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2.2Newtonâs laws of motion
Newtonâs first law of motion
Suppose you have two identical train carriages. Both are equipped with all the apparatus you need to do physics experiments. One train carriage is at rest at the train station. The other moves in a straight line with constant speedâthe ride is perfectly smooth, there are no bumps, there is no noise and there are no windows to look outside. Every physics experiment conducted in the train at rest will give identical results to similar experiments made in the moving train. We have no way of determining whether a carriage is âreally at restâ or âreally movingâ. We find it perfectly natural to believe, correctly, that no net force is present in the case of the carriage at rest. Therefore no net force is required in the case of the carriage moving in a straight line with constant speed.
Newtonâs first law of motion states that:
KEY POINT
When the net force on a body is zero, the body will move with constant velocity (which may be zero); in other words, it will move on a straight line with constant speed (which may be zero).
In effect, Newtonâs first law defines what a force is. A force is what changes a bodyâs velocity. A force is not what is required to keep something moving as Aristotle thought.
Using the law in reverse allows us to conclude that if a body is not moving with constant velocity (which may mean not moving in a straight line, or not moving with constant speed or both) then a force must be acting on the body. So, since the earth revolves around the sun we know that a force must be acting on the earth.
Newtonâs first law is also called the law of inertia. Inertia is what keeps the body in the same state of motion when no forces act on the body. When a car accelerates forward, the passengers are thrown back into their seats because their original state of motion was motion with low speed. If a car brakes abruptly, the passengers are thrown forward, Figure 2.31. This implies that a mass tends to stay in the state of motion it was in before the force acted on it. The reaction of a body to a change in its state of motion (acceleration) is inertia.

Figure 2.31: The car was originally travelling at high speed. When it hits the wall the car stops, but the passenger wants to stay in the original high speed state of motion. This results in him hitting the steering wheel and the windshield (which is why it is a good idea to have safety belts and air bags).

Newtonâs third law of motion
Newtonâs third law of motion states that:
KEY POINT
If body X exerts a force on body Y, then body Y will exert an equal and opposite force on body X.
Make sure you understand that these equal and opposite forces act on different bodies. Thus, you cannot use this law to claim that it is impossible to ever have a net force on a body because for every force on it there is also an equal and opposite force. Here are four examples of this law:
1You stand on roller-skates facing a wall. You push on the wall, and you move away from it. This is because you exerted a force on the wall, and in turn the wall exerted an equal and opposite force on you, making you move away (Figure 2.32).
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Figure 2.32: The girl pushes on the wall, so the wall pushes on her in the opposite direction.

2You step on a scale. The scale exerts an upward force on you, and so you exert a downward force on the scale. This is the force that the scale reads (Figure 2.33).

Figure 2.33: The familiar bathroom scale does not measure mass. It measures the force that you exert on it. This force is the weight only when the scale is at rest.

3A helicopter hovers in air, Figure 2.34. Its rotors exert a force downward on the air. Thus, the air exerts the upward force on the helicopter that keeps it from falling.

Figure 2.34: The upward force on the rotor is due to the force the rotor exerts on the air downward.

4A book of mass 2 kg is allowed to fall feely. The earth exerts a force on the book, namely the weight of the book of about 20 N. Thus, the book exerts an equal and opposite force on the earthâa force upward equal to 20 N.
You must be careful with situations with two forces that are equal and opposite; they do not always have to do with the third law. For example, a block of mass 3 kg resting on a horizontal table has two forces acting on it. Its weight of about 30 N, and the normal force from the table that is also 30 N. These two forces are equal and opposite, but they are acting on the same body and so have nothing to do with Newtonâs third law. (The force that pairs with the weight of the block is an upward force on the earth, and the one that pairs with the normal force is a downward force on the table.)
Newtonâs third law also applies to cases where there is no contact between the bodies. Examples are the electric force between two electrically charged particles or the gravitational force between any two massive particles. These forces must be equal and opposite. (See Figure 2.35.)

Figure 2.35: The two charges and the two masses are different, but the forces they exert on each other are equal and opposite.

