A force acts at an angle to the direction a particle can move. What do its two components do?

The component parallel to the direction of motion is the part that has an effect on the particle. The component perpendicular to it has no effect in that direction, and the two together have exactly the same effect as the original force.

Two forces act on a particle at an angle to each other. How can their resultant be found without splitting them into components?

Draw them nose to tail so that they form a triangle, in which the third side is the resultant. Its magnitude and direction then come from trigonometry on that triangle, using the cosine and sine rules when the angle between the forces is not a right angle.

On an inclined plane, why is it easier to resolve parallel and perpendicular to the slope than horizontally and vertically?

Because the normal reaction already acts perpendicular to the plane, so it does not have to be resolved at all, and any motion is along the slope. That leaves the weight as the only force pointing in some other direction, so it is the only one that has to be split into components.

Where does the slope's angle reappear when you resolve the weight?

It reappears as the angle between the weight and the line perpendicular to the plane . Spotting that angle is what tells you which component takes the cosine and which takes the sine, and getting the two the wrong way round is the commonest error in these problems.

True or False? An object sliding down a slope has no acceleration in the direction perpendicular to the slope.

True. The object stays on the surface, neither lifting away from the plane nor sinking into it, so the perpendicular forces must balance exactly. That is what fixes the size of the normal reaction, and it is why all of the acceleration is along the line of greatest slope .

True or False? A stationary object can have a frictional force acting on it.

True. Friction acts whenever something is trying to push or pull the object along the surface, whether or not it succeeds in moving it. With nothing but the weight and the normal reaction acting, though, there is no friction at all, because there is nothing for it to oppose.

An object is stationary, but a force is trying to push it along a rough surface. In which direction does the friction act?

Opposite to the resultant of the other forces acting parallel to the surface, which is the direction the object would move in if friction were not there. For a stationary object friction opposes the tendency to move rather than any actual motion, and it always acts parallel to the surface.

How do you decide whether a block placed on a rough slope will slide?

Compare the pull of gravity along the slope with the greatest friction the surface can supply. If gravity's pull along the slope is the larger of the two the block slides; if it is not, friction matches it exactly and the block stays where it is.

True or False? A block that is stationary on a rough slope must be in limiting equilibrium.

False. It is in limiting equilibrium only if it is on the point of moving , which is a special case rather than the general one. A block sitting comfortably on a gentle slope is in ordinary equilibrium, with its friction well below the maximum available.

Two connected particles lie on rough slopes either side of a pulley. On which particles does friction act, and in which directions?

On both , because both rest on rough surfaces; a particle on a smooth surface would have none. Each particle's friction acts along its own slope, opposing that particle's own motion. So the one sliding down has friction acting up its slope, while the one being dragged up has friction acting down its slope.

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

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A force acts at an angle to the direction a particle can move. What do its two components do?

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Cards in this collection (37)

