State Newton's third law of motion.

For two bodies, the force exerted on the second body by the first is equal in magnitude and opposite in direction to the force exerted on the first body by the second. The two forces always act on different bodies, which is why they never cancel each other out on a single object. It is the law that lets you say the tension pulling a trailer forwards is the same size as the tension pulling back on the car.

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.

What makes a mechanics problem a "lift problem", and what does that tell you about the directions involved?

A lift problem is one where the objects are in direct contact with each other, such as a person or a load standing on the floor of a lift, rather than being joined by a rope. The motion is vertical only , so every equation is a vertical one and the weight of each object is always involved. The same model covers anything of that shape: a crane raising a loaded pallet, or a fairground ride carrying its passengers.

A lift carries a load on its floor. Which treatment gives you the tension in the cable , and which gives you the force between the lift and the load ?

For the tension in the cable , treat the lift and the load as a single particle of combined mass; the force between them cancels, leaving the cable tension and the total weight. For the force between them , you must treat them separately . That force is internal to the system, so it does not appear at all in the combined equation and cannot be found from it.

Why can a pulley system not be treated as a single particle, when a car and trailer can be?

Because the two particles move in different directions : one goes up while the other goes down, or one moves horizontally while the other falls. Treating bodies as one particle only works when they all move the same way, so that a single acceleration in a single direction describes the whole system. In a pulley problem there is no such direction, so each particle must be given its own equation of motion.

What does a pulley do in a mechanics problem, and how is a peg different?

A pulley lets an inextensible string change direction without changing what it does to the particles at its ends, so one particle can move horizontally while another moves vertically. A peg is a fixed point that a string passes over or a particle hangs from, like a nail in a wall. It does the same job of redirecting the string but does not rotate.

A block on a smooth horizontal table is connected over a pulley to a hanging mass. Why does the block's weight not appear in its equation of motion?

The block moves horizontally , and its weight acts vertically , so the weight is perpendicular to the motion and contributes nothing to the horizontal equation. Vertically the block is not accelerating, so its weight is balanced by the normal reaction from the table. The weight only becomes relevant to the horizontal motion if the table is rough , because friction depends on that normal reaction, which in turn depends on the weight.

A LevelMathsEdexcelMechanicsFlashcardsForces & Newton's LawsNewton's Second Law (F=ma)

Newton's Second Law (F=ma) (Edexcel A Level Maths: Mechanics): Flashcards

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FrontF = ma
State Newton's second law of motion.
BackF = ma
For two bodies, the force exerted on the second body by the first is equal in magnitude and opposite in direction to the force exerted on the first body by the second.
The two forces always act on different bodies, which is why they never cancel each other out on a single object.
It is the law that lets you say the tension pulling a trailer forwards is the same size as the tension pulling back on the car.

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Cards in this collection (25)