SCIENCE IN CONTEXT
Helicopters, planes and rockets function because of Newtonâs third law.
Equilibrium
When there are angles between the various forces, solving equilibrium problems is a bit more cumbersome and will involve getting components of forces using vector methods. We choose a set of axes whose origin is the body in question and find the components of all the forces on the body. Figure 2.36 shows three forces acting at the same point. We have equilibrium, which means the net force acting at the point is zero. We need to find the unknown magnitude and direction of force F1. This situation could represent three people pulling on three ropes that are tied at a point.
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Figure 2.36: Force diagram of three forces in equilibrium pulling a common point. Notice that the three vectors representing the three forces form a triangle.

Getting components along the horizontal (x) and vertical (y) directions for the known forces, we have
F2x=0
F2y=â22.0Â NÂ (put minus sign by hand)
F3x=â29.0cos37Â°=â23.16Â NÂ (put minus sign by hand)
F3y=29.0sin37Â°=17.45Â N
Equilibrium demands that âFx=0 and âFy=0.
âFx=0Â impliesÂ F1.x+0â23.16=0âF1x=23.16Â N
âFy=0Â impliesÂ F1yâ22+17.45âF1y=4.55Â N
Therefore, F1=23.162+4.552ââââââââââââ=23.6Â N. The angle is found from tanÎ¸=F1yF1xâÎ¸=tanâ1(4.5523.16)=11.1Â°.
EXAM TIP
If we know the x and y components of a force, we can find the magnitude of the force from F=F2x+F2yââââââââ
SCIENCE IN CONTEXT

Figure 2.37: The temple of Hephaestus in Athens.

The temple of Hephaestus in Athens (see Figure 2.37) still stands after 2500 years because it was built according to the laws of equilibrium.
WORKED EXAMPLE 2.6
A body of weight 98.0 N hangs from two strings that are attached to the ceiling as shown in Figure 2.38. Determine the tension in each string.

Figure 2.38: For worked example 2.6.

Answer
The three forces acting on the body are as shown, with T and S being the tensions in the two strings. Taking components about horizontal and vertical axes through the body we find:
Tx=âTcos30Â°Ty=Tsin30Â°Sx=Scos50Â°Sy=Ssin50Â°Wx=0Wy=â98.0Â N
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Equilibrium demands âFx=0 and âFy=0:
âFx=0Â impliesÂ âTcos30Â°+Scos50Â°=0Î£Fy=0Â impliesÂ Tsin30Â°+Ssin50Â°â98=0
From the first equation we find that S=Tcos30Â°cos50Â°=1.3473ÃT. Substituting this in the second equation gives
T(sin30Â°+1.3473sin50Â°)=98
which solves to =63.96â64.0Â N. Hence S=1.3473Ã63.96=86.17â86.2Â N.
WORKED EXAMPLE 2.7
A mass of 125 g is attached to a spring of spring constant k = 58 N mâ1 that is hanging vertically.
aFind the extension of the spring.
bIf the mass and the spring are placed on the moon, will there be any change in the extension of the spring?
Answer
aThe forces on the hanging mass are its weight and the tension of the spring. Since we have equilibrium, the two forces are equal in magnitude. Therefore
kxx=mg=mgk=0.125Ã9.8158=0.021Â m=2.1Â cm
bThe extension will be less, since the acceleration of gravity is less.
Newtonâs second law of motion
Newtonâs second law of motion states that:
KEY POINT
The net force on a body of constant mass is proportional to that bodyâs acceleration and is in the same direction as the acceleration.
Mathematically
FâÂ =maâÂ 
where the constant of proportionality, m, is the mass of the body.
Figure 2.39 shows the net force on a freely falling body, which happens to be its weight, W = mg. By Newtonâs second law, the net force equals the mass times the acceleration, and so
mga=ma=g
The acceleration of the freely falling body is exactly g. Experiments going back to Galileo show that indeed all bodies fall with the same acceleration in a vacuum irrespective of their density, their mass, their shape and the material from which they are made. Look for David Scottâs demonstration dropping a hammer and a feather on the moon in Apollo 15âs mission in 1971. You can do the same demonstration without going to the moon by placing a hammer and a feather on a book and dropping the book. If the heavy and the light object fell with different accelerations, the one with the smaller acceleration would lift off the bookâbut it doesnât.

Figure 2.39: A mass falling to the ground acted upon by gravity.

