What does Newton's third law say about two bodies that act on each other?

The force the first body exerts on the second is equal in magnitude and opposite in direction to the force the second exerts on the first. The two forces act on different bodies, which is why they never cancel each other out on a single force diagram.

True or False? The resultant force on a moving object always acts in the direction the object is moving.

False. The resultant force acts in the direction of the acceleration , which is not always the direction of motion. A car braking while travelling forwards has a resultant force acting backwards , which is exactly why it slows down.

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.

True or False? The reaction force on a load standing in a moving lift is always equal to the load's weight.

False. The two are equal only when the lift has zero acceleration , whether it is at rest or moving at a constant speed. While the lift is accelerating the load must have a resultant force acting on it, so the reaction and the weight cannot be equal.

A lift is accelerating downwards. Is the reaction force on a load on its floor greater or less than the load's weight?

The reaction force is less than the weight. To accelerate downwards the load needs a resultant force acting downwards, and the only downward force on it is its weight, so the weight has to be the larger of the two.

A pallet of bricks is being raised by a crane. Why is this the same problem as a person standing in a lift?

The force structure is identical: the bricks rest on a platform, the platform hangs from a cable, and the two accelerate together. The crane cable plays the part of the lift cable and the reaction between bricks and platform plays the part of the lift floor, so exactly the same equations apply.

Two particles are connected by a string over a pulley. Why is treating them as a single particle not the right approach here?

Because the two particles move in different directions , so there is no single direction in which the whole system can be said to accelerate. Write a separate equation of motion for each particle instead, taking each one's own direction of motion as positive.

In a pulley problem, must the two particles always move in opposite directions?

No : they need only move in different directions. When both particles hang from the pulley they do move in opposite directions, but when one sits on a table it moves horizontally while the other falls vertically , so those two directions are at right angles.

A block on a smooth horizontal table is pulled by a string passing over a pulley at the table's edge. Why does the block's weight not appear in its equation of motion?

Because the block moves horizontally , while its weight and the normal reaction both act vertically , at right angles to the motion. Only forces with a component along the direction of motion enter that equation, which here leaves just the tension .

What is a peg , and how does it differ from a pulley?

A peg is a fixed point, such as a nail in a wall, that a particle can be suspended from or that a string can pass over. Unlike a pulley it does not rotate, but both are modelled as smooth, so a peg behaves in exactly the same way when the equations are written.

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

Newton's Second Law (F=ma) (AQA AS Maths: Mechanics): Flashcards

Exam code: 7356

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FrontF = ma
Fill in the missing units in Newton's second law:
Resultant force in $______$ equals mass in $______$ times acceleration in $______$ .

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Cards in this collection (34)

