Why is a force acting at an angle usually split into two components?

Because only the component along the direction you care about has any effect in that direction, and the component perpendicular to it has none at all. Splitting the force lets each direction be handled on its own, and the two components together have exactly the same effect as the original force.

True or False? Once a force has been split into components, the diagram carries two extra forces.

False. The two components replace the original force: they are two ways of describing the same push or pull, not additions to it. Counting the original force as well as its components puts the same force into the equations twice.

Two forces act on a particle at an angle to each other. How can you find their resultant without splitting either into components?

Draw them nose to tail to form a triangle: the third side, from the start of the first to the end of the second, is the resultant. Its magnitude and direction then come from trigonometry on that triangle.

Define line of greatest slope .

The line of greatest slope is the most direct route from the bottom of an inclined plane to the top. Problems on an inclined plane are always set along this line, so the motion is one-dimensional even though the plane is a surface.

A block slides along an inclined plane. In which direction can it accelerate?

Only parallel to the plane, along the line of greatest slope. The block stays on the surface, so it has no acceleration perpendicular to the plane, which means the perpendicular forces must balance exactly and give an equation of their own.

Define contact force .

The contact force is the single force a surface exerts on an object, made up of the normal reaction and the friction together. In working it is almost always kept as those two perpendicular parts; only when its magnitude is asked for are they combined back into one vector.

A stationary object is pushed along a rough surface by a force smaller than the maximum friction available. What is the frictional force?

It is exactly equal and opposite to the push, so the two cancel and the object stays where it is. Friction takes whatever value it needs to prevent motion, up to its maximum, which is why the relationship is an inequality rather than an equation.

Why is the normal reaction not simply the weight when a force acts at an angle?

Because the angled force has a vertical component, which joins the weight and the normal reaction in the vertical balance. A force pulling partly upwards reduces the normal reaction, and one pushing partly downwards increases it.

Why must the normal reaction be found before the acceleration can be?

Because the greatest friction available depends on the normal reaction, and the friction is one of the forces that make up the resultant. Without the normal reaction there is no way to know how big the friction is, and so no way to find the resultant that the acceleration comes from.

In which direction does friction act on a block on a slope, and when does that direction change?

Always opposite to the way the block is moving, or is on the point of moving, along the line of greatest slope. So friction acts up the slope on a block sliding down, and down the slope on a block being pushed up: the same block on the same plane can have friction either way.

True or False? Knowing that a block stays at rest on a rough slope tells you the coefficient of friction exactly.

False. It gives only a lower bound : the coefficient must be at least large enough to hold the block, and any larger value would hold it just as well. An exact value follows only if the block is on the point of slipping, which a question has to state.

Two particles are connected by a string over a pulley, one on a rough plane and one hanging freely. Which of them experiences friction?

Only the particle resting on the rough plane. Friction acts between an object and a surface it is in contact with, so a particle hanging in the air has none, however rough the other plane is.

Two connected particles on rough slopes are in limiting equilibrium. Why does friction act up the slope on one and down the slope on the other?

Because friction always opposes the motion that particular particle is on the point of making, and the two are about to move in opposite senses along the string. The particle about to slide down has friction acting up its slope, while the one about to be dragged up has friction acting down its slope.

A LevelMathsCambridge (CIE)MechanicsFlashcardsForces & EquilibriumResolving Forces, Inclined Planes & Friction

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

Exam code: 9709

1/33

0Still learning

Know0

FrontResolving Forces & Inclined Planes
Why is a force acting at an angle usually split into two components?

Was this flashcard helpful?

Cards in this collection (33)