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The equation F = ma defines the unit of force, the newton (symbol N). One newton is the force required to accelerate a mass of 1 kg by 1 m sâ2 in the direction of the force.
It is important to realise that the force in the second law is the net force âF on the body.
EXAM TIP
How to solve an F = ma problem:
Step 1: Make a diagram and choose the direction that you will take as positive.
Step 2: Make a free-body diagram.
Step 3: Find the net force by looking at the free-body diagram.
Step 4: Set the net force equal to ma.
Step 5: Solve for what is unknown.
Letâs apply the Exam tip steps to a simple problem (Figure 2.40): a block of mass 5.0 kg is accelerated to the right on a horizontal road by a horizontal force F. The acceleration is 0.80 m sâ2. A frictional force of 12 N acts on the block. What is the magnitude of F?

Figure 2.40: The diagram shows the horizontal forces acting on the block.

In step 3 we find the net force by looking at the free-body diagram: it is simply F â 12. (Because the positive direction is to the right.)
In step 4 we equate the net force to ma: F â 12 = 5.0 Ã 0.8.
Finally, in step 5 we solve for the only unknown quantity, F: F â 12 = 4.0 â F = 16 N.
CHECK YOURSELF 5
An elevator moving upwards is slowing down to a stop. Which diagram in Figure 2.41âA, B or Câshows the forces on the elevator?

Figure 2.41: For check yourself question 5.

WORKED EXAMPLE 2.8
A person of mass m = 70 kg stands on the floor of an elevator. Find the normal force they experience from the elevator floor when the elevator:
ais standing still
bmoves up at constant speed 3.0 m sâ1
cmoves up with acceleration 4.0 m sâ2
dmoves down with acceleration 4.0 m sâ2.
For simplicity, take g = 10 m sâ2.
Answer
Two forces act on the person: their weight mg vertically down and the normal force R from the floor vertically up.
aThere is no acceleration, and so by Newtonâs second law the net force on the person must be zero. Hence
R=mg=7.0Ã102Â N
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bThere is no acceleration and so again
R=mg=7.0Ã102Â N
cTake the up direction to be positive; then the net force is âF=RupâmgÂ downÂ .
So R â mg = ma and
R=mg+ma=700Â N+280Â N=9.8Ã102Â N
dTake the down direction to be positive: âF=mgâR. So
mgâR=ma and
R=mgâma=700Â Nâ280Â N=4.2Ã102Â N
WORKED EXAMPLE 2.9
A man of mass 70 kg is standing in an elevator. The elevator is moving upward at a speed of 3.0 m sâ1. The elevator comes to rest in a time of 2.0 s. Determine the normal force on the man from the elevator floor during the period of deceleration. Repeat for the case where the elevator goes down and then stops.
Answer
Let us take the up direction to be positive. The acceleration is
a=vâut=0â3.02.0=â1.5Â mÂ sâ2
So, the net force is
âF=Râmg
Hence:
RRâmg=ma=ma+mg=70Ã(â1.5)+70Ã10=â105+700=595Â Nâ6.0Ã102Â N
What if, instead, the man was moving downward and then decelerated to rest? Take the up direction to be positive again. The net force is âF=Râmg and a=vâut=0â(â3.0)2.0=+1.5Â m sâ2. So,
Râmg=ma
R=ma+mg=105+700=805Â Nâ8.0Ã102Â N
Both cases are easily experienced in daily life. When the elevator goes up and then stops we feel âlighterâ during the deceleration period. When going down and about to stop, we feel âheavierâ during the deceleration period. The feeling of âlightnessâ or âheavinessâ has to do with the normal force we experience from the floor.
The inclined plane
Many problems in mechanics involve an inclined plane. Figure 2.42 shows a block at rest on an inclined plane that makes an angle Î¸ to the horizontal.

Figure 2.42: A block at rest on an inclined plane.