Fill in the missing units in Newton's second law:
Resultant force in $______$ equals mass in $______$ times acceleration in $______$ .
Resultant force in newtons equals mass in kilograms times acceleration in ${m s}^{- 2}$ .
A mass given in tonnes must therefore be converted first: 5 tonnes is $5000$ kg, and using 5 would make the force a thousand times too small.
A question calls the friction on a block $F$ N. Why can this $F$ not simply be substituted into $F = m a$ ?
Because the $F$ in $F = m a$ means the resultant force, the total of all the forces acting on the block, not any single one of them.
Friction is only one of those forces, so writing the resultant as $F_{net}$ instead keeps the two meanings apart.
What does Newton's third law say about two bodies that act on each other?
The force the first body exerts on the second is equal in magnitude and opposite in direction to the force the second exerts on the first.
The two forces act on different bodies, which is why they never cancel each other out on a single force diagram.
True or False?
The resultant force on a moving object always acts in the direction the object is moving.
False.
The resultant force acts in the direction of the acceleration, which is not always the direction of motion.
A car braking while travelling forwards has a resultant force acting backwards, which is exactly why it slows down.
How can you tell whether a problem needs $F = m a$ or one of the constant acceleration formulae?
Look for force and mass: neither of them appears in any of the constant acceleration formulae, so a problem mentioning either one needs $F = m a$ .
Acceleration appears in both, which is what lets you use one to find $a$ and then feed that value into the other.
An object of mass $m$ kg falls with its weight as the only force acting on it. What does $F = m a$ give?
It gives $W = m g$ , the formula for weight.
The only force is the weight $W$ , and with downwards taken as positive the acceleration is $g$ , so substituting straight into $F = m a$ produces $W = m g$ .
How does $F = m a$ describe a particle that is in equilibrium?
Equilibrium is the case $a = 0$ , which makes the right-hand side zero and so gives a zero resultant force.
Equilibrium is therefore not a separate rule at all: it is Newton's second law with no acceleration.
A train of mass 5000 kg accelerates at $0.9375 {m s}^{- 2}$ under a driving force of 6250 N. Find the total resistive force.
The total resistive force is 1562.5 N.
The resultant is $F_{net} = 5000 \times 0.9375 = 4687.5$ N, and the resultant is the driving force minus the resistance, so the resistance is $6250 - 4687.5 = 1562.5$ N.
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.
Which two forces act on a crate resting on the floor of a lift?
Its weight $m g$ N acting downwards, and the normal reaction $R$ N from the lift floor acting upwards.
The tension in the lift cable acts on the lift, not on the crate, so it does not appear on the crate's own diagram at all.
True or False?
The reaction force on a load standing in a moving lift is always equal to the load's weight.
False.
The two are equal only when the lift has zero acceleration, whether it is at rest or moving at a constant speed.
While the lift is accelerating the load must have a resultant force acting on it, so the reaction and the weight cannot be equal.
A lift is accelerating downwards. Is the reaction force on a load on its floor greater or less than the load's weight?
The reaction force is less than the weight.
To accelerate downwards the load needs a resultant force acting downwards, and the only downward force on it is its weight, so the weight has to be the larger of the two.
The lift floor pushes up on a load with a force of $R$ N. What force does the load exert on the lift, and where does it show up?
The load pushes down on the lift floor with a force of exactly the same size, $R$ N, by Newton's third law.
That is why $R$ appears upwards in the load's equation of motion but downwards in the lift's, alongside the lift's own weight.
Fill in the two missing terms in the equation of motion for a load of mass $m$ kg on the floor of a lift, taking downwards as positive:
$m g -______=______$
The completed equation is:
$m g - R = m a$
The weight acts downwards and the reaction from the floor acts upwards, so their difference is the resultant force that produces the load's acceleration.
A load of mass $m$ kg rests on the floor of a lift accelerating downwards at $0.3 {m s}^{- 2}$ , and the reaction on it is 399 N. Find $m$ , taking $g = 9.8$ .
The mass of the load is $m = 42$ kg.
Taking downwards as positive, the load's equation is $m g - 399 = 0.3 m$ , so $m (9.8 - 0.3) = 399$ and $m = \frac{399}{9.5} = 42$ .
A pallet of bricks is being raised by a crane. Why is this the same problem as a person standing in a lift?
The force structure is identical: the bricks rest on a platform, the platform hangs from a cable, and the two accelerate together.
