Resultants from components | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Add components with their signs

Definition

What is meant by the resultant force?

Model answer: The vector sum of all the forces acting; it can replace those forces without changing their combined effect.

Choose positive horizontal and vertical directions.
Resolve every angled force in those same directions.
Add the signed horizontal components to get $R_x$ and the signed vertical components to get $R_y$ .

Common mistake

Do not add component magnitudes. A leftward or downward component must enter the sum with a negative sign if right and up were chosen positive.

Use the signs to choose the quadrant

Swipe left or right to see the whole diagram →

The component signs place the resultant before any inverse tangent is used. Here Rx is negative and Ry is positive, so the direction must be in the second quadrant.

R=\sqrt{R_x^2+R_y^2}

The calculator value $\tan^{-1}(|R_y/R_x|)$ is a reference angle. Use the signs of $R_x$ and $R_y$ to place the resultant in the correct quadrant before stating its direction.

The four quadrants are numbered anticlockwise, starting with quadrant I in the upper-right region. Therefore left and up is quadrant II.

Key idea

A magnitude is never negative. A negative component changes the direction, not the size of the resultant.

Finish with an unambiguous direction

Worked example

A resultant in the second quadrant

A 10 N force acts at 120° anticlockwise from the positive horizontal. A 2 N force acts right and a 3 N force acts down.

R_x=10\cos120^\circ+2=-3

R_y=10\sin120^\circ-3=5.660

R^2=(-3)^2+5.660^2

R=6.41\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{5.660}{3}\right)=62.1^\circ

Since $R_x<0$ and $R_y>0$ , the resultant is in the second quadrant. Its direction from positive horizontal is $180^\circ-62.1^\circ=117.9^\circ$ .

Answer

R=6.41\text{ N at }117.9^\circ

Examiner note

In a force problem with an unknown magnitude and angle, resolving in two perpendicular directions usually supplies the two equations you need. (9709/42/M/J/24 Q3)

Your turnIndependent check [5 marks]

A 14 N force acts at 150° anticlockwise from the positive horizontal. A 5 N force acts right and a 2 N force acts down. Find the magnitude and direction of the resultant.

Show worked answer

R_x=14\cos150^\circ+5=-7.124

R_y=14\sin150^\circ-2=5

R^2=(-7.124)^2+5^2

R=8.70\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{5}{7.124}\right)=35.1^\circ

Second quadrant, so the direction is 180° − 35.1° = 144.9°.

Answer

R=8.70\text{ N at }144.9^\circ

Add components with their signs

Definition

What is meant by the resultant force?

Model answer: The vector sum of all the forces acting; it can replace those forces without changing their combined effect.

Choose positive horizontal and vertical directions.
Resolve every angled force in those same directions.
Add the signed horizontal components to get $R_x$ and the signed vertical components to get $R_y$ .

Common mistake

Do not add component magnitudes. A leftward or downward component must enter the sum with a negative sign if right and up were chosen positive.

Use the signs to choose the quadrant

Swipe left or right to see the whole diagram →

The component signs place the resultant before any inverse tangent is used. Here Rx is negative and Ry is positive, so the direction must be in the second quadrant.

R=\sqrt{R_x^2+R_y^2}

The calculator value $\tan^{-1}(|R_y/R_x|)$ is a reference angle. Use the signs of $R_x$ and $R_y$ to place the resultant in the correct quadrant before stating its direction.

The four quadrants are numbered anticlockwise, starting with quadrant I in the upper-right region. Therefore left and up is quadrant II.

Key idea

A magnitude is never negative. A negative component changes the direction, not the size of the resultant.

Finish with an unambiguous direction

Worked example

A resultant in the second quadrant

A 10 N force acts at 120° anticlockwise from the positive horizontal. A 2 N force acts right and a 3 N force acts down.

R_x=10\cos120^\circ+2=-3

R_y=10\sin120^\circ-3=5.660

R^2=(-3)^2+5.660^2

R=6.41\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{5.660}{3}\right)=62.1^\circ

Since $R_x<0$ and $R_y>0$ , the resultant is in the second quadrant. Its direction from positive horizontal is $180^\circ-62.1^\circ=117.9^\circ$ .

Answer

R=6.41\text{ N at }117.9^\circ

Examiner note

In a force problem with an unknown magnitude and angle, resolving in two perpendicular directions usually supplies the two equations you need. (9709/42/M/J/24 Q3)

Your turnIndependent check [5 marks]

A 14 N force acts at 150° anticlockwise from the positive horizontal. A 5 N force acts right and a 2 N force acts down. Find the magnitude and direction of the resultant.

Show worked answer

R_x=14\cos150^\circ+5=-7.124

R_y=14\sin150^\circ-2=5

R^2=(-7.124)^2+5^2

R=8.70\text{ N}

\alpha=\tan^{-1}\!\left(\dfrac{5}{7.124}\right)=35.1^\circ

Second quadrant, so the direction is 180° − 35.1° = 144.9°.

Answer

R=8.70\text{ N at }144.9^\circ