A book rests on a table and presses down on it. What does Newton's third law say about the table?

The table pushes up on the book with a force equal in magnitude and opposite in direction. The two forces of such a pair always act on different bodies, which is why they never cancel each other out: one acts on the book and the other on the table.

Two particles are connected by a rope and are moving in the same direction. What is gained by treating them as a single particle, and what happens to the tension?

Treating them as one particle gives a single equation of motion for the whole system, using the total mass . The tension disappears from it: it pulls one particle forwards and the other backwards with equal magnitude, so the two contributions cancel. That makes it the quickest route to the acceleration or to an external force; to find the tension itself, go back to one of the particles on its own.

A car tows a caravan with a tow bar, modelled as a light rod . When is the rod in tension and when is it in thrust?

A rod is in tension when it is being stretched, which happens while the car is accelerating and the rod has to drag the caravan along. It is in thrust , or compression, when it is being squashed, which happens while the car is braking and the caravan tends to catch up with the car and push against it. So the test is which way the caravan tends to move relative to the car: falling behind stretches the rod, catching up compresses it.

What can a rope do that a rod cannot, and what does that mean for the forces in it?

A rope can go slack . A rope only ever pulls, so it can only ever be in tension ; if it would ever need to push, it goes slack and the tension becomes zero. A rod can be in tension or in thrust, so it can push as well as pull and never goes slack.

For two connected particles, how many equations of motion can you write, and how many do you actually need?

Three are available: one for each particle on its own, and one for the two treated as a single system. Only two of them are independent, because the system equation is just the two separate equations added together, with the tension cancelling. So choose whichever two make the unknown you want easiest to reach, rather than writing all three and hoping.

A trailer is towed along level ground. There is no vertical motion, so is there any point in writing a vertical equation?

Yes: with no vertical motion the vertical resultant is zero , which still gives a usable equation. For the trailer that equation says the normal reaction equals the weight, and the normal reaction is exactly what you need if friction is involved later, since friction depends on it. No motion in a direction does not mean no information in that direction; it means the acceleration there is zero, which is a value like any other.

Why does the normal reaction appear as a downward force in the lift's equation?

Because the upward reaction on the load is matched by an equal and opposite force from the load pressing down on the lift floor. That downward force acts on the lift, so it belongs in the lift's equation alongside the lift's own weight, opposing the cable tension.

A pulley is a wheel that turns as a string passes over it, allowing the string to change direction while the particles attached to it move. In these models a pulley is always smooth and light. A peg does the same job but is a fixed point, such as a nail in a wall, rather than a wheel.

Two particles joined by a string over a pulley are in motion. Why is each one handled separately rather than as a single particle?

Because they are moving in different directions : one descends while the other rises or slides along a table. Each particle therefore gets its own equation of motion, with its own positive direction chosen to match the way that particle is actually moving.

One of the two particles slides along a smooth horizontal table. Why is no vertical equation needed for it?

Because the table is smooth , so the only horizontal force on that block is the tension and the vertical forces play no part in its motion. They would matter only if the table were rough , since friction depends on the normal reaction, and finding that reaction is what the vertical equation is for.

True or False? A particle hanging over a pulley and accelerating downwards has a string tension equal to its weight.

False. If the tension equalled the weight there would be no resultant force on the particle, and it would not accelerate at all. Because it is accelerating downwards, the tension must be less than its weight.

ASMathsCambridge (CIE)MechanicsFlashcardsForces & EquilibriumNewton's Second Law

Newton's Second Law (Cambridge (CIE) AS Maths: Mechanics): Flashcards

Exam code: 9709

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FrontF = ma
State Newton's second law of motion.

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Cards in this collection (23)