The crane cable plays the part of the lift cable and the reaction between bricks and platform plays the part of the lift floor, so exactly the same equations apply.
Two particles are connected by a string over a pulley. Why is treating them as a single particle not the right approach here?
Because the two particles move in different directions, so there is no single direction in which the whole system can be said to accelerate.
Write a separate equation of motion for each particle instead, taking each one's own direction of motion as positive.
A pulley system has one particle moving up and another moving down. How is the positive direction chosen for each equation?
Each particle takes its own direction of motion as positive, so one equation is written upwards and the other downwards.
That is what allows both equations to use the same acceleration $a$ , even though the particles are travelling in opposite directions.
True or False?
A particle hanging from a string that passes over a pulley to a second particle falls with acceleration $g$ .
False.
As well as its weight, the particle has the tension in the string pulling upwards on it, so the resultant force on it is smaller than its weight.
Its acceleration is therefore always less than $g$ , and it comes closer to $g$ the heavier the falling particle is compared with the one it is pulling.
In a pulley problem, must the two particles always move in opposite directions?
No: they need only move in different directions.
When both particles hang from the pulley they do move in opposite directions, but when one sits on a table it moves horizontally while the other falls vertically, so those two directions are at right angles.
A block on a smooth horizontal table is pulled by a string passing over a pulley at the table's edge. Why does the block's weight not appear in its equation of motion?
Because the block moves horizontally, while its weight and the normal reaction both act vertically, at right angles to the motion.
Only forces with a component along the direction of motion enter that equation, which here leaves just the tension.
What is a peg, and how does it differ from a pulley?
A peg is a fixed point, such as a nail in a wall, that a particle can be suspended from or that a string can pass over.
Unlike a pulley it does not rotate, but both are modelled as smooth, so a peg behaves in exactly the same way when the equations are written.
A 25 kg block on a smooth table is connected over a pulley to a 60 kg mass hanging freely. Find the acceleration, taking $g = 9.8$ .
The acceleration is $6.9 {m s}^{- 2}$ , to 2 significant figures.
The block gives $T = 25 a$ and the hanging mass gives $60 g - T = 60 a$ ; substituting removes $T$ and leaves $85 a = 60 g$ , so $a = \frac{60 \times 9.8}{85} = 6.917 \dots$
In $F = m a$ , how are the directions of $F$ and $a$ related?
They are always the same: the acceleration is parallel to the resultant force and points the same way along it.
Multiplying by the mass changes only the size of the vector and never its direction, because a mass is a positive scalar.
Fill in the missing entries in $F = m a$ written with column vectors:
$(\begin{matrix} F_{x} \\ F_{y} \end{matrix}) = m (\begin{matrix} ______ \\ ______ \end{matrix})$
The completed equation is:
$(\begin{matrix} F_{x} \\ F_{y} \end{matrix}) = m (\begin{matrix} a_{x} \\ a_{y} \end{matrix})$
The same relationship written in $i$ , $j$ notation is $F_{x} i + F_{y} j = m (a_{x} i + a_{y} j)$ .
How do you use $F = m a$ when the force and the acceleration are given in $i$ , $j$ form?
Treat the $i$ and $j$ parts as two separate ordinary equations, applying $F = m a$ to each component on its own.
Solve those two equations independently, then put the two answers back together into a single vector.
True or False?
In many two-dimensional force problems using $i$ and $j$ , the weight of the particle appears in neither component equation.
True.
Most such problems are set on a horizontal plane seen from above, like a snooker table, so $i$ and $j$ are two horizontal directions.
Weight acts vertically, which is a third direction the model is not using at all, so it never enters either equation.
A problem uses $i$ horizontally and $j$ vertically upwards. Write the weight of a particle of mass $m$ kg as a vector.
The weight is $W = - m g j$ N.
Weight always acts vertically downwards, so it has no $i$ component at all, and the minus sign is what places it opposite to $j$ .
How are the units written on a force given in $i$ , $j$ form?
The unit is written once, outside the bracket: $(3 i - 5 j)$ N.
Each component is already a number of newtons, so repeating the N inside the bracket would be saying the same thing twice.
A yacht of mass 3500 kg has acceleration $(0.8 i + 0.3 j) {m s}^{- 2}$ and meets a water resistance of $(- 14 i - 19 j)$ N. Find the driving force $D$ .
The driving force is $D = (2814 i + 1069 j)$ N.
Applying $F = m a$ to each component gives $D_{x} - 14 = 3500 \times 0.8$ and $D_{y} - 19 = 3500 \times 0.3$ , so $D_{x} = 2814$ and $D_{y} = 1069$ .

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