A force acts at an angle to the direction an object can move in. Why is it useful to resolve it into components?

Because only part of the force does anything useful, and splitting it lets you deal with that part on its own. The component parallel to the direction of motion is the part that drives the object along, and the component perpendicular to it has no effect in that direction at all. The two components together have exactly the same effect as the original force, so nothing is lost by replacing one with the other.

True or False? You can assume that once a force has been resolved, the diagram now shows three forces acting on the object.

False. The two components replace the original force; they are not extra forces acting alongside it. Counting all three would include the same push or pull twice and give a resultant that is too big. On a diagram it helps to draw the components with a different style of line, or to cross out the original arrow once it has been resolved.

Two forces act on a particle at an angle to each other. What is the alternative to resolving them into components?

Use the triangle law for vector addition : draw the two forces nose to tail, and the resultant is the third side of the triangle. Its magnitude and direction then come from trigonometry on that triangle, using the sine rule or the cosine rule if it is not right-angled. This is often quicker for exactly two forces; components are usually better when there are three or more, since they can all be collected in each direction at once.

What is meant by the line of greatest slope of an inclined plane?

It is the most direct route from the bottom of the plane to the top: the steepest path up the surface. An object sliding freely down a plane travels along a line of greatest slope, and inclined plane problems at this level always take the motion to be along one. A force described as acting in the same vertical plane as the line of greatest slope is acting along that direct route rather than across the slope at an angle.

On an inclined plane, why resolve parallel and perpendicular to the plane rather than horizontally and vertically?

Because it leaves the least work to do: the normal reaction is already perpendicular to the plane, and the acceleration is along it, so neither has to be resolved. That leaves the weight as the only force that has to be broken up, since it is the one force that insists on acting vertically whatever the slope is doing. Resolving horizontally and vertically instead would mean splitting both the reaction and the acceleration, which is more work for the same answer.

A box sits on a rough floor and nobody is pushing it. What is the frictional force on it, and what does that tell you about friction in general?

The frictional force is zero . Friction is not always present: it appears only in response to something trying to move the object along the surface, and it takes whatever value is needed to oppose that, up to a limit. So pushing gently with 5 N gives 5 N of friction back, pushing with 10 N gives 10 N back, and nothing moves until the limit is reached.

True or False? You can assume that the normal reaction on an object resting on a horizontal surface is equal to its weight.

False. It is equal to the weight only when the weight and the reaction are the only vertical forces. If another force has a vertical component, the reaction adjusts to whatever makes the total vertical force zero: a force pulling partly upwards reduces it, one pushing partly downwards increases it. The reaction is never simply read off the mass; it is always found from the vertical equation.

Two particles are connected by a string over a pulley. On which of them does friction act?

Only on a particle that is resting on a rough surface . A particle hanging freely on the end of the string has nothing in contact with it, so there is no normal reaction and therefore no friction. So in the common set-up of one particle on a rough table and one hanging, friction appears in the equation for the particle on the table and nowhere else.

You have worked out which way a connected two-particle system tends to move. In which direction does friction act on each particle?

Friction opposes the motion of each particle , so it acts on each one against the direction that particle would move. The two particles are connected, so they share a single sense of motion: if one is about to slide down its slope, the other is about to be dragged up its own. So the friction on the first acts up its slope and on the second down its slope, each opposing its own particle's motion even though the two look opposed.

A LevelMathsEdexcelMechanicsFlashcardsForces & Newton's LawsResolving Forces, Inclined Planes & Friction

Resolving Forces, Inclined Planes & Friction (Edexcel A Level Maths: Mechanics): Flashcards

Exam code: 9MA0

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FrontResolving Forces & Inclined Planes
A force acts at an angle to the direction an object can move in. Why is it useful to resolve it into components?
BackResolving Forces & Inclined Planes
Because only part of the force does anything useful, and splitting it lets you deal with that part on its own.
The component parallel to the direction of motion is the part that drives the object along, and the component perpendicular to it has no effect in that direction at all.
The two components together have exactly the same effect as the original force, so nothing is lost by replacing one with the other.
BackResolving Forces & Inclined Planes
Draw a right-angled triangle with the original force as the hypotenuse and the two components as the other two sides, then use trigonometry.
The component adjacent to the angle is the force multiplied by $\cos$ of the angle, and the component opposite the angle is the force multiplied by $\sin$ of the angle.
Which is which depends on where the angle is measured from, so it is worth marking the angle on the triangle rather than reaching for a remembered formula.