The free-body diagram shows the three forces acting on the block (the block is now shown as a point; it is convenient to always do that so that the forces are shown acting at the same point). The forces are the normal force R at right angles to the plane, the weight mg vertically down and a frictional force f acting up the plane. We have also drawn a set of x and y axes. The x-axis is chosen to be along the inclined plane. This is convenient because if the block were to slide down, it would move along the x axis. In other words it is convenient to choose an axis along which motion takes or could take place.
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We now need to find the components of the weight along the two axes. Along the x-axis the component is mg sin Î¸ and along the y-axis it is mg cos Î¸. Since we have equilibrium we have two equations:
f=mgsinÎ¸Â andÂ R=mgcosÎ¸
Now suppose that the largest angle of the inclined plane at which we can have equilibrium is Î¸max. In other words if the angle of the inclined plane were to increase ever so slightly above Î¸max, the block would slide down. This means that at this angle the frictional force is the largest possible and so is given by fmax = Î¼s R, i.e., fmax = Î¼s mg cos Î¸max. So our first equilibrium equation becomes
Î¼smgcosÎ¸max=mgsinÎ¸max
And so
Î¼s=tanÎ¸max
Consider now the case where the inclined plane is frictionless. The normal force is still R = mg cos Î¸, but along the x-direction we have an unbalanced force mg sin Î¸, and so the block will accelerate down the plane. The acceleration is found from Newtonâs second law:
mgsinÎ¸=maâa=gsinÎ¸.
CHECK YOURSELF 6
A block of mass 2.0 kg slides down a frictionless inclined plane with acceleration 4.0 m sâ2. What is the acceleration of a 5.0 kg block sliding down the same plane?
Cases with more than one body present
We now examine cases where more than one body are present. We simply have to apply the five steps of the previous exam tip to each body in the problem. So consider the following situation:
Two blocks of mass 4.0 and 6.0 kg are joined by a string and rest on a frictionless horizontal table (see Figure 2.43). A force of 100 N is applied horizontally on one of the blocks. We want to find the acceleration of each block and the tension of the string joining the two blocks.

Figure 2.43: A force pulling two blocks joined by a string.

Figure A2.43 shows the forces on each block. The net force on the 6.0 kg block is 100 â T and on the 4.0 kg mass just T. Physics strings are assumed to be inextensible, which means that the two bodies move with the same acceleration. Thus, applying Newtonâs second law separately on each block,
100âTT=6a=4a
We have to unknown quantities and two equations. Solving for a (by adding the two equations side by side or using your calculator) gives a = 10 m sâ2. The tension is
T=4.0Ã10=40Â N.
Note that the free-body force diagram makes it clear that the 100 N force acts only on the body to the right. It is a common mistake to say that the body to the left is also acted upon by the 100 N force.
However, in problems with more than one body it is often a simplification to treat the bodies as one. In our problem here, we may consider the two bodies as one of mass 10 kg. The net force on the body is 100 N. Note that the tensions are irrelevant now since they cancel out. (They have become internal forces, and these are irrelevant.) Applying Newtonâs second law on the single body we have 100 = 10a â a = 10 m sâ2. But to find the tension we must break up the combined body into the original two bodies. Newtonâs second law on the 4.0 kg body gives T = 4a = 40 N.
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CHECK YOURSELF 7
Two blocks of mass 3 kg and 5 kg are joined by a string and are allowed to fall freely, as shown in Figure 2.44.
What is the tension in the string?

Figure 2.44: For check yourself question 7.

WORKED EXAMPLE 2.10
(Atwoodâs machine) Two blocks of m = 4.0 kg and M = 6.0 kg are joined together by a string that passes over a pulley. The blocks are held stationary and then released. Determine the acceleration of each block.
Answer
Intuition tells us that the larger mass will start moving downward and the small mass will go up. The acceleration of each block will be the same in magnitude since the string is assumed to be inextensible.
Method 1
Figure 2.45 shows the free-body diagrams: weight mg and tension T on m and weight Mg and tension T on M.

Figure 2.45: For worked example 2.10.

Newtonâs second law applied to each mass states
Tâmg=ma (1)
MgâT=Ma (2)
Note these equations carefully. Each says that the net force on the mass in question is equal to that mass times that massâs acceleration. In the first equation we find the net force in the upward direction, because that is the direction of acceleration. In the second we find the net force in the downward direction, since that is the direction of acceleration in that case. We want to find the acceleration, so we simply add up these two equations side by side to find
Mgâmg=(m+M)a
Hence
a=MâmM+mg
(Note that if M â« m, the acceleration tends to g (because in that case we can neglect m compared to M). This shows clearly that if the two masses are equal then there is no acceleration. This is a convenient method for measuring g: Atwoodâs machine effectively âslows downâ g so the falling mass has a much smaller acceleration from which g can then be determined. Putting in the numbers for our example we find a = 2.0 m sâ2. Having found the acceleration we may, if we wish, also find the tension in the string, T. Putting the value for a in formula (1) we find
T=mMâmM+mg+mg=2MmM+mg=48Â N
(If M â« m, the tension tends to 2mg (we can neglect m compared to M in the denominator in which case M cancels out).
Method 2
We treat the two masses as one body and apply Newtonâs second law on this body (but this is trickier than in the previous example)âsee Figure 2.46.
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Figure 2.46: The forces on the system of the two blocks.

