Resolving forces and equilibrium

Resolving a force means replacing it by perpendicular components that have exactly the same combined effect. This turns a two dimensional force problem into two one dimensional equations.

For a particle in equilibrium, the resultant force is zero, so the components balance in every direction:

$$\boxed{\sum F_x=0,\qquad \sum F_y=0.}$$

The difficult part is rarely the algebra. It is choosing useful axes and deciding whether each component uses sine or cosine.

Prerequisites

You should be able to:

• use sine, cosine and Pythagoras in right angled triangles;
• work with vectors and directed quantities;
• identify weight, reaction, tension and other forces using free body diagrams;
• use $W=mg$ and SI units from quantities and units.

Unless a question gives another value, use $g=9.8\ \mathrm{m\,s^{-2}}$.

Components of a force

Suppose a force of magnitude $F$ makes an angle $\theta$ above the positive horizontal direction. Its components are

$$\boxed{F_x=F\cos\theta,\qquad F_y=F\sin\theta.}$$

This follows from the right angled component triangle:

$$\cos\theta=\frac{F_x}{F}, \qquad \sin\theta=\frac{F_y}{F}.$$

The component adjacent to the marked angle uses cosine. The component opposite the angle uses sine. This is safer than memorising that horizontal always means cosine, because the angle might instead be measured from the vertical.

The projection rule

The component of a force $F$ along an axis is $F\cos\alpha$, where $\alpha$ is the angle between the force and the positive direction of that axis. The sign records whether the component points with or against the axis.

Worked example 1: resolving from the horizontal

A force of $50\ \mathrm N$ acts at $35^\circ$ above the horizontal. Find its horizontal and vertical components.

The horizontal component is adjacent to $35^\circ$:

$$F_x=50\cos35^\circ=40.96\ldots\ \mathrm N.$$

The vertical component is opposite $35^\circ$:

$$F_y=50\sin35^\circ=28.68\ldots\ \mathrm N.$$

Therefore, to $3$ significant figures, the components are

$$\boxed{41.0\ \mathrm N\text{ horizontally and }28.7\ \mathrm N\text{ vertically upwards}.}$$

A quick check is

$$\sqrt{(50\cos35^\circ)^2+(50\sin35^\circ)^2}=50.$$

The two components are not extra forces. Together they replace the original $50\ \mathrm N$ force.

Worked example 2: an angle from the vertical

A cable pulls with tension $120\ \mathrm N$ at $20^\circ$ to the vertical. Find the horizontal and upward components.

The upward component is adjacent to the given angle, so

$$F_y=120\cos20^\circ=112.8\ldots\ \mathrm N.$$

The horizontal component is opposite the angle, so

$$F_x=120\sin20^\circ=41.0\ldots\ \mathrm N.$$

Thus the components are

$$\boxed{41.0\ \mathrm N\text{ horizontally and }113\ \mathrm N\text{ upwards}}$$

to $3$ significant figures. Using $120\cos20^\circ$ horizontally would silently treat $20^\circ$ as an angle from the horizontal.

Self-check 1

A force of $80\ \mathrm N$ acts downwards and to the right at $30^\circ$ below the horizontal. Taking right and up as positive, find its components.

Answer

The horizontal component is positive and the vertical component is negative:

$$F_x=80\cos30^\circ=40\sqrt3\ \mathrm N,$$ $$F_y=-80\sin30^\circ=-40\ \mathrm N.$$

So the components are $\boxed{69.3\ \mathrm N}$ right and $\boxed{40\ \mathrm N}$ down.

Resolving several forces

Choose two perpendicular axes, resolve every relevant force along them, then add signed components. For horizontal and vertical axes,

$$R_x=\sum F_x, \qquad R_y=\sum F_y.$$

The resultant has magnitude and direction

$$R=\sqrt{R_x^2+R_y^2}, \qquad \tan\phi=\frac{R_y}{R_x},$$

but the signs of $R_x$ and $R_y$ must be used to identify the correct quadrant. A calculator’s inverse tangent alone can give an ambiguous direction.

Worked example 3: finding a resultant

Two forces act on a particle. One is $18\ \mathrm N$ east. The other is $12\ \mathrm N$ at $60^\circ$ north of east. Find the resultant.

Take east as positive $x$ and north as positive $y$. Then

$$R_x=18+12\cos60^\circ=24,$$ $$R_y=12\sin60^\circ=6\sqrt3.$$

Hence

$$R=\sqrt{24^2+(6\sqrt3)^2} =\sqrt{684} =6\sqrt{19}\ \mathrm N.$$

If $\phi$ is measured north of east,

$$\tan\phi=\frac{6\sqrt3}{24}=\frac{\sqrt3}{4},$$

so $\phi=23.4^\circ$ to $3$ significant figures. The resultant is

$$\boxed{26.2\ \mathrm N\text{ at }23.4^\circ\text{ north of east}.}$$

Its direction lies between the directions of the two original forces, which is a useful reasonableness check.