A force acts at an angle to the direction a particle can move. What do its two components do?
The component parallel to the direction of motion is the part that has an effect on the particle.
The component perpendicular to it has no effect in that direction, and the two together have exactly the same effect as the original force.
Fill in the two missing components of a force of magnitude $F$ N acting at an angle $θ$ above the horizontal:
The horizontal component is $______$ and the vertical component is $______$ .
The horizontal component is $F \cos θ$ and the vertical component is $F \sin θ$ .
The force is the hypotenuse of a right-angled triangle whose other two sides are the components, so the side next to the angle is the one that takes the cosine.
A box of mass 2 kg is pulled along a smooth floor by a 10 N force at $30 °$ above the horizontal. Find the normal reaction, taking $g = 9.8$ .
The normal reaction is $15$ N, to 2 significant figures.
The pull has an upward component of $10 \sin 30 ° = 5$ N, so resolving vertically gives $R + 5 = 2 g$ and therefore $R = 19.6 - 5 = 14.6$ .
The reaction is less than the weight, because part of the pull is helping to hold the box up.
Two forces act on a particle at an angle to each other. How can their resultant be found without splitting them into components?
Draw them nose to tail so that they form a triangle, in which the third side is the resultant.
Its magnitude and direction then come from trigonometry on that triangle, using the cosine and sine rules when the angle between the forces is not a right angle.
A 2 kg box on a smooth floor is pulled by a 10 N force at $30 °$ above the horizontal. Find its acceleration.
The acceleration is $\frac{5 \sqrt{3}}{2} {m s}^{- 2}$ , which is about $4.33 {m s}^{- 2}$ .
Only the horizontal component $10 \cos 30 ° = 5 \sqrt{3}$ N acts along the direction of motion, and the floor is smooth, so $F = m a$ gives $5 \sqrt{3} = 2 a$ .
On an inclined plane, why is it easier to resolve parallel and perpendicular to the slope than horizontally and vertically?
Because the normal reaction already acts perpendicular to the plane, so it does not have to be resolved at all, and any motion is along the slope.
That leaves the weight as the only force pointing in some other direction, so it is the only one that has to be split into components.
A block of mass $m$ rests on a plane inclined at $θ$ to the horizontal. Find the normal reaction, and say why it is not $m g$ .
The normal reaction is $R = m g \cos θ$ , which is less than $m g$ for any tilted plane.
The weight is vertical, so the part of it pressing into the plane is only $m g \cos θ$ ; the rest of it, $m g \sin θ$ , acts down the slope instead.
Where does the slope's angle reappear when you resolve the weight?
It reappears as the angle between the weight and the line perpendicular to the plane.
Spotting that angle is what tells you which component takes the cosine and which takes the sine, and getting the two the wrong way round is the commonest error in these problems.
True or False?
An object sliding down a slope has no acceleration in the direction perpendicular to the slope.
True.
The object stays on the surface, neither lifting away from the plane nor sinking into it, so the perpendicular forces must balance exactly.
That is what fixes the size of the normal reaction, and it is why all of the acceleration is along the line of greatest slope.
A 5 kg box is pushed up a smooth ramp inclined at $10 °$ by a force of 12 N up the slope. Find its acceleration, taking $g = 9.8$ .
The acceleration is $0.70 {m s}^{- 2}$ , to 2 significant figures.
Resolving up the slope gives $12 - 5 g \sin 10 ° = 5 a$ , and since $5 g \sin 10 ° = 8.509 \dots$ this leaves $5 a = 3.491 \dots$ and $a = 0.698 \dots$
Fill in the two missing quantities:
The frictional force satisfies $F \leq μ R$ , where $μ$ is the $______$ and $R$ is the $______$ .
The frictional force satisfies $F \leq μ R$ , where $μ$ is the coefficient of friction between the two surfaces and $R$ is the normal reaction between them.
The coefficient $μ$ has no units, because it is a ratio of one force to another.
Define limiting equilibrium.
A body is in limiting equilibrium when it is still stationary but the frictional force on it has reached its greatest possible value, $μ R$ .
It is then on the point of moving: any increase at all in the force trying to shift it will start it sliding.
A box of mass 12 kg rests on a floor with $μ = 0.7$ , and an 80 N horizontal force is applied to it. Find the frictional force, taking $g = 9.8$ .
The frictional force is 80 N, not the maximum of $82.32$ N.
Here $R = 12 g = 117.6$ N, so the greatest possible friction is $0.7 \times 117.6 = 82.32$ N, and since $80 < 82.32$ the box stays still.
Friction only ever supplies as much force as is needed to prevent motion, up to that maximum, so here it exactly matches the 80 N.
True or False?
A stationary object can have a frictional force acting on it.
True.
Friction acts whenever something is trying to push or pull the object along the surface, whether or not it succeeds in moving it.
With nothing but the weight and the normal reaction acting, though, there is no friction at all, because there is nothing for it to oppose.
An object is stationary, but a force is trying to push it along a rough surface. In which direction does the friction act?
Opposite to the resultant of the other forces acting parallel to the surface, which is the direction the object would move in if friction were not there.
For a stationary object friction opposes the tendency to move rather than any actual motion, and it always acts parallel to the surface.
An object is sliding across a rough surface. What is the frictional force on it?
It is $μ R$ , its greatest possible value, whatever other forces happen to be acting.
Once the object is actually moving, friction becomes a force of constant magnitude opposing the motion, so it no longer adjusts itself the way it does while the object is at rest.
For a surface described as smooth, what is the value of $μ$ ?
For a smooth surface $μ = 0$ .
Then $F \leq μ R$ becomes $F \leq 0$ , so the friction is zero and can be left out of the problem altogether, whereas a rough surface has $μ > 0$ and friction must be included.
Pulling a box along a rough floor at an angle above the horizontal reduces the normal reaction. What does that do to the friction?
It reduces the greatest possible friction as well, because that maximum is $μ R$ and $R$ has gone down.
The coefficient $μ$ is a property of the two surfaces and does not change; it is $R$ that carries the effect of the angle.
Fill in the two missing terms in the equation of motion for a block being dragged along a rough horizontal floor by a horizontal force $P$ N:
$P -______=______$
The completed equation is:
$P - μ R = m a$