In this case the net force is Mg â mg and, since this force acts on a body of mass M + m, the acceleration is found as before from F = mass Ã acceleration. Note that the tension T does not appear in this case, being now an internal force.
WORKED EXAMPLE 2.11
In Figure 2.47, a block of mass M is connected to a smaller block of mass m through a string that goes over a pulley. Ignoring friction, find the acceleration of each block and the tension in the string.

Figure 2.47: For worked example 2.11.

Answer
Method 1
The forces are shown in Figure 2.47. Convince yourself that you agree with this diagram. Thus
mgâT=ma
T=Ma
from which (adding the two equations side by side)
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a=mgm+M
(If M â« m the acceleration tends to zero.) If M = 8.0 kg and m = 2.0 kg, this gives a = 2.0 m sâ2.
HenceÂ T=Mmgm+M=16Â N
Method 2
Treating the two bodies as one results in the situation shown in Figure 2.48.

Figure 2.48: The forces on the two blocks taken as one.

The net force on the mass M + m is mg. Hence mg = (M + m)a
mg=(M+m)a
âa=mgm+M
The tension can then be found as before.
WORKED EXAMPLE 2.12
A block of 4.0 kg is placed on top of a block of mass 6.0 kg, as shown in Figure 2.49. The blocks are placed on a frictionless horizontal surface. A horizontal force of magnitude F is applied to the bottom block. The static coefficient of friction between the blocks is 0.60. What is the largest force F so that both blocks move together without sliding on each other?

Figure 2.49: Free-body diagram for two blocks, one on top of the other and moving together.

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Answer
As the bottom block accelerates to the right, because of inertia, the top block wants to stay where it is, i.e., slide backwards relative to the bottom block. Thus a friction force will develop that will push the top block to the right. This gives the free-body diagrams in Figure 2.49. Forces in blue and red represent pairs according to Newtonâs third law.
Since we want the maximum value of F we will have to use the maximum frictional force that can develop between the two blocks, and that is fmax = Î¼s N = 0.60 Ã 40 = 24 N. Applying Newtonâs second law to each body,
Fâ2424=6a=4a
This means that a = 6.0 m sâ2 and so F = 24 + 36 = 60 N.
WORKED EXAMPLE 2.13
The force F in the previous example is 80 N. What is the acceleration of each block? The coefficient of dynamic friction between the blocks is 0.50.
Answer
We know that there will be sliding since the force is now larger than the maximum of 60 N found in the previous example. The force of friction will now be f = Î¼d N = 0.50 Ã 40 = 20 N. Thus,
80â20=6a1âa1=10Â mÂ sâ2
20=4a2âa2=5.0Â mÂ sâ2
Both blocks are accelerating forward, but relative to the bottom block the top block is sliding backward.
Links
Newtonâs second law (for a particle of constant mass) is written as F = ma. In this form, this equation does not seem particularly powerful. However, if we realise that in calculus acceleration is a=dvdt=d2xdt2 the second law becomes d2xdt2âFm=0. This is a differential equation. Given the dependence of the force on time or position, the equation can be solved to give the actual path that the particle will move on under the action of the force. For example, Newton showed that if the force depends on position as F~1x2 then the motion has to be along a conic section (ellipse, circle, and so on). Similarly, if the force depends on position as F ~ x (as in a spring) then the motion will be oscillations about a fixed point. So the same equation applied to different contexts allows for the determination of the type of motion that results under the action of a force.
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TEST YOUR UNDERSTANDING
15A mass is hanging from a string that is attached to the ceiling. A second piece of string (identical to the first) hangs from the lower end of the mass (see Figure 2.50).

Figure 2.50: For question 15.

State and explain which string will break if:
athe bottom string is slowly pulled with ever increasing force
bthe bottom string is very abruptly pulled down.
16A block of mass 15.0 kg rests on a horizontal table. A force of 50.0 N is applied vertically downward on the block. Calculate the force that the block exerts on the table.
17A block of mass M = 12 kg is connected with a string to a smaller block of mass m = 4.0 kg. The big block is resting on a smooth inclined plane as shown in Figure 2.51. Determine the angle Î¸ of the plane so that neither mass moves.