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 measured in newtons, $m$ in kilograms and $a$ in ${m s}^{- 2}$ . The law only balances if all three are in those units.
True or False?
You can assume that the $F$ in $F = m a$ is the driving force acting on the object.
False.
The $F$ in $F = m a$ is the resultant force: everything acting on the object, added together with their directions taken into account.
For a car being driven along a road, that means the driving force minus any resistance, not the driving force on its own.
The two happen to agree only when the driving force is the sole force acting in that direction.
State Newton's third law of motion.
For two bodies, the force exerted on the second body by the first is equal in magnitude and opposite in direction to the force exerted on the first body by the second.
The two forces always act on different bodies, which is why they never cancel each other out on a single object.
It is the law that lets you say the tension pulling a trailer forwards is the same size as the tension pulling back on the car.
A problem can be attacked with the constant-acceleration formulae or with $F = m a$ . What tells you which one you need, and what connects them?
Look for force and mass. Neither appears anywhere in the constant-acceleration formulae, so a question mentioning either of them needs $F = m a$ .
The two are connected by the acceleration, which appears in both; a typical problem uses one to find the acceleration and the other to find what was asked for.
How does the weight of an object enter an equation of motion?
Weight is a force, so it goes into the resultant like any other force, with magnitude $m g$ newtons.
It always acts vertically downwards, so it only appears in a vertical equation, and its sign follows the direction chosen as positive: taking upwards as positive it enters as $- m g$ .
If the weight is the only force acting, $F = m a$ becomes $m g = m a$ , so the acceleration is $g$ whatever the mass.
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.
What makes a mechanics problem a "lift problem", and what does that tell you about the directions involved?
A lift problem is one where the objects are in direct contact with each other, such as a person or a load standing on the floor of a lift, rather than being joined by a rope.
The motion is vertical only, so every equation is a vertical one and the weight of each object is always involved.
The same model covers anything of that shape: a crane raising a loaded pallet, or a fairground ride carrying its passengers.
A lift carries a load on its floor. Which treatment gives you the tension in the cable, and which gives you the force between the lift and the load?
For the tension in the cable, treat the lift and the load as a single particle of combined mass; the force between them cancels, leaving the cable tension and the total weight.
For the force between them, you must treat them separately.
That force is internal to the system, so it does not appear at all in the combined equation and cannot be found from it.
The floor of a lift pushes up on a load with a force of $R$ newtons. What else does that tell you, and why does it matter?
The load pushes down on the floor of the lift with a force of the same magnitude, $R$ newtons.
The two forces act on different bodies, so they do not cancel on either one; $R$ appears upwards in the equation for the load and downwards in the equation for the lift.
That is exactly why $R$ disappears when the two are added together to treat the system as one.
A lift accelerates downwards. Is the force from the floor on a person inside greater or less than their weight?
Less than their weight.
The person is accelerating downwards, so the resultant force on them must act downwards, which means the upward push from the floor must be smaller than the downward weight.
Taking downwards as positive, $m g - R = m a$ gives $R = m (g - a)$ , which is less than $m g$ whenever the downward acceleration $a$ is positive.
True or False?
You can assume that the force between a lift and the load on its floor is equal to the weight of the load.
False.
They are equal only when the lift is not accelerating, that is when it is at rest or moving at constant velocity.
While the lift accelerates upwards the force is greater than the weight, and while it accelerates downwards it is less.
The weight never changes: it is $m g$ throughout, and it is the reaction that varies, because that is the force which has to supply the acceleration.
Why can a pulley system not be treated as a single particle, when a car and trailer can be?
Because the two particles move in different directions: one goes up while the other goes down, or one moves horizontally while the other falls.
Treating bodies as one particle only works when they all move the same way, so that a single acceleration in a single direction describes the whole system.
In a pulley problem there is no such direction, so each particle must be given its own equation of motion.
What does a pulley do in a mechanics problem, and how is a peg different?
A pulley lets an inextensible string change direction without changing what it does to the particles at its ends, so one particle can move horizontally while another moves vertically.
A peg is a fixed point that a string passes over or a particle hangs from, like a nail in a wall.
It does the same job of redirecting the string but does not rotate.
Two particles hang either side of a pulley and are released. What is the relationship between their accelerations, and how do you handle it when writing equations?
The accelerations have the same magnitude, because the string does not stretch, but the particles move in opposite senses: as one rises the other falls by the same amount.
So take each particle's own direction of motion as positive when writing its equation; then both equations contain the same positive $a$ and no sign has to be tracked between them.
That is why the standard pair comes out as $T - m_{1} g = m_{1} a$ for the rising particle and $m_{2} g - T = m_{2} a$ for the falling one.
You have written $T - m_{1} g = m_{1} a$ and $m_{2} g - T = m_{2} a$ for a pulley system. What is the quickest way to find the acceleration?
Add the two equations together.
The tension appears as $+ T$ in one and $- T$ in the other, so adding eliminates it in a single step and leaves an equation in $a$ alone:
$m_{2} g - m_{1} g = (m_{1} + m_{2}) a$
Substituting that acceleration back into either original equation then gives the tension.
A block on a smooth horizontal table is connected over a pulley to a hanging mass. Why does the block's weight not appear in its equation of motion?
The block moves horizontally, and its weight acts vertically, so the weight is perpendicular to the motion and contributes nothing to the horizontal equation.
Vertically the block is not accelerating, so its weight is balanced by the normal reaction from the table.
The weight only becomes relevant to the horizontal motion if the table is rough, because friction depends on that normal reaction, which in turn depends on the weight.
How is $F = m a$ used when the force and acceleration are given as vectors?
Apply it to each component separately, as two ordinary equations in numbers: the horizontal components on their own and the vertical components on their own.
The mass is a scalar, so it multiplies both components equally and does not change any direction, which is why the acceleration always points the same way as the resultant force.
Solve the two component equations, then put the answers back into vector form if that is what the question asked for.
Fill in the blanks to complete $F = m a$ written in each of the two vector notations:
$(\begin{matrix} F_{x} \\ F_{y} \end{matrix}) =______(\begin{matrix} a_{x} \\ a_{y} \end{matrix})$
$F_{x} i + F_{y} j = m (______)$
The completed forms are:
$(\begin{matrix} F_{x} \\ F_{y} \end{matrix}) = m (\begin{matrix} a_{x} \\ a_{y} \end{matrix})$
$F_{x} i + F_{y} j = m (a_{x} i + a_{y} j)$
The mass sits outside as a multiplier in both, because it is a scalar. Reading either line across its two components gives the same pair of equations.
How is the weight of a particle written as a vector?
Weight is a force, so it is a vector; it always acts vertically downwards, so it has no horizontal component at all:
$W = - m g j N$
The minus sign is there because $j$ points upwards, so downwards is the negative $j$ direction.
Since the two directions are handled separately, weight only ever enters the vertical equation.
True or False?
You can assume that weight has to be included in every two-dimensional $F = m a$ problem.
False.
Most two-dimensional vector problems are set in a horizontal plane, viewed from above: a boat on water, a puck on ice, a ball on a table.
In those, the two modelled directions are both horizontal, and gravity acts perpendicular to the whole plane, in a third direction that is not part of the model.
The weight therefore never appears; it is included only when one of the two directions is genuinely vertical.
For a particle in equilibrium the forces acting on it add to zero. What do they add to when the particle is accelerating?
They add to $m a$ : the mass multiplied by the acceleration vector.
Equilibrium is the special case of this in which the acceleration is zero, so the forces add to the zero vector.
That gives you the same technique in both cases: to find an unknown force, work out what the forces must total, then subtract the forces you already know from it.

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