Why is a force acting at an angle usually split into two components?
Because only the component along the direction you care about has any effect in that direction, and the component perpendicular to it has none at all.
Splitting the force lets each direction be handled on its own, and the two components together have exactly the same effect as the original force.
A force of magnitude $P$ acts at an angle $θ$ above the horizontal. Fill in its two components:
horizontal component $= P______$ , vertical component $= P______$
horizontal component $= P \cos θ$ , vertical component $= P \sin θ$
The original force is the hypotenuse of a right-angled triangle whose other two sides are the components, so the angle decides which of $\sin$ and $\cos$ goes with which.
A box of mass $2 kg$ on a horizontal floor is pulled by a force of $10 N$ at $30 °$ above the horizontal. Taking $g = 10 {m s}^{- 2}$ , find the normal reaction.
The pull has an upward component of $10 \sin 30 ° = 5 N$ , and the weight is $20 N$ downwards.
Balancing vertically gives $R + 5 = 20$ , so $R = 15 N$ , which is less than the weight because part of the pull is helping to hold the box up.
True or False?
Once a force has been split into components, the diagram carries two extra forces.
False.
The two components replace the original force: they are two ways of describing the same push or pull, not additions to it.
Counting the original force as well as its components puts the same force into the equations twice.
Two forces act on a particle at an angle to each other. How can you find their resultant without splitting either into components?
Draw them nose to tail to form a triangle: the third side, from the start of the first to the end of the second, is the resultant.
Its magnitude and direction then come from trigonometry on that triangle.
Define line of greatest slope.
The line of greatest slope is the most direct route from the bottom of an inclined plane to the top.
Problems on an inclined plane are always set along this line, so the motion is one-dimensional even though the plane is a surface.
A block of mass $m$ rests on a plane inclined at $θ$ . What are the components of its weight parallel and perpendicular to the plane?
Down the slope the component is $m g \sin θ$ , and into the plane it is $m g \cos θ$ .
The angle between the weight and the perpendicular to the plane is equal to the angle of the slope itself, which is where the $θ$ in both components comes from.
A block slides along an inclined plane. In which direction can it accelerate?
Only parallel to the plane, along the line of greatest slope.
The block stays on the surface, so it has no acceleration perpendicular to the plane, which means the perpendicular forces must balance exactly and give an equation of their own.
Define contact force.
The contact force is the single force a surface exerts on an object, made up of the normal reaction and the friction together.
In working it is almost always kept as those two perpendicular parts; only when its magnitude is asked for are they combined back into one vector.
Define coefficient of friction.
The coefficient of friction, written $μ$ , is the number linking the frictional force between two surfaces to the normal reaction between them.
It depends on what the two surfaces are made of, and it is never negative.
Complete the inequality satisfied by the frictional force:
$F \leq______$
where $R$ is the $______$ force between the object and the surface.
The completed inequality is:
$F \leq μ R$
Here $R$ is the normal reaction force between the object and the surface.
A stationary object is pushed along a rough surface by a force smaller than the maximum friction available. What is the frictional force?
It is exactly equal and opposite to the push, so the two cancel and the object stays where it is.
Friction takes whatever value it needs to prevent motion, up to its maximum, which is why the relationship is an inequality rather than an equation.
Define limiting equilibrium.
An object is in limiting equilibrium when it is still stationary but on the point of moving.
This is the one stationary case in which the friction has reached its maximum, so $F = μ R$ exactly.
True or False?
The coefficient of friction is measured in newtons.
False.
$μ$ has no units at all, because it is a ratio of one force to another and the newtons cancel.
That is why a coefficient of friction is quoted as a bare number such as $0.2$ or $0.7$ .
A box of mass $12 kg$ rests on a floor with $μ = 0.7$ , and a horizontal force of $80 N$ is applied. Taking $g = 10 {m s}^{- 2}$ , does it move?
No. The normal reaction is $120 N$ , so the greatest friction available is $0.7 \times 120 = 84 N$ .
The applied $80 N$ is smaller than that, so the box stays still and the friction acting on it is $80 N$ , not $84 N$ .
An object is already sliding along a rough surface. What is the frictional force on it?
It is $μ R$ , its maximum value, whatever the other forces happen to be.
Pushing harder does not increase the friction once the object is moving: it simply leaves a bigger resultant force and so a bigger acceleration.
Why is the normal reaction not simply the weight when a force acts at an angle?
Because the angled force has a vertical component, which joins the weight and the normal reaction in the vertical balance.
A force pulling partly upwards reduces the normal reaction, and one pushing partly downwards increases it.
A box of mass $m$ on a horizontal floor is acted on by a force $P$ at an angle $θ$ to the horizontal. Fill in the two signs:
if $P$ pulls upwards at that angle, $R = m g______P \sin θ$
if $P$ pushes downwards at that angle, $R = m g______P \sin θ$
If $P$ pulls upwards at that angle, $R = m g - P \sin θ$ .
If $P$ pushes downwards at that angle, $R = m g + P \sin θ$ .
In each case $R$ takes whatever value makes the total vertical force zero.
