Newton's law on inclined planes | CAIE 9709 Mechanics Notes

Mechanics

Mathematics9709

Resolve weight into the two useful directions

Swipe left or right to see the whole diagram →

First draw the actual forces. Resolve weight in a separate step so components are never mistaken for extra forces.

Let $\theta$ be the angle between the plane and the horizontal. Resolving means replacing one force by perpendicular components that have the same combined effect.

The component down the plane is $mg\sin\theta$ .
The component into the plane is $mg\cos\theta$ .
With no acceleration perpendicular to the plane and no other perpendicular force, $R=mg\cos\theta$ .
If another force has a perpendicular component, include it before finding $R$ .

These sine and cosine positions come from the right-angled component triangle: the down-slope component is opposite $\theta$ , while the into-plane component is adjacent to $\theta$ .

Common mistake

Never write

R=mg\cos\theta

as a memorised rule before checking every force perpendicular to the plane.

Friction reverses when the direction of motion reverses

For a 4 kg particle on a rough plane inclined at $30^\circ$ , let $\mu=0.2$ . Here $\mu$ is the coefficient of friction and $F$ is the friction magnitude. Because the particle is sliding, $F=\mu R$ :

R=40\cos30^\circ=20\sqrt3

F=\mu R=4\sqrt3

If the particle moves down the plane, friction acts up the plane:

20-4\sqrt3=4a

a=5-\sqrt3

If it moves up the plane, both weight's component and friction act down the plane. Taking up the plane as positive:

-20-4\sqrt3=4a

a=-5-\sqrt3

Key idea

A particle moving up and later sliding down the same rough plane usually has two different acceleration magnitudes. Redraw the force direction at the turning point.

Common mistake

For a stationary particle not stated to be in limiting equilibrium, friction adjusts up to its maximum:

F\leq\mu R

. Do not automatically replace this inequality by equality.

Separate the force stage from the motion stage

Mark the actual direction of motion or impending motion.
Draw friction opposite that direction.
Resolve perpendicular to find the normal reaction.
Use $F=\mu R$ when the particle is sliding, or when limiting equilibrium is stated.
Resolve parallel to the plane to find $a$ .
Use kinematics only after the acceleration is known.

Examiner note

In a current connected-particle question, the mark scheme separately rewarded finding the reaction, using

F=\mu R

and writing Newton's law with the correct force terms. Show these stages rather than compressing them into one calculator line. (9709/42/M/J/24 Q7(a))

Your turnIndependent check [6 marks]

A 5 kg particle is on a rough plane where $\sin\theta=0.6$ , $\cos\theta=0.8$ and $\mu=0.25$ . Find its acceleration (a) while it slides down the plane and (b) while it moves up the plane.

Show worked answer

R=50(0.8)=40

F=0.25(40)=10

30-10=5a

a=4\text{ m s}^{-2}

The first acceleration is 4 m s⁻² down the plane.

-30-10=5a

a=-8\text{ m s}^{-2}

While moving up, the acceleration is 8 m s⁻² down the plane.

Answer

R=40\text{ N},\ F=10\text{ N}

4\text{ m s}^{-2}\text{ down the plane}

8\text{ m s}^{-2}\text{ down the plane}

Resolve weight into the two useful directions

Swipe left or right to see the whole diagram →

First draw the actual forces. Resolve weight in a separate step so components are never mistaken for extra forces.

Let $\theta$ be the angle between the plane and the horizontal. Resolving means replacing one force by perpendicular components that have the same combined effect.

The component down the plane is $mg\sin\theta$ .
The component into the plane is $mg\cos\theta$ .
With no acceleration perpendicular to the plane and no other perpendicular force, $R=mg\cos\theta$ .
If another force has a perpendicular component, include it before finding $R$ .

These sine and cosine positions come from the right-angled component triangle: the down-slope component is opposite $\theta$ , while the into-plane component is adjacent to $\theta$ .

Common mistake

Never write

R=mg\cos\theta

as a memorised rule before checking every force perpendicular to the plane.

Friction reverses when the direction of motion reverses

For a 4 kg particle on a rough plane inclined at $30^\circ$ , let $\mu=0.2$ . Here $\mu$ is the coefficient of friction and $F$ is the friction magnitude. Because the particle is sliding, $F=\mu R$ :

R=40\cos30^\circ=20\sqrt3

F=\mu R=4\sqrt3

If the particle moves down the plane, friction acts up the plane:

20-4\sqrt3=4a

a=5-\sqrt3

If it moves up the plane, both weight's component and friction act down the plane. Taking up the plane as positive:

-20-4\sqrt3=4a

a=-5-\sqrt3

Key idea

A particle moving up and later sliding down the same rough plane usually has two different acceleration magnitudes. Redraw the force direction at the turning point.

Common mistake

For a stationary particle not stated to be in limiting equilibrium, friction adjusts up to its maximum:

F\leq\mu R

. Do not automatically replace this inequality by equality.

Separate the force stage from the motion stage

Mark the actual direction of motion or impending motion.
Draw friction opposite that direction.
Resolve perpendicular to find the normal reaction.
Use $F=\mu R$ when the particle is sliding, or when limiting equilibrium is stated.
Resolve parallel to the plane to find $a$ .
Use kinematics only after the acceleration is known.

Examiner note

In a current connected-particle question, the mark scheme separately rewarded finding the reaction, using

F=\mu R

and writing Newton's law with the correct force terms. Show these stages rather than compressing them into one calculator line. (9709/42/M/J/24 Q7(a))

Your turnIndependent check [6 marks]

A 5 kg particle is on a rough plane where $\sin\theta=0.6$ , $\cos\theta=0.8$ and $\mu=0.25$ . Find its acceleration (a) while it slides down the plane and (b) while it moves up the plane.

Show worked answer

R=50(0.8)=40

F=0.25(40)=10

30-10=5a

a=4\text{ m s}^{-2}

The first acceleration is 4 m s⁻² down the plane.

-30-10=5a

a=-8\text{ m s}^{-2}

While moving up, the acceleration is 8 m s⁻² down the plane.

Answer

R=40\text{ N},\ F=10\text{ N}

4\text{ m s}^{-2}\text{ down the plane}

8\text{ m s}^{-2}\text{ down the plane}