3.03m, 3.03n - Particle equilibrium and polygons of forces

3.03m, 3.03n - Particle equilibrium and polygons of forces

Use zero resultant to solve particle-equilibrium problems with vectors, strings, inclines and simple ideal connections. You will select effective axes, interpret closed force polygons, and check whether the calculated forces can exist in the stated model.

Zero resultant and equilibrium equations

Can supporting a 60 N60\,\mathrm N weight need more than 60 N60\,\mathrm N of tension in each string? Yes: two almost horizontal strings supply only small upward components, even when their tensions are large. First establish what it means for all the force vectors to balance.

You already know

The resultant is a vector sum. For a particle, zero resultant gives zero acceleration, allowing either rest or constant velocity in a straight line.

The word particle matters: its size and turning effects are ignored. Concurrent forces on a body act through one point, so their balance can also be treated there. General rigid-body turning effects need further conditions studied later.

Particle equilibrium

For coplanar forces on a particle,

∑F=0⟺∑Fx=0 and ∑Fy=0\sum\mathbf F=\mathbf0 \quad\Longleftrightarrow\quad \sum F_x=0\text{ and }\sum F_y=0

for two perpendicular axes. Equivalently, the resolved sum is zero in every direction. Not in the formula sheet: learn it.

“One direction” must not mean “only one direction checked”. A horizontal balance can coexist with an unbalanced vertical force.

Worked example 1: an unknown balancing force

Question. A particle is in equilibrium under only (3i+4j) N(3\mathbf i+4\mathbf j)\,\mathrm N, (−5i+j) N(-5\mathbf i+\mathbf j)\,\mathrm N and F\mathbf F. Find F\mathbf F and its magnitude exactly.

Plan. Cancel the sum of the two known vectors.

S=(3−5)i+(4+1)j=−2i+5jknown sumF=2i−5jopposite vector∣F∣=22+(−5)2=29length\begin{aligned} \mathbf S&=(3-5)\mathbf i+(4+1)\mathbf j=-2\mathbf i+5\mathbf j &&\text{known sum}\\ \mathbf F&=2\mathbf i-5\mathbf j &&\text{opposite vector}\\ |\mathbf F|&=\sqrt{2^2+(-5)^2}=\sqrt{29} &&\text{length} \end{aligned}

Why reverse both signs? Each perpendicular sum must be zero.

Answer. F=(2i−5j) N\mathbf F=(2\mathbf i-5\mathbf j)\,\mathrm N, magnitude 29 N\sqrt{29}\,\mathrm N.

Check. 3−5+2=03-5+2=0, 4+1−5=04+1-5=0.

Change only the second known force's vertical component. The unknown vector must adapt to that change alone.

Worked example 2: changing one component

Question. A particle is in equilibrium under only (3i+4j) N(3\mathbf i+4\mathbf j)\,\mathrm N, (−5i−4j) N(-5\mathbf i-4\mathbf j)\,\mathrm N and F\mathbf F. Find F\mathbf F and its magnitude exactly.

Plan. Recalculate the known sum before reversing it.

S=(3−5)i+(4−4)j=−2ivertical parts cancelF=2izero resultant∣F∣=22+02=2length\begin{aligned} \mathbf S&=(3-5)\mathbf i+(4-4)\mathbf j=-2\mathbf i &&\text{vertical parts cancel}\\ \mathbf F&=2\mathbf i &&\text{zero resultant}\\ |\mathbf F|&=\sqrt{2^2+0^2}=2 &&\text{length} \end{aligned}

Why is no vertical balancing force needed? The given vertical components already sum to zero.

Answer. F=2i N\mathbf F=2\mathbf i\,\mathrm N, magnitude 2 N2\,\mathrm N.

Check. The final horizontal sum is 3−5+2=03-5+2=0, while the vertical sum remains zero.

An unconstrained unknown force can have either component sign. A string tension instead has a known direction and non-negative magnitude; a string cannot push.

Watch out

“Equilibrium, therefore velocity zero” adds an unsupported conclusion. Equilibrium determines acceleration zero; initial conditions determine whether the constant velocity is zero.

Use a new pair of known forces and interpret what equilibrium tells you.

A particle suspended by two strings

String tensions point away from the particle along the strings. Their vertical components support the weight; their horizontal components cancel each other.

Two-string suspension

For weight WW, a left string at α\alpha above horizontal and a right string at β\beta above horizontal,

−TLcos⁡α+TRcos⁡β=0,TLsin⁡α+TRsin⁡β=W.-T_L\cos\alpha+T_R\cos\beta=0, \qquad T_L\sin\alpha+T_R\sin\beta=W.