State Newton's second law of motion.
The resultant force acting on a body equals the mass of the body multiplied by its acceleration:
$F = m a$
Here $F$ is in newtons, $m$ is in kilograms and $a$ is in ${m s}^{- 2}$ .
Fill in which of Newton's three laws each of these belongs to:
An object with no resultant force on it carries on exactly as it was: Newton's $______$ law.
For a given mass, a bigger resultant force produces a bigger acceleration: Newton's $______$ law.
Forces between two bodies always come in equal and opposite pairs: Newton's $______$ law.
An object with no resultant force on it carries on exactly as it was: Newton's first law.
For a given mass, a bigger resultant force produces a bigger acceleration: Newton's second law.
Forces between two bodies always come in equal and opposite pairs: Newton's third law.
A book rests on a table and presses down on it. What does Newton's third law say about the table?
The table pushes up on the book with a force equal in magnitude and opposite in direction.
The two forces of such a pair always act on different bodies, which is why they never cancel each other out: one acts on the book and the other on the table.
In $F = m a$ , which force does $F$ stand for?
$F$ is the resultant force, the sum of all the forces acting on the body, and not any single one of them.
This matters because $F$ is also the usual letter for friction, so a diagram can carry an $F$ that must not be substituted into $F = m a$ .
How do you tell whether a problem needs $F = m a$ or only the constant-acceleration formulae?
If mass or force appears anywhere in the problem, you need $F = m a$ , because neither of them appears in the constant-acceleration formulae at all.
Acceleration is the quantity the two share, so a problem often uses the formulae to find $a$ and then $F = m a$ to reach a force.
A train of mass 5 tonnes accelerates at $0.9375 {m s}^{- 2}$ against a driving force of $6250 N$ . Find the total resistance to its motion.
A tonne is $1000 kg$ , so the mass is $5000 kg$ and the resultant force is $F = m a = 5000 \times 0.9375 = 4687.5 N$ .
The driving force and the resistance act in opposite directions, so the resistance is $6250 - 4687.5 = 1562.5 N$ .
True or False?
In $F = m a$ , a negative value of $F$ means the object is slowing down.
False.
A negative $F$ means the resultant force acts in whichever direction was chosen as negative, which is not the same as decelerating.
If the object is already travelling in that negative direction, a negative resultant force speeds it up.
Forces act on a moving particle in two perpendicular directions. How do you apply $F = m a$ ?
Apply it separately in each direction, using only the forces along that direction and the acceleration along that direction.
A particle can therefore have an acceleration in one of the two directions and none at all in the other, which is exactly what happens to a block sliding along a horizontal table.
Two particles are connected by a rope and are moving in the same direction. What is gained by treating them as a single particle, and what happens to the tension?
Treating them as one particle gives a single equation of motion for the whole system, using the total mass.
The tension disappears from it: it pulls one particle forwards and the other backwards with equal magnitude, so the two contributions cancel.
That makes it the quickest route to the acceleration or to an external force; to find the tension itself, go back to one of the particles on its own.
A car tows a caravan with a tow bar, modelled as a light rod. When is the rod in tension and when is it in thrust?
A rod is in tension when it is being stretched, which happens while the car is accelerating and the rod has to drag the caravan along.
It is in thrust, or compression, when it is being squashed, which happens while the car is braking and the caravan tends to catch up with the car and push against it.
So the test is which way the caravan tends to move relative to the car: falling behind stretches the rod, catching up compresses it.
What can a rope do that a rod cannot, and what does that mean for the forces in it?
A rope can go slack.
A rope only ever pulls, so it can only ever be in tension; if it would ever need to push, it goes slack and the tension becomes zero.
A rod can be in tension or in thrust, so it can push as well as pull and never goes slack.
For two connected particles, how many equations of motion can you write, and how many do you actually need?
Three are available: one for each particle on its own, and one for the two treated as a single system.
Only two of them are independent, because the system equation is just the two separate equations added together, with the tension cancelling.
So choose whichever two make the unknown you want easiest to reach, rather than writing all three and hoping.
A trailer is towed along level ground. There is no vertical motion, so is there any point in writing a vertical equation?
Yes: with no vertical motion the vertical resultant is zero, which still gives a usable equation.
For the trailer that equation says the normal reaction equals the weight, and the normal reaction is exactly what you need if friction is involved later, since friction depends on it.
No motion in a direction does not mean no information in that direction; it means the acceleration there is zero, which is a value like any other.
A person of mass $m$ stands in a lift. When is the normal reaction from the floor equal to their weight?
Only when the lift is not accelerating, whether it is at rest or moving at a constant speed.
While the lift accelerates upwards the reaction is greater than the weight, and while it accelerates downwards the reaction is smaller, which is why a lift briefly makes you feel heavier or lighter.
A load of mass $m$ rests on the floor of a lift accelerating downwards with acceleration $a$ . Write its equation of motion.
Taking downwards as positive, the forces on the load are its weight $m g$ downwards and the normal reaction $R$ upwards, so
$m g - R = m a$
This equation involves the load alone, so the cable tension never enters it.
Why does the normal reaction appear as a downward force in the lift's equation?
Because the upward reaction on the load is matched by an equal and opposite force from the load pressing down on the lift floor.
That downward force acts on the lift, so it belongs in the lift's equation alongside the lift's own weight, opposing the cable tension.
A load on the floor of a lift accelerating downwards at $0.3 {m s}^{- 2}$ receives a normal reaction of $399 N$ . Taking $g = 10 {m s}^{- 2}$ , find its mass.
Taking downwards as positive, $m g - R = m a$ becomes $10 m - 399 = 0.3 m$ .
That rearranges to $9.7 m = 399$ , so $m = 41.1 kg$ to three significant figures.
True or False?
In a lift problem the lift and the load always have the same acceleration.
True.
The load rests on the lift floor and stays in contact with it, so the two move together and share a single acceleration.
It would fail only if the lift accelerated downwards faster than $g$ , when the load would lift off the floor and no longer be in contact.
Define pulley.
A pulley is a wheel that turns as a string passes over it, allowing the string to change direction while the particles attached to it move.
In these models a pulley is always smooth and light.
A peg does the same job but is a fixed point, such as a nail in a wall, rather than a wheel.
Two particles joined by a string over a pulley are in motion. Why is each one handled separately rather than as a single particle?
Because they are moving in different directions: one descends while the other rises or slides along a table.
Each particle therefore gets its own equation of motion, with its own positive direction chosen to match the way that particle is actually moving.
A $25 kg$ block on a smooth horizontal table is joined over a pulley to a $60 kg$ block hanging freely. Taking $g = 10 {m s}^{- 2}$ , find the acceleration.
For the hanging block $600 - T = 60 a$ , and for the block on the table $T = 25 a$ .
Substituting the second into the first gives $600 - 25 a = 60 a$ , so $85 a = 600$ and $a = 7.06 {m s}^{- 2}$ to three significant figures.
One of the two particles slides along a smooth horizontal table. Why is no vertical equation needed for it?
Because the table is smooth, so the only horizontal force on that block is the tension and the vertical forces play no part in its motion.
They would matter only if the table were rough, since friction depends on the normal reaction, and finding that reaction is what the vertical equation is for.
True or False?
A particle hanging over a pulley and accelerating downwards has a string tension equal to its weight.
False.
If the tension equalled the weight there would be no resultant force on the particle, and it would not accelerate at all.
Because it is accelerating downwards, the tension must be less than its weight.

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