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Cards in this collection (29)

A force acts at an angle to the direction an object can move in. Why is it useful to resolve it into components?
Because only part of the force does anything useful, and splitting it lets you deal with that part on its own.
The component parallel to the direction of motion is the part that drives the object along, and the component perpendicular to it has no effect in that direction at all.
The two components together have exactly the same effect as the original force, so nothing is lost by replacing one with the other.
How do you find the components of a force acting at an angle?
Draw a right-angled triangle with the original force as the hypotenuse and the two components as the other two sides, then use trigonometry.
The component adjacent to the angle is the force multiplied by $\cos$ of the angle, and the component opposite the angle is the force multiplied by $\sin$ of the angle.
Which is which depends on where the angle is measured from, so it is worth marking the angle on the triangle rather than reaching for a remembered formula.
True or False?
You can assume that once a force has been resolved, the diagram now shows three forces acting on the object.
False.
The two components replace the original force; they are not extra forces acting alongside it.
Counting all three would include the same push or pull twice and give a resultant that is too big.
On a diagram it helps to draw the components with a different style of line, or to cross out the original arrow once it has been resolved.
Two forces act on a particle at an angle to each other. What is the alternative to resolving them into components?
Use the triangle law for vector addition: draw the two forces nose to tail, and the resultant is the third side of the triangle.
Its magnitude and direction then come from trigonometry on that triangle, using the sine rule or the cosine rule if it is not right-angled.
This is often quicker for exactly two forces; components are usually better when there are three or more, since they can all be collected in each direction at once.
What is meant by the line of greatest slope of an inclined plane?
It is the most direct route from the bottom of the plane to the top: the steepest path up the surface.
An object sliding freely down a plane travels along a line of greatest slope, and inclined plane problems at this level always take the motion to be along one.
A force described as acting in the same vertical plane as the line of greatest slope is acting along that direct route rather than across the slope at an angle.
On an inclined plane, why resolve parallel and perpendicular to the plane rather than horizontally and vertically?
Because it leaves the least work to do: the normal reaction is already perpendicular to the plane, and the acceleration is along it, so neither has to be resolved.
That leaves the weight as the only force that has to be broken up, since it is the one force that insists on acting vertically whatever the slope is doing.
Resolving horizontally and vertically instead would mean splitting both the reaction and the acceleration, which is more work for the same answer.
A particle of mass $m$ rests on a plane inclined at $θ$ to the horizontal. Fill in the blanks with the components of its weight:
component parallel to the plane, down the slope $=______$
component perpendicular to the plane, into it $=______$
The completed components are:
component parallel to the plane, down the slope $= m g \sin θ$
component perpendicular to the plane, into it $= m g \cos θ$
A useful check on which way round they go: as the plane flattens towards $θ = 0$ , the parallel component goes to zero and the perpendicular one goes to the full weight $m g$ , which is what a horizontal surface should give.
An object slides up or down an inclined plane. What can you say about the forces in each of the two directions?
Parallel to the plane the object is accelerating, so the forces there do not balance and $F = m a$ applies.
Perpendicular to the plane the object neither lifts off the surface nor sinks into it, so it has no acceleration in that direction and the forces there balance exactly.
That perpendicular balance is what gives you the normal reaction, and the parallel equation is what gives you the acceleration.
A box sits on a rough floor and nobody is pushing it. What is the frictional force on it, and what does that tell you about friction in general?
The frictional force is zero.
Friction is not always present: it appears only in response to something trying to move the object along the surface, and it takes whatever value is needed to oppose that, up to a limit.
So pushing gently with 5 N gives 5 N of friction back, pushing with 10 N gives 10 N back, and nothing moves until the limit is reached.
Fill in the blanks in the friction inequality and say what each letter means:
$F______μ R$