Equilibrium

A particle is in equilibrium when the vector sum of all forces acting on it is zero:

$$\boxed{\sum\mathbf F=\mathbf 0.}$$

Resolving along any two perpendicular axes gives two scalar equations. With horizontal and vertical axes,

$$\sum F_x=0, \qquad \sum F_y=0.$$

Equilibrium means zero acceleration. The particle may be at rest or moving with constant velocity. It does not mean that no forces act.

Worked example 4: a suspended sign

A sign of mass $15\ \mathrm{kg}$ hangs in equilibrium from two cables. The left cable has tension $T_1$ and is at $30^\circ$ above the horizontal. The right cable has tension $T_2$ and is at $50^\circ$ above the horizontal. Find both tensions.

The sign has three forces: $T_1$, $T_2$, and weight $15g$ vertically downwards.

Resolve horizontally, taking right as positive:

$$T_2\cos50^\circ-T_1\cos30^\circ=0.$$

Therefore

$$T_1=\frac{T_2\cos50^\circ}{\cos30^\circ}.$$

Resolve vertically:

$$T_1\sin30^\circ+T_2\sin50^\circ=15g.$$

Substitute for $T_1$:

$$T_2\left(\frac{\cos50^\circ\sin30^\circ}{\cos30^\circ}+\sin50^\circ\right)=147.$$

Hence

$$T_2=129.3\ldots\ \mathrm N,$$

and

$$T_1=\frac{129.3\ldots\cos50^\circ}{\cos30^\circ}=96.0\ldots\ \mathrm N.$$

Thus

$$\boxed{T_1=96.0\ \mathrm N,\qquad T_2=129\ \mathrm N}$$

to $3$ significant figures. The right cable is more nearly vertical, so it provides vertical support more efficiently, but the unequal angles also require unequal tensions to balance horizontally.

Self-check 2

A particle of weight $40\ \mathrm N$ is held in equilibrium by a horizontal force $P$ and a tension $T$ acting at $35^\circ$ above the horizontal. Find $P$ and $T$.

Answer

Vertical equilibrium gives

$$T\sin35^\circ=40,$$

so

$$T=\frac{40}{\sin35^\circ}=69.7\ \mathrm N.$$

Horizontal equilibrium gives

$$P=T\cos35^\circ=57.1\ \mathrm N.$$

Therefore $\boxed{T=69.7\ \mathrm N}$ and $\boxed{P=57.1\ \mathrm N}$ to $3$ significant figures.

Choosing useful axes

Horizontal and vertical axes are often convenient, but there is no requirement to use them. Any perpendicular pair works. Choose axes that reduce the number of forces needing resolution.

For a particle on an inclined plane, use axes:

• parallel to the plane;
• perpendicular to the plane.

The normal reaction then lies entirely on the perpendicular axis, while friction lies entirely on the parallel axis.

Resolving weight on an inclined plane

Let a plane make an angle $\theta$ with the horizontal. Weight $mg$ acts vertically downwards. Relative to axes parallel and perpendicular to the plane, its components are

$$\boxed{mg\sin\theta\text{ down the plane},}$$ $$\boxed{mg\cos\theta\text{ into the plane}.}$$

This is a common source of confusion because the diagram’s angle is not initially obvious. The angle between the downward vertical and the direction perpendicular into the plane is also $\theta$. Therefore the perpendicular component is adjacent to $\theta$ and uses cosine.

Two limiting cases confirm the formula:

• if $\theta=0^\circ$, the plane is horizontal, so the down slope component is $0$ and the perpendicular component is $mg$;
• if $\theta$ approaches $90^\circ$, almost all the weight acts down the plane and almost none acts into it.

Worked example 5: a smooth inclined plane

A $6\ \mathrm{kg}$ particle rests in equilibrium on a smooth plane inclined at $25^\circ$ to the horizontal. It is held by a force $P$ acting up the plane. Find $P$ and the normal reaction $R$.

Smooth means there is no friction. Resolve parallel to the plane:

$$P-6g\sin25^\circ=0.$$

Therefore

$$P=58.8\sin25^\circ=24.85\ldots\ \mathrm N.$$

Resolve perpendicular to the plane:

$$R-6g\cos25^\circ=0.$$

Therefore

$$R=58.8\cos25^\circ=53.29\ldots\ \mathrm N.$$

Thus, to $3$ significant figures,

$$\boxed{P=24.9\ \mathrm N,\qquad R=53.3\ \mathrm N.}$$

The reaction is not $mg$ because only the component of weight perpendicular to the plane presses the particle into it.

Worked example 6: an angled force on a slope

A $10\ \mathrm{kg}$ block is in equilibrium on a smooth plane inclined at $20^\circ$ to the horizontal. A force $P$ acts at $30^\circ$ above the plane, directed partly up the slope. Find $P$ and the normal reaction $R$.