The friction is subtracted because it acts backwards, against the direction the block is being dragged.
Why must the normal reaction be found before the frictional force in an $F = m a$ problem?
Because the greatest possible friction is $μ R$ , so it cannot be worked out until $R$ is known.
$R$ comes from resolving perpendicular to the surface, which is why that direction is always dealt with first, even when the motion is entirely along the surface.
True or False?
On a horizontal surface the normal reaction is always equal to the object's weight.
False.
It equals the weight only when the weight and the reaction are the only vertical forces acting on the object.
As soon as another force has a vertical component, $R$ has to change so that the vertical forces still balance.
A box on a rough floor is pushed by a force angled downwards. What happens to the greatest possible friction?
It increases, because the downward component of the push adds to the force pressing the box into the floor, making the normal reaction larger than the weight.
Since the maximum friction is $μ R$ , pushing downwards at an angle makes the box harder to slide, not easier.
A 2 kg block on a rough floor is pulled by a 15 N force at $30 °$ , giving a normal reaction of 12.1 N, and $μ = 0.6$ . Find the acceleration.
The acceleration is $2.9 {m s}^{- 2}$ , to 2 significant figures.
The greatest possible friction is $0.6 \times 12.1 = 7.26$ N and the horizontal component of the pull is $15 \cos 30 ° = 12.99$ N, which is larger, so the block moves.
The resultant along the floor is $12.99 - 7.26 = 5.73$ N, and $F = m a$ gives $a = \frac{5.73}{2} = 2.865$ .
A block of mass $m$ rests on a rough plane inclined at $θ$ . Write down the greatest possible frictional force on it.
The greatest possible friction is $F_{MAX} = μ m g \cos θ$ .
The maximum is always $μ R$ , and on a slope with no other perpendicular forces the normal reaction is $m g \cos θ$ rather than $m g$ .
How do you decide whether a block placed on a rough slope will slide?
Compare the pull of gravity along the slope with the greatest friction the surface can supply.
If gravity's pull along the slope is the larger of the two the block slides; if it is not, friction matches it exactly and the block stays where it is.
The same block is once sliding down a rough slope and once being pushed up it. How does the friction differ between the two cases?
Its direction reverses: up the slope while the block slides down, and down the slope while the block is pushed up.
Its magnitude is the same $μ m g \cos θ$ in both cases, because the normal reaction has not changed.
True or False?
Making a slope steeper increases both the force pulling a block down it and the greatest friction holding it back.
False.
Steepening the slope does increase the component of the weight down it, $m g \sin θ$ , but it decreases the normal reaction $m g \cos θ$ .
The greatest possible friction therefore falls at the very moment the pull increases, which is why any block will slide once the slope is steep enough.
Fill in the two missing terms in the equation of motion for a block of mass $m$ sliding down a rough plane inclined at $θ$ , taking down the slope as positive:
$m g \sin θ -______=______$
The completed equation is:
$m g \sin θ - μ m g \cos θ = m a$
The friction is subtracted because it opposes the motion down the slope.
A 1 kg block is released from rest on a rough slope of $20 °$ with $μ = 0.2$ . Find its acceleration, taking $g = 9.8$ .
The acceleration is $1.5 {m s}^{- 2}$ down the slope, to 2 significant figures.
The weight component down the slope is $g \sin 20 ° = 3.352$ N and the friction is $0.2 \times g \cos 20 ° = 1.842$ N, so the resultant down the slope is $3.352 - 1.842 = 1.510$ N.
The mass is 1 kg, so $F = m a$ gives $a = 1.510$ .
A block rests without slipping on a rough plane inclined at $θ$ . Show that $μ \geq \tan θ$ .
Parallel to the plane the friction must balance $m g \sin θ$ , and friction can never exceed $μ R = μ m g \cos θ$ , so $m g \sin θ \leq μ m g \cos θ$ .
Dividing through by $m g \cos θ$ leaves $\tan θ \leq μ$ , which is the required condition.
Why does a friction question about a stationary block ask for the least possible value of $μ$ rather than its value?
Because friction obeys an inequality rather than an equation, so knowing that the block has not moved only tells you that $μ$ is large enough.
Any larger value would hold the block just as well, so the information fixes a range of possible values and the least member of that range is the answer.
True or False?
A block that is stationary on a rough slope must be in limiting equilibrium.
False.
It is in limiting equilibrium only if it is on the point of moving, which is a special case rather than the general one.
A block sitting comfortably on a gentle slope is in ordinary equilibrium, with its friction well below the maximum available.
A block is stationary on a rough plane inclined at $25 °$ . Find the least possible value of $μ$ .
The least possible value is $μ = \tan 25 ° = 0.466$ , to 3 significant figures.
Neither the mass of the block nor the value of $g$ is needed, because both of them cancel out of the condition entirely.
Two particles of equal mass sit on rough slopes of different angles, joined by a string over a pulley at the top. Which way does the system tend to move?
The particle on the steeper slope moves down, pulling the other one up its own slope.
That is because the component of weight along a slope is $m g \sin θ$ , and $\sin θ$ increases with the angle, so the steeper slope pulls harder.
True or False?
In a connected-particle problem on two slopes, the heavier particle always moves down its slope.
False.
What decides the motion is the component of weight along each slope, $m g \sin θ$ , and not the weight on its own.
A lighter particle on a much steeper slope can easily win, so the angles have to be taken into account alongside the masses.
Two connected particles lie on rough slopes either side of a pulley. On which particles does friction act, and in which directions?
On both, because both rest on rough surfaces; a particle on a smooth surface would have none.
Each particle's friction acts along its own slope, opposing that particle's own motion.
So the one sliding down has friction acting up its slope, while the one being dragged up has friction acting down its slope.
Two connected particles on rough slopes are in limiting equilibrium. What does that tell you about the friction on each of them?
Each one's friction has reached its own maximum, $μ R$ , worked out separately for that particle on its own slope.
The two coefficients need not be equal, since they describe different pairs of surfaces, and the two normal reactions differ as well because the slopes differ.

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