where $R$ is the $______$ between the object and the surface
The completed inequality is:
$F \leq μ R$
where $R$ is the normal reaction between the object and the surface.
$F$ is the frictional force and $μ$ is the coefficient of friction for that particular pair of surfaces.
It is an inequality rather than an equation precisely because friction varies: $μ R$ is the most it can ever be, not the value it always takes.
What is meant by limiting equilibrium?
A body is in limiting equilibrium when it is still at rest but the friction has reached its maximum value, so that $F = μ R$ .
It is the boundary case: any more force along the surface and the object would start to move.
A question describing an object as on the point of moving is telling you it is in limiting equilibrium, and that is what lets you replace the inequality with an equation.
An object is already sliding along a rough surface. What is the frictional force on it?
It is at its maximum, $F = μ R$ , whatever else is happening.
Once the object is moving, friction no longer adjusts itself to balance the other forces; it takes its full value and acts opposite to the motion.
That is the useful difference between a moving object and a stationary one: for a moving object you always have an equation for the friction, not just an upper bound.
A surface is described as smooth. What value does that give the coefficient of friction, and what does it do to the working?
It gives $μ = 0$ .
Since $F \leq μ R$ , the frictional force is then zero no matter how large the normal reaction is, so friction drops out of every equation.
A surface described as rough has $μ > 0$ , and friction has to be carried through the working.
True or False?
You can assume that the normal reaction on an object resting on a horizontal surface is equal to its weight.
False.
It is equal to the weight only when the weight and the reaction are the only vertical forces.
If another force has a vertical component, the reaction adjusts to whatever makes the total vertical force zero: a force pulling partly upwards reduces it, one pushing partly downwards increases it.
The reaction is never simply read off the mass; it is always found from the vertical equation.
A crate is dragged along rough ground. Why is it easier to pull it with a rope held at an angle above the horizontal than with a rope held level?
Pulling at an angle gives the force an upward component, which reduces the normal reaction.
Because the maximum friction is $μ R$ , a smaller reaction means less friction to overcome.
The cost is that only the horizontal component now drives the crate forwards, so some of the force is spent lifting rather than pulling; the best angle balances the two effects.
In a friction problem on a horizontal surface, why must you find the normal reaction before you can find the acceleration?
Because the friction depends on the reaction, and the acceleration depends on the friction.
The reaction comes from the vertical equation, where the object is not accelerating, so the vertical forces balance; only then can the maximum friction $μ R$ be worked out.
That maximum is compared with the resultant of the horizontal forces to decide whether the object moves at all, and only if it does is $F = m a$ used horizontally to get the acceleration.
You have found the maximum friction and the resultant of the other horizontal forces. What do you conclude in each case?
If the horizontal resultant is less than or equal to the maximum friction, the object stays at rest: friction takes exactly the value needed to balance it, and the acceleration is zero.
If the horizontal resultant is greater than the maximum friction, the object accelerates: friction takes its full value $μ R$ , and the resultant driving the motion is the applied force minus that friction.
So the comparison decides which situation you are in before any acceleration is calculated.
A $2 kg$ block on rough ground with $μ = 0.6$ is pulled by a $15 N$ force at $30 °$ above the horizontal. Why is the normal reaction not $2 g$ ?
The pull has an upward component of $15 \sin 30 \circ = 7.5 N$ , which supports part of the weight.
Vertically the forces balance, so:
$R = 2 g - 7.5 = 19.6 - 7.5 = 12.1 N$
The maximum friction is therefore $0.6 \times 12.1 = 7.26 N$ , not $0.6 \times 19.6 = 11.76 N$ .
A block of mass $m$ rests on a rough plane inclined at $θ$ to the horizontal, with no other forces acting. What is the normal reaction, and why is it not $m g$ ?
The normal reaction is $R = m g \cos θ$ : it acts perpendicular to the plane, so it has to balance only the part of the weight acting in that direction.
The rest of the weight, $m g \sin θ$ , acts along the plane, where the reaction does nothing to oppose it, and that is the part that tries to slide the block down.
Since $\cos θ < 1$ for any real slope, the reaction on a slope is always less than the weight.
True or False?