Figure 2.51: For question 17.

18Describe under what circumstances a constant force would result in:
aan increasing acceleration on a body
ba decreasing acceleration on a body.
19A block of mass 2.6 kg is projected with initial speed 18 m sâ1 along a rough horizontal plane. The dynamic coefficient of friction between the block and the plane is 0.35.
aAfter how much time will the block stop?
bWhat distance will the block cover?
cWithout further calculations, state how the answers to a and b would change, if at all, when the experiment is repeated with a block of double the mass.
20A block of mass 4.2 kg is projected with speed 14 m sâ1 up an inclined plane making an angle 28Â° with the horizontal. The dynamic coefficient of friction between the block and the inclined plane is 0.42, and the static coefficient is 0.55.
aAfter how much time will the block stop?
bWhat distance will the block cover?
cWhen the block stops, will it then slide down the plane, or will it stay at rest?
dIn another experiment the same block is projected down the plane with the same speed as before. Will the block stop?
21A man of mass m stands in an elevator.
aFind the normal force from the elevator floor on the man when:
ithe elevator is standing still
iithe elevator moves up at constant speed v
iiithe elevator accelerates down with acceleration a
ivthe elevator accelerates down with acceleration a = g.
bWhat happens when a > g?
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22Get in an elevator and stretch out your arm holding your heavy physics book. Press the button to go up. Describe and explain what is happening to your stretched arm. Repeat as the elevator comes to a stop at the top floor. What happens when you press the button to go down, and what happens when the elevator again stops? Explain your observations carefully using the second law of motion.
23Figure 2.52 shows a person in an elevator pulling on a rope that goes over a pulley and is attached to the top of the elevator. The mass of the elevator is 30.0 kg and that of the person is 70.0 kg. (g = 10 m sâ2).
aOn a copy of the diagram, draw the forces on the person.
bDraw the forces on the elevator.
cThe elevator accelerates upwards at 0.50 m sâ2. Find the normal force on the person from the elevator floor.
dThe force the person exerts on the elevator floor is 200 N. Find the acceleration of the elevator.

Figure 2.52: For question 23.

24Two blocks X (mass 2.0 kg) and Y (mass 6.0 kg) are in contact on a frictionless horizontal table. A force of 24 N acts horizontally on block X as shown in Figure 2.53.

Figure 2.53: For question 24.

aWhat is the acceleration of each block?
bDetermine the force X exerts on Y.
25aCalculate the tension in the string joining the two masses in Figure 2.54.
bIf the position of the masses is interchanged, will the tension change?

Figure 2.54: For question 25.

26A mass of 3.0 kg is acted upon by three forces of 4.0 N, 6.0 N and 9.0 N and is in equilibrium. Convince yourself that these forces can indeed be in equilibrium. The 9.0 N force is suddenly removed. Determine the acceleration of the mass.
27A block of mass M = 15 kg is connected with a string to a smaller block of mass m. The big block is on a rough inclined plane as shown in Figure 2.55. The coefficient of static friction is 0.40. The angle of the plane is Î¸ = 25Â°. Determine the largest and smallest mass m that can hang from the string and still have equilibrium.

Figure 2.55: For question 27.

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28A block of mass 25 kg and density 2800 kg mâ3 is fully submerged in a liquid of density 850 kg mâ3. The block hangs from a vertical string. What is the tension in the string?
29A rectangular block of density 750 kg mâ3 floats in water of density 1000 kg mâ3.
aWhat fraction of the vertical height of the block is under water?
bThe block is pushed downward so it is fully submerged and is then released. What is the initial acceleration of the block?
30A rectangular raft of mass 1200 kg and density 800 kg mâ3 floats in water of density 1000 kg mâ3. How many people of average mass 50 kg can stand on the raft so that it is just about to be fully submerged?
31A ball of density 3500 kg mâ3 and radius 6.0 mm is released from rest in a liquid of density 920 kg mâ3 and viscosity 0.70 Pa s.
aWhat is the terminal speed of the ball as it falls through the liquid? Ignore the buoyant force.
bHow do you expect the answer to a to change if the liquid temperature is increased?
cRepeat a taking the buoyant force into account.
