Resolving a force | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Measure from the positive direction

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The component beside the marked angle uses cosine; the component opposite that angle uses sine.

To find the component in any chosen positive direction, measure the angle $\phi$ from that positive direction to the force.

C=F\cos\phi

Symbols

$C$ = signed component in the chosen direction (N)
$F$ = magnitude of the original force (N)
$\phi$ = angle from the chosen positive direction to the force (degrees)

If a force makes an angle $\theta$ anticlockwise from the positive horizontal, then $F_x=F\cos\theta$ and $F_y=F\sin\theta$ . An angle greater than 90° gives a negative horizontal component automatically.

Key idea

A positive component points with the chosen positive direction. A negative component points against it. Keep the sign when the component enters an equation.

Force12 NAngle from horizontal35°

horizontal = 9.83 N

vertical = 6.88 N

check = 12.00 N

Positive horizontal is right and positive vertical is up. The signs change with direction, but the component resultant still has magnitude F.

Common mistake

“Horizontal is cosine” is only true when the marked angle is measured from the horizontal. Find the angle from the chosen positive direction, then multiply the force by the cosine of that angle.

Choose axes that remove unnecessary work

An axis is a chosen direction in which components are given signs. Horizontal and vertical axes are natural for forces around a point. On an inclined plane, axes parallel and perpendicular to the plane are usually cleaner.

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On a slope, rotate the axes. Resolving parallel and perpendicular to the plane usually leaves fewer components in each equation.

\text{down the plane: }mg\sin\theta

\text{into the plane: }mg\cos\theta

Key idea

Weight is still vertical. Only the chosen axes have rotated. Normal reaction and friction already lie along the new axes, so they do not need resolving. A force perpendicular to a chosen direction has zero component in that direction; this can remove an unknown from an equation.

Reverse the component calculation when needed

Sometimes a component is known and the original force is unknown. Use the same equation and rearrange it; do not swap sine and cosine.

Worked example

Find a force from one component

A force $P$ has a component of 8 N in a direction that makes an angle of 60° with $P$ . Find $P$ .

P\cos60^\circ=8

P=\dfrac{8}{\cos60^\circ}=16\text{ N}

Answer

P=16\text{ N}

Your turnIndependent check [3 marks]

An 18 N force acts at 130° anticlockwise from the positive horizontal. Find its signed horizontal and vertical components.

Show worked answer

F_x=18\cos130^\circ=-11.6\text{ N}

F_y=18\sin130^\circ=13.8\text{ N}

The negative horizontal sign means 11.6 N to the left.

Answer

F_x=-11.6\text{ N}

F_y=13.8\text{ N}

Measure from the positive direction

Swipe left or right to see the whole diagram →

The component beside the marked angle uses cosine; the component opposite that angle uses sine.

To find the component in any chosen positive direction, measure the angle $\phi$ from that positive direction to the force.

C=F\cos\phi

Symbols

$C$ = signed component in the chosen direction (N)
$F$ = magnitude of the original force (N)
$\phi$ = angle from the chosen positive direction to the force (degrees)

Key idea

A positive component points with the chosen positive direction. A negative component points against it. Keep the sign when the component enters an equation.

Force12 NAngle from horizontal35°

horizontal = 9.83 N

vertical = 6.88 N

check = 12.00 N

Positive horizontal is right and positive vertical is up. The signs change with direction, but the component resultant still has magnitude F.

Common mistake

“Horizontal is cosine” is only true when the marked angle is measured from the horizontal. Find the angle from the chosen positive direction, then multiply the force by the cosine of that angle.

Choose axes that remove unnecessary work

Swipe left or right to see the whole diagram →

On a slope, rotate the axes. Resolving parallel and perpendicular to the plane usually leaves fewer components in each equation.

\text{down the plane: }mg\sin\theta

\text{into the plane: }mg\cos\theta

Key idea

Reverse the component calculation when needed

Sometimes a component is known and the original force is unknown. Use the same equation and rearrange it; do not swap sine and cosine.

Worked example

Find a force from one component

A force $P$ has a component of 8 N in a direction that makes an angle of 60° with $P$ . Find $P$ .

P\cos60^\circ=8

P=\dfrac{8}{\cos60^\circ}=16\text{ N}

Answer

P=16\text{ N}

Your turnIndependent check [3 marks]

An 18 N force acts at 130° anticlockwise from the positive horizontal. Find its signed horizontal and vertical components.

Show worked answer

F_x=18\cos130^\circ=-11.6\text{ N}

F_y=18\sin130^\circ=13.8\text{ N}

The negative horizontal sign means 11.6 N to the left.

Answer

F_x=-11.6\text{ N}

F_y=13.8\text{ N}