You can assume that the normal reaction on a block resting on a plane inclined at $θ$ is $m g \cos θ$ .
False.
That is right only when the weight is the only force with a component perpendicular to the plane.
If anything else pushes into the plane or pulls away from it, such as a rope at an angle to the surface, the reaction changes to whatever makes the total perpendicular force zero.
The reaction is always found from the perpendicular equation, never quoted from a formula.
A block is placed on a rough slope and released. What decides whether it slides?
Compare the part of the weight pulling it down the slope with the largest friction available.
Down the slope the weight contributes $m g \sin θ$ , and the maximum friction is $μ R = μ m g \cos θ$ .
If $m g \sin θ$ is greater, the block slides; if it is less than or equal, friction balances it exactly and the block stays put.
A block slides down a rough plane inclined at $θ$ . Why does its acceleration not depend on its mass?
Down the slope the resultant force is $m g \sin θ - μ m g \cos θ$ , and $F = m a$ gives:
$a = g (\sin θ - μ \cos θ)$
The mass appears in every term of the resultant, because both the driving force and the friction come from the weight, so it cancels when you divide by $m$ .
So two blocks of different masses, made of the same material, slide down the same slope with the same acceleration.
A block is projected up a rough slope, slows down, and then slides back down. Why is its deceleration going up bigger than its acceleration coming down?
Because friction reverses when the direction of motion reverses, while the weight does not.
Going up, both the weight component and friction act down the slope, so they add: the deceleration is $g (\sin θ + μ \cos θ)$ .
Coming down, friction acts up the slope against the weight component, so they subtract: the acceleration is $g (\sin θ - μ \cos θ)$ .
True or False?
You can assume that a body resting in equilibrium on a rough surface has a frictional force of $μ R$ acting on it.
False.
A body at rest is somewhere in the range $F \leq μ R$ : the friction takes whatever value balances the other forces, and that is usually less than the maximum.
It equals $μ R$ only in limiting equilibrium, when the body is on the point of moving.
So a question saying a body is stationary on a rough slope does not let you write $F = μ R$ , but a question saying it is on the point of slipping does.
Why is being told a body is in limiting equilibrium so much more useful than being told it is simply in equilibrium?
Because it replaces an inequality with an equation.
Ordinary equilibrium gives only $F \leq μ R$ , which bounds the answer but does not pin it down; limiting equilibrium gives $F = μ R$ exactly.
That single equation is what makes an unknown solvable, which is why questions asking for a specific value of $μ$ , or for a least or greatest value, almost always describe the body as on the point of moving.
A block rests on a rough plane inclined at $θ$ to the horizontal. What is the least coefficient of friction that keeps it there?
The least value is $μ = \tan θ$ .
At the point of slipping the friction is at its maximum and balances the weight component down the slope:
$m g \sin θ = μ m g \cos θ$
The mass and $g$ cancel, leaving $μ = \frac{\sin θ}{\cos θ} = \tan θ$ , so the block stays put provided $μ \geq \tan θ$ , whatever its mass.
Two particles are connected by a string over a pulley. On which of them does friction act?
Only on a particle that is resting on a rough surface.
A particle hanging freely on the end of the string has nothing in contact with it, so there is no normal reaction and therefore no friction.
So in the common set-up of one particle on a rough table and one hanging, friction appears in the equation for the particle on the table and nowhere else.
Two particles rest on rough planes of different angles, connected by a string over a pulley at the top. How do you decide which way the system tends to move?
Compare the components of the two weights along their own slopes, that is $m g \sin θ$ for each particle; the larger one tends to pull its particle down.
Do not simply compare the masses: a lighter particle on a much steeper slope can have the greater component and win.
With equal masses it comes down to the angles alone, so the particle on the steeper slope is the one that tends to descend.
You have worked out which way a connected two-particle system tends to move. In which direction does friction act on each particle?
Friction opposes the motion of each particle, so it acts on each one against the direction that particle would move.
The two particles are connected, so they share a single sense of motion: if one is about to slide down its slope, the other is about to be dragged up its own.
So the friction on the first acts up its slope and on the second down its slope, each opposing its own particle's motion even though the two look opposed.

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