Md6–Md10 - Resolving forces and particle equilibrium

Md6–Md10 - Resolving forces and particle equilibrium

Md6–Md7 resolve forces in useful directions and add concurrent forces as vectors. Md8–Md10 use a zero resultant to solve particle-equilibrium problems, either through component equations or a closed triangle or polygon of forces.

Resolve first, then add components

Can two unequal tensions support a weight while their horizontal effects cancel? Yes: a nearly horizontal string can supply a substantial sideways force but little upward force. Resolving makes those different jobs visible.

You already know

Forces are vectors: direction matters as well as magnitude. With a force at angle θ\theta above a positive horizontal axis, its components are Fcos⁡θF\cos\theta horizontally and Fsin⁡θF\sin\theta vertically. Use signed components outside the first quadrant.

Choose perpendicular directions that simplify the geometry. Horizontal and vertical suit hanging strings; parallel and perpendicular to a slope suit contact with an inclined plane. The force itself does not change when you change your axes.

Concurrent forces and resultant

Forces are concurrent when their lines of action pass through one point. Their resultant is their vector sum, with components obtained by adding the corresponding signed components.

For example, (6,0) N(6,0)\,\mathrm N and (0,4) N(0,4)\,\mathrm N have resultant (6,4) N(6,4)\,\mathrm N, not a force of magnitude 10 N10\,\mathrm N.

For perpendicular resultant components X,YX,Y, the magnitude is X2+Y2\sqrt{X^2+Y^2}. Use the signs of both components to identify the direction; inverse tangent alone can point into the wrong quadrant. These component and resultant rules are not in the formula booklet: learn them.

Worked example 1: a first-quadrant force added to a horizontal force

Question. Two concurrent forces act on a particle: 6 N6\,\mathrm N horizontally right and 4 N4\,\mathrm N at 60∘60^\circ above the rightward horizontal. Find the magnitude and direction of their resultant.

Plan. Add horizontal and vertical components, then use Pythagoras and the component ratio.

X=6+4cos⁡60∘=8rightward componentsY=4sin⁡60∘=23upward components∣R∣=82+(23)2=219 NPythagorasθ=arctan⁡238=23.413…∘X,Y>0\begin{aligned} X&=6+4\cos60^\circ=8 && \text{rightward components} \\ Y&=4\sin60^\circ=2\sqrt3 && \text{upward components} \\ |\mathbf R|&=\sqrt{8^2+(2\sqrt3)^2}=2\sqrt{19}\,\mathrm N && \text{Pythagoras} \\ \theta&=\arctan\frac{2\sqrt3}{8}=23.413\ldots^\circ && X,Y>0 \end{aligned}

Why this step? Adding magnitudes would give 10 N10\,\mathrm N, which is possible only if both forces acted in the same direction.

Answer. 219 N2\sqrt{19}\,\mathrm N, at 23.4∘23.4^\circ above the rightward horizontal.

Check. Its magnitude is approximately 8.72 N8.72\,\mathrm N, between 6−4=26-4=2 and 6+4=106+4=10, as the triangle of vector addition requires.

Change only the direction of the 4 N4\,\mathrm N force, from 60∘60^\circ to 120∘120^\circ. Its horizontal component now opposes the 6 N6\,\mathrm N force.

Worked example 2: a component changes sign

Question. Find the resultant when the second force acts at 120∘120^\circ anticlockwise from the rightward horizontal.

Plan. Retain the component signs before calculating the magnitude and direction.

X=6+4cos⁡120∘=4cos⁡120∘=−1/2Y=4sin⁡120∘=23upward component unchanged∣R∣=42+(23)2=27 Nθ=arctan⁡234=40.893…∘X,Y>0\begin{aligned} X&=6+4\cos120^\circ=4 && \cos120^\circ=-1/2 \\ Y&=4\sin120^\circ=2\sqrt3 && \text{upward component unchanged} \\ |\mathbf R|&=\sqrt{4^2+(2\sqrt3)^2}=2\sqrt7\,\mathrm N \\ \theta&=\arctan\frac{2\sqrt3}{4}=40.893\ldots^\circ && X,Y>0 \end{aligned}

Why this step? A negative horizontal component does not make the force's magnitude negative. It records which way that component acts.