These are equilibrium equations for taut strings with no other forces. Not in the formula sheet: learn it.

Separate strings need not have equal tensions. Symmetry can justify equality; the word “equilibrium” alone cannot.

Worked example 3: unequal string angles

Question. A particle of weight 60 N60\,\mathrm N hangs at rest from two separate light strings, left at 30∘30^\circ and right at 60∘60^\circ above horizontal. Find both tensions exactly.

Plan. Express one tension using horizontal balance, then substitute vertically.

TRcos⁡60∘=TLcos⁡30∘horizontal balanceTR=3 TLrearrange12TL+32(3 TL)=60vertical balance2TL=60simplifyTL=30,TR=303solve both tensions\begin{aligned} T_R\cos60^\circ&=T_L\cos30^\circ &&\text{horizontal balance}\\ T_R&=\sqrt3\,T_L &&\text{rearrange}\\ \tfrac12T_L+\tfrac{\sqrt3}{2}(\sqrt3\,T_L)&=60 &&\text{vertical balance}\\ 2T_L&=60 &&\text{simplify}\\ T_L&=30,\qquad T_R=30\sqrt3 &&\text{solve both tensions} \end{aligned}

Why eliminate horizontally first? There is no weight term there, giving a direct ratio of tensions.

Answer. Left tension 30 N30\,\mathrm N, right tension 303 N30\sqrt3\,\mathrm N.

Check. Upward components are 15 N15\,\mathrm N and 45 N45\,\mathrm N, totalling 60 N60\,\mathrm N; both tensions are positive.

Change only the weight to 90 N90\,\mathrm N. The angles, equations and requested outputs remain the same.

Worked example 4: scaling the weight

Question. A particle of weight 90 N90\,\mathrm N hangs at rest from two separate light strings, left at 30∘30^\circ and right at 60∘60^\circ above horizontal. Find both tensions exactly.

Plan. Reuse the horizontal ratio and solve the new vertical balance.

TR=3 TLhorizontal balance12TL+32(3 TL)=90vertical balance2TL=90collect termsTL=45,TR=453solve\begin{aligned} T_R&=\sqrt3\,T_L &&\text{horizontal balance}\\ \tfrac12T_L+\tfrac{\sqrt3}{2}(\sqrt3\,T_L)&=90 &&\text{vertical balance}\\ 2T_L&=90 &&\text{collect terms}\\ T_L&=45,\qquad T_R=45\sqrt3 &&\text{solve} \end{aligned}

Why do both tensions scale equally? Their directions stayed fixed, so only the force scale changed.

Answer. Left tension 45 N45\,\mathrm N, right tension 453 N45\sqrt3\,\mathrm N.

Check. 22.5+67.5=90 N22.5+67.5=90\,\mathrm N upwards; each tension is 3/23/2 of its previous value.

For equal angles θ\theta, horizontal balance gives equal tensions and vertical balance gives 2Tsin⁡θ=W2T\sin\theta=W. With W=60 NW=60\,\mathrm N, θ=5∘\theta=5^\circ needs T=60/(2sin⁡5∘)≈344 NT=60/(2\sin5^\circ)\approx344\,\mathrm N in each string. That answers the opening puzzle. Exactly horizontal strings have zero upward component, so no finite tension can support a positive weight in this model.

Watch out

Writing TL+TR=WT_L+T_R=W adds magnitudes in different directions. It is the upward components that sum to weight.

Now change the string geometry to a symmetric arrangement.

Closed triangles and polygons of forces

Equilibrium can also be seen geometrically: force arrows placed head to tail return to the starting point.

Closed force polygon

A force polygon is a head-to-tail addition diagram for force vectors. Closure means the vector sum is zero. Its arrows keep the directions and lengths of the forces, although they are translated away from the particle.

For example, arrows from (0,0)(0,0) to (6,0)(6,0), then to (6,8)(6,8), need a closing vector (−6,−8) N(-6,-8)\,\mathrm N, magnitude 10 N10\,\mathrm N. Reading it the other way would give the opposite force and fail to close the polygon.

[DIAGRAM: asset_name: One equilibrium represented in two ways; asset_slug: suspension_force_triangle; description: Separate equal-scale panels show the three forces at the particle and their closed head-to-tail triangle for weight 60 N and string angles 30 and 60 degrees; caption: The same three equilibrium forces at a particle and in a closed triangle]
The same three equilibrium forces at a particle and in a closed triangle

In the diagram, vectors (−153,15)(-15\sqrt3,15), (153,45)(15\sqrt3,45) and (0,−60)(0,-60) newtons close. A triangle of forces lets you use the sine rule, which is not in the formula sheet and must be learnt.

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