Resolve parallel to the plane. The component of $P$ up the plane balances the component of weight down the plane:

$$P\cos30^\circ=10g\sin20^\circ.$$

Hence

$$P=\frac{98\sin20^\circ}{\cos30^\circ}=38.70\ldots\ \mathrm N.$$

Resolve perpendicular to the plane, taking away from the plane as positive. The force $P$ partly lifts the block away from the surface:

$$R+P\sin30^\circ-10g\cos20^\circ=0.$$

Thus

$$R=98\cos20^\circ-P\sin30^\circ =72.70\ldots\ \mathrm N.$$

Therefore

$$\boxed{P=38.7\ \mathrm N,\qquad R=72.7\ \mathrm N.}$$

Notice that $R<98\cos20^\circ$ because the applied force has a component lifting the block. A force directed into the plane would increase $R$ instead.

Self-check 3

An $8\ \mathrm{kg}$ particle is held at rest on a smooth plane inclined at $40^\circ$ by a horizontal force $H$. Find $H$ and the normal reaction.

Answer

The angle between the horizontal force and the direction up the plane is $40^\circ$. Resolving parallel to the plane gives

$$H\cos40^\circ=8g\sin40^\circ,$$

so

$$H=8g\tan40^\circ=65.8\ \mathrm N.$$

The horizontal force pushes into the plane. Perpendicular equilibrium gives

$$R=8g\cos40^\circ+H\sin40^\circ.$$

Substituting $H=8g\tan40^\circ$ gives

$$R=\frac{8g}{\cos40^\circ}=102\ \mathrm N$$

to $3$ significant figures.

Equilibrium with three forces

If exactly three non-parallel forces keep a particle in equilibrium, their vector sum is zero. They can be drawn head to tail as a closed triangle. This gives an alternative method using the sine rule or cosine rule.

Resolving is usually the more systematic approach. A force triangle can be quicker when all three directions are known, but it must represent force vectors, not the physical layout of strings or surfaces.

For three forces of magnitudes $P$, $Q$ and $R$ in equilibrium, Lami’s theorem states

$$\frac{P}{\sin\alpha} =\frac{Q}{\sin\beta} =\frac{R}{\sin\gamma},$$

where $\alpha$ is the angle between the other two forces, and similarly for $\beta$ and $\gamma$. Use this only when the opposite angles have been identified carefully. Resolving avoids that indexing trap and generalises immediately to more than three forces.

Common misconceptions

Sine is not always vertical

Sine gives the component opposite the marked angle. If the angle is measured from the vertical, the horizontal component uses sine and the vertical component uses cosine.

Components do not act as well as the original force

Resolve a force for calculation, but do not count the original force and its components together. That would double its effect.

Reaction is perpendicular to the surface

On a slope, the normal reaction is not vertical and is not automatically equal to $mg$. Resolve perpendicular to the surface to find it.

Equilibrium is not the same as rest

Equilibrium means $\mathbf a=\mathbf0$. A particle moving in a straight line at constant speed is also in equilibrium.

Do not round components early

Keep expressions such as $50\cos35^\circ$ until the final calculation. Early rounding can noticeably alter a later tension or angle.

A reliable method

1. Isolate the particle and draw every external force.
2. Choose perpendicular axes that align with as many forces as possible.
3. Mark every relevant angle on the diagram.
4. Resolve each angled force using the angle between that force and the axis.
5. Attach signs from the component’s direction.
6. Use $\sum F=0$ for equilibrium, or $\sum F=ma$ when there is acceleration.
7. Solve without premature rounding, then check magnitude, direction and units.

Mixed self-check

1. A force has components $-12\ \mathrm N$ horizontally and $5\ \mathrm N$ vertically. Find its magnitude and describe its quadrant.
2. A $4\ \mathrm{kg}$ particle lies on a smooth plane at $30^\circ$ to the horizontal. Find the force parallel to the plane needed for equilibrium and the normal reaction.
3. Explain why the equations $\sum F_x=0$ and $\sum F_y=0$ are sufficient for equilibrium in a plane.

Answers

1. Its magnitude is

   $$\sqrt{(-12)^2+5^2}=\boxed{13\ \mathrm N}.$$

   It points left and upwards, in the second quadrant.

2. Parallel to the plane,

   $$P=4g\sin30^\circ=\boxed{19.6\ \mathrm N}$$

   up the plane. Perpendicular to the plane,

   $$R=4g\cos30^\circ=\boxed{19.6\sqrt3\ \mathrm N}$$

   or $34.0\ \mathrm N$ to $3$ significant figures.

3. Every force in the plane is completely determined by its components along two perpendicular axes. If the total component is zero along both axes, the resultant vector has zero magnitude, so the particle has zero acceleration.

Next steps

Use resolved components in Newton’s laws of motion when the resultant is $m\mathbf a$ rather than zero. Then apply the same slope axes to friction and limiting equilibrium, and combine force balance with turning effects in moments.