Answer. 27 N2\sqrt7\,\mathrm N, at 40.9∘40.9^\circ above the rightward horizontal.

Check. The vertical component is unchanged while the horizontal component is smaller, so the resultant should be shorter and steeper. Both are true.

In vector notation the same calculation is simply R=F1+F2\mathbf R=\mathbf F_1+\mathbf F_2. Components are a way to carry out vector addition, not a separate physical force system.

Watch out

Wrong: use 4sin⁡120∘4\sin120^\circ as the horizontal component because the diagram looks steep.

Right: identify the angle's reference axis. The component along that axis uses cosine; the perpendicular component uses sine.

The second force now points vertically upwards, making a useful exact-value check.

Equilibrium gives two component equations

A particle is in equilibrium under concurrent forces if and only if their resultant is zero. In a plane, that means zero resultant in each of two perpendicular directions. A particle has no size, so there is no separate turning condition in this model.

Particle equilibrium

For coplanar concurrent forces,

∑Fx=0,∑Fy=0\sum F_x=0,\qquad \sum F_y=0

These equations are equivalent to a zero resultant. For an initially resting particle, they describe continued rest; more generally zero resultant means zero acceleration.

Not in the formula booklet: learn it.

Draw one free-body diagram, choose your axes, then label each equation with its direction. Unknown string tensions must be non-negative: a string pulls along its length and cannot push. A negative calculated tension indicates that the assumed taut-string equilibrium is impossible.

[DIAGRAM: asset_name: A weight supported by two unequal tensions; asset_slug: h645_b057_strings; caption: The horizontal components cancel while the vertical components support the weight; description: A particle has downward weight 100 N, a left string directed 30 degrees above the leftward horizontal and a right string directed 60 degrees above the rightward horizontal. Tension arrows act away from the particle along each string. Labels state left tension 50 N and right tension 50 square root three N. Horizontal reference rays identify both given angles.]
The horizontal components cancel while the vertical components support the weight

Worked example 3: two supporting strings

Question. A particle of weight 100 N100\,\mathrm N hangs in equilibrium from two light strings. The left string is 30∘30^\circ above the horizontal and the right string 60∘60^\circ above the horizontal. Find the tensions LL and TT, respectively.

Plan. Balance horizontal components to relate the tensions, then use vertical balance to find their sizes.

Tcos⁡60∘−Lcos⁡30∘=0horizontal equilibriumT=3 LrearrangeLsin⁡30∘+Tsin⁡60∘=100vertical equilibrium12L+32(3 L)=100substitute2L=100,L=50,T=503\begin{aligned} T\cos60^\circ-L\cos30^\circ&=0 && \text{horizontal equilibrium} \\ T&=\sqrt3\,L && \text{rearrange} \\ L\sin30^\circ+T\sin60^\circ&=100 && \text{vertical equilibrium} \\ \frac12L+\frac{\sqrt3}{2}(\sqrt3\,L)&=100 && \text{substitute} \\ 2L&=100,\quad L=50,\quad T=50\sqrt3 \end{aligned}

Why this step? The strings' angles differ, so their tensions need not be equal. It is their horizontal components, rather than the full tensions, that cancel.

Answer. L=50 NL=50\,\mathrm N, T=503 NT=50\sqrt3\,\mathrm N.

Check. The horizontal magnitudes are both 253 N25\sqrt3\,\mathrm N. The vertical contributions are 2525 and 75 N75\,\mathrm N, summing to 100 N100\,\mathrm N.

Change only the weight to 160 N160\,\mathrm N. The string geometry stays fixed, so each required tension scales in the same proportion.

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