm N up give vector (6i+8j) N(6\mathbf i+8\mathbf j)\,\mathrm N, magnitude 10 N10\,\mathrm N. Component-form data use those same sums; angled data first need resolution.

Resultant of concurrent forces

The resultant is the single force having the same vector effect as all the given forces acting at a point. Concurrent forces have lines of action meeting at that point.

Two perpendicular 10 N10\,\mathrm N forces have resultant 102 N10\sqrt2\,\mathrm N, not 20 N20\,\mathrm N. More surprisingly, two equal 10 N10\,\mathrm N forces separated by 120∘120^\circ have resultant just 10 N10\,\mathrm N: their sideways components cancel and their components along the bisector are 5 N5\,\mathrm N each.

Resultant from components

For coplanar forces resolved along the same perpendicular axes,

S=(∑Fx)i+(∑Fy)j,∣S∣=(∑Fx)2+(∑Fy)2.\mathbf S=\Bigl(\sum F_x\Bigr)\mathbf i+\Bigl(\sum F_y\Bigr)\mathbf j, \qquad |\mathbf S|=\sqrt{\Bigl(\sum F_x\Bigr)^2+\Bigl(\sum F_y\Bigr)^2}.

Use the component signs to determine direction. Not in the formula sheet: learn it.

The diagram translates one force head to tail without changing its direction. You can choose horizontal and vertical components, or rotate both axes together: the resultant vector remains the same physical force. Use exact component calculation to find its length and direction.

[DIAGRAM: asset_name: Adding perpendicular angled forces; asset_slug: angled_force_resultant; description: Equal-scale force-component axes show 10 N at 30 degrees and 10 N at 120 degrees, their head-to-tail parallelogram and the resultant at 75 degrees; caption: Two perpendicular 10 N forces give a resultant at 75 degrees]
Two perpendicular 10 N forces give a resultant at 75 degrees

Worked example 3: summing two angled forces

Question. Two 10 N10\,\mathrm N forces act at 30∘30^\circ and 120∘120^\circ anticlockwise from positive horizontal. Find their resultant vector and magnitude exactly, and its direction to one decimal place if a direction exists.

Plan. Resolve, sum each coordinate, then find length and quadrant.

Sx=10cos⁡30∘+10cos⁡120∘=5(3−1)horizontal sumSy=10sin⁡30∘+10sin⁡120∘=5(1+3)vertical sum∣S∣=25(3−1)2+25(1+3)2=102Pythagorasθ=tan⁡−1 ⁣(1+33−1)=75.0∘first quadrant\begin{aligned} S_x&=10\cos30^\circ+10\cos120^\circ=5(\sqrt3-1) &&\text{horizontal sum}\\ S_y&=10\sin30^\circ+10\sin120^\circ=5(1+\sqrt3) &&\text{vertical sum}\\ |\mathbf S|&=\sqrt{25(\sqrt3-1)^2+25(1+\sqrt3)^2}=10\sqrt2 &&\text{Pythagoras}\\ \theta&=\tan^{-1}\!\left(\frac{1+\sqrt3}{\sqrt3-1}\right)=75.0^\circ &&\text{first quadrant} \end{aligned}

Why check the quadrant? The ratio alone cannot distinguish opposite directions.

Answer. S=[5(3−1)i+5(1+3)j] N\mathbf S=[5(\sqrt3-1)\mathbf i+5(1+\sqrt3)\mathbf j]\,\mathrm N, magnitude 102 N10\sqrt2\,\mathrm N, direction 75.0∘75.0^\circ anticlockwise from positive horizontal.

Check. Equal forces give a resultant along their angular bisector; (30+120)/2=75(30+120)/2=75.

Change only the second force's direction to 210∘210^\circ. The forces now oppose exactly.

Worked example 4: a zero resultant

Question. Two 10 N10\,\mathrm N forces act at 30∘30^\circ and 210∘210^\circ anticlockwise from positive horizontal. Find their resultant vector and magnitude exactly, and its direction to one decimal place if a direction exists.

Plan. Sum the components before trying any inverse tangent.

Sx=53−53=0horizontal cancellationSy=5−5=0vertical cancellationS=0,∣S∣=0zero vector\begin{aligned} S_x&=5\sqrt3-5\sqrt3=0 &&\text{horizontal cancellation}\\ S_y&=5-5=0 &&\text{vertical cancellation}\\ \mathbf S&=\mathbf0,\qquad |\mathbf S|=0 &&\text{zero vector} \end{aligned}

Why stop before finding an angle? A zero vector has no direction; an inverse-tangent ratio would be 0/00/0.

Answer. Resultant 0 N\mathbf0\,\mathrm N, magnitude 0 N0\,\mathrm N; its direction is undefined.

Check. The two equal forces differ in direction by 180∘180^\circ, so they cancel.

For a non-zero resultant on mass mm, acceleration is S/m\mathbf S/m. Positive mass preserves direction. Velocity points along the motion; acceleration points along the resultant. For example, (−4i+3j) N(-4\mathbf i+3\mathbf j)\,\mathrm N on 2 kg2\,\mathrm{kg} gives (−2i+1.5j) m s−2(-2\mathbf i+1.5\mathbf j)\,\mathrm{m\,s^{-2}}, even if the initial velocity is rightwards.

Watch out

A negative horizontal and positive vertical resultant points up-left. Giving only an acute inverse-tangent answer without saying “above left” loses the direction information.

Try a non-zero resultant in that quadrant.

Angled forces on a horizontal surface

The opening pull changes two things: horizontal acceleration and the reaction required to maintain contact.

A smooth horizontal surface supplies normal reaction but no friction. For a particle staying on it, vertical acceleration is zero. Weight acts downwards, reaction upwards, and the applied force has its own two components.

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