Why must the normal reaction be found before the acceleration can be?
Because the greatest friction available depends on the normal reaction, and the friction is one of the forces that make up the resultant.
Without the normal reaction there is no way to know how big the friction is, and so no way to find the resultant that the acceleration comes from.
A block on a rough horizontal floor is being dragged along. What goes into the resultant force used in $F = m a$ ?
The horizontal component of the applied force minus the friction, since the two act in opposite directions.
The friction is one of the forces being combined, not the resultant itself, which matters here because both are commonly written as $F$ .
True or False?
Doubling the mass of a box on a rough floor doubles the horizontal force needed to start it sliding.
True.
Doubling the mass doubles the weight, so it doubles the normal reaction and therefore doubles the maximum friction $μ R$ .
This holds while the applied force is horizontal: a force at an angle changes the normal reaction itself, and the doubling no longer follows.
On a plane inclined at $θ$ the weight of a block has components $m g \sin θ$ down the slope and $m g \cos θ$ into it. What is the greatest friction available?
The block does not accelerate into or out of the plane, so the perpendicular forces balance and $R = m g \cos θ$ .
The greatest friction available is therefore $μ R = μ m g \cos θ$ , which is smaller than $μ m g$ would be on level ground.
True or False?
Making a slope steeper reduces the greatest frictional force available to hold a block on it.
True.
The normal reaction is $m g \cos θ$ , and $\cos θ$ gets smaller as the slope gets steeper, so the maximum friction falls with it.
At the same time the component pulling the block down the slope, $m g \sin θ$ , grows, which is why a block slides once the slope is steep enough.
A block of mass $1 kg$ is released from rest on a rough plane inclined at $20 °$ , with $μ = 0.2$ . Taking $g = 10 {m s}^{- 2}$ , find its acceleration.
The normal reaction is $10 \cos 20 ° = 9.40 N$ , so the friction is $0.2 \times 9.40 = 1.88 N$ up the slope.
The weight component down the slope is $10 \sin 20 ° = 3.42 N$ , so the resultant is $1.54 N$ and the acceleration is $1.54 {m s}^{- 2}$ down the slope.
In which direction does friction act on a block on a slope, and when does that direction change?
Always opposite to the way the block is moving, or is on the point of moving, along the line of greatest slope.
So friction acts up the slope on a block sliding down, and down the slope on a block being pushed up: the same block on the same plane can have friction either way.
What decides whether a block released on a rough slope slides or stays put?
Compare the component of the weight down the slope with the greatest friction available: it slides only if $m g \sin θ$ exceeds $μ m g \cos θ$ .
Both sides grow in proportion to the block's weight, so a heavier block of the same material on the same slope is no more likely to slide than a lighter one.
When may you write $F = μ R$ for a block stationary on a rough slope?
Only when the block is on the point of moving, which is limiting equilibrium; then the friction has reached its maximum.
For a block that is simply stationary you know only that $F \leq μ R$ , so assuming equality would be assuming something the question has not told you.
A block of unknown mass stays at rest on a plane inclined at $25 °$ . Find the least possible value of the coefficient of friction.
For the block not to slide, the friction must be able to match the weight component down the slope, so $μ m g \cos 25 ° \geq m g \sin 25 °$ .
The $m g$ cancels from both sides, leaving $μ \geq \tan 25 °$ , so the least possible value is $0.466$ to three significant figures.
Because the mass cancels, the answer does not depend on how heavy the block is.
True or False?
Knowing that a block stays at rest on a rough slope tells you the coefficient of friction exactly.
False.
It gives only a lower bound: the coefficient must be at least large enough to hold the block, and any larger value would hold it just as well.
An exact value follows only if the block is on the point of slipping, which a question has to state.
Two particles are connected by a string over a pulley, one on a rough plane and one hanging freely. Which of them experiences friction?
Only the particle resting on the rough plane.
Friction acts between an object and a surface it is in contact with, so a particle hanging in the air has none, however rough the other plane is.
Two connected particles rest on rough planes of different slopes. How do you decide which way the system will move?
Compare the components of the weights along the slopes, $m g \sin θ$ for each, rather than the weights themselves.
The particle with the larger component is the one that moves down, so a lighter particle on a steep plane can pull a heavier one up a gentle plane.
Particles $A$ and $B$ of equal mass sit on planes inclined at $60 °$ and $30 °$ , joined by a string over a pulley at the top. Which one is on the point of moving down?
Particle $A$ , on the steeper plane.
Its weight component down the slope is $m g \sin 60 °$ , which is larger than $B$ 's $m g \sin 30 °$ , and the masses are equal so the comparison rests on the angles alone.
Two connected particles on rough slopes are in limiting equilibrium. Why does friction act up the slope on one and down the slope on the other?
Because friction always opposes the motion that particular particle is on the point of making, and the two are about to move in opposite senses along the string.
The particle about to slide down has friction acting up its slope, while the one about to be dragged up has friction acting down its slope.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Newton's Second LawNext:Work & Energy