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www.furthermaths.org.uk
The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
Cymru gan
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
REVISION SHEET – M1 (WJEC)
Dynamics of a particle, Newton’s Laws of motion
Forces acting on a mass initiating motion Forces parallel to the direction of motion It is important to remember that the motion of a mass acted upon by a force can be modeled by Newton’s 2
nd
Law of Motion, the well-known F=ma, where F is the force, m the mass of the object and a is its
acceleration.
The following diagrams will make it clear what is the F in Newton’s 2nd
Law. Imagine just one single force acting on this car causing it to accelerate. Then the equation of motion of this
car is P = ma, as P is the force in the direction of motion.
P = ma
Before the exam you should know
• How to draw diagram of forces such as weight, friction, normal reaction, tension and thrust acting
on a mass
• Use Newton’s 2nd
Law of motion (F = ma) to write equations of motion of a moving mass.
• Solve problems involving lifts going up and down.
• Resolve forces parallel and perpendicular to an inclined plane.
• Solve problems involving motion on an incline plane.
• Clearly draw forces acting on particles connected
by a particle one of which is freely hanging and the
other freely hanging or lying on a horizontal plane
or on an incline plane.
• Find the acceleration of particles connected by an inextensible string.
• Find the tension in the string connecting two
particles.
The main ideas are
WJ
EC
Newton’s 2nd Law of motion
M1
Constant forces acting of masses M1
Lift problems M1
Motion on inclined plane M1
Connected particles M1
P
www.furthermaths.org.uk
The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
Cymru gan
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Now, imagine there is a resistance force also acting on the car and the car is moving from left to right. This
time the F in Newton’s 2nd
Law is P- R and the required equation is P – R=ma.
P- R = ma
Next, imagine there are two forces opposing the action of P as shown below. This time the F in Newton’s 2nd
Law is P – R-F and the required equation is P- R - F=ma.
P- R - F=ma.
Example 1
A tug tows a barge of mass 10000kg with acceleration 0.2ms-2
. The water resistance to the movement of the
bare is 900 N. Given that the tow rope is horizontal, calculate the tension in this rope.
Always draw a diagram showing all the forces acting of the mass. Model the mass by a simple shape as I’ve
done below.
Direction of motion
By Newton’s 2nd
Law of motion maF = . Thus 2.010000900 ×=−T , giving NT 29002000900 =+=
For the WJEC syllabus, the forces will be constant in this Section. So, the acceleration will be constant too.
Now problems could arise that require you to use the equations of motion you learnt in Section 1. Here is an
example.
Example 2
A constant retarding force of 2500N is applied to a car of mass 900 kg moving on a level road so as to reduce
its speed from 25 ms-1
to 15 ms-1
. Find, for the interval during which the car reduces speed,
(a) the time taken
(b) the distance covered.
To answer this question , you have to find the constant retardation so that you can apply the equation of
motion for uniform acceleration to solve the problem. Draw a diagram as usual.
Motion in this direction Therefore F = - 2500N
By Newton’s 2nd
Law, 2278.0
9000
250090002500
−
−=−=⇒=− msaa .
P R
P R
F
T 900
2500
a = 0.2 ms-2
a
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
Cymru gan
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Now,
?
?
278.0
15
25
2
1
1
=
=
=
=
=
−
−
−
t
s
msa
msv
msu
When forces are parallel to the direction of motion, it is fairly straightforward to for the equation of motion
as the above examples have shown.
When forces are at an angle to the direction of motion then we have to resolve the forces parallel and
perpendicular to the direction of motion.
Only forces acting in the direction of motion contribute to the equation of motion.
Forces perpendicular to the direction of motion do not contribute to the equation of motion unless frictional
forces are involved.
Forces at an angle to the direction of motion Consider the following situation:
A single force is applied to the object of mass m as shown on the left. On the right the force P has been
resolved horizontally and vertically.
Direction of motion
So the equation of motion of this mass is maPo
=35cos assuming it is moving to the right with acceleration
a.
If there are several forces at various angles, then each one has to be resolved parallel and perpendicular to the
direction of motion. The equation of motion the follows from Newton’s 2nd
Law of motion.
(a) Using atuv +=
t278.02515 −+=
st 97.35278.0
10
278.0
2515=
−
−
=
−
−
=
(b) Using tvu
s ×
+=
2
ms 4.71997.352097.352
1525=×=×
+=
a
35o
P
35o
P
Psin35o
Pcos35o
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
Cymru gan
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Consider the following situation:
Direction of motion
So the equation of motion of this mass is maPQ oo
=− 35cos50cos assuming it is moving to the right with
acceleration a.
Examples 4
Find the force T and the angle θ in the following
situation, given that a = 3 ms-2.
Once again, resolve the force T parallel and
perpendicular to the direction of motion.
Form two equations, which have to be solved
simultaneously.
3152cos ×=−θT ……(1)
147100sin =+θT ..….(2)
47cos =θT ………… ..(3)
47sin =θT ……………(4)
)3()4( ÷ gives o
451tan =⇒= θθ
Sub for θ in (3) to get NT 5.6645cos
47==
100 N
Examples 3
Find the forces R and T in the following
situation.
Resolve the force T parallel and perpendicular
to direction of motion.
Therefore, Tcos30o – 1 = 10 x 4, giving
NNT 473.4730cos
41≈==
Also, R + Tsin30o = 98 (no vertical motion)
NR 7435.7430sin3.4798 ≈=−=
30o
T m = 10 kg
1N
a = 4ms-2
98 N
R
Psin35o
35o 50
o
Q P
35o 50
o
Q P
Qcos50o Pcos35
o
Qsin50o
a
Tcos 30
o
R Tsin30o
1N
98N
Tcosθ
Tsinθ 100N
2N
147 N
θ
T m = 15 kg
2N
a 147 N
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Lift Problems Lift problems can be tackled using the same techniques you’ve used above. Remember that you have to form
equations of motion using F = ma. An example will clarify this point.
Example 5
The motion of a lift of mass 300 kg, when ascending from rest, is given in three stages.
• First, it accelerates at 0.7 ms-2
until it reaches a certain velocity.
• It then maintains this velocity for a period of time.
• Finally it slows down with retardation 0.5 ms-2.
(a) Find the tension in the lift cable while it is accelerating upwards.
(b) Find the reaction between the floor of the lift and a man of mass 90kg during each of these three stages.
Tackle each stage at a time, drawing a sketch.
Notice that during acceleration the reaction between the man and the
lift is greatest.
During deceleration, the reaction between the man and the lift is least.
At constant speed, when acceleration is zero, the reaction is in
between the greatest and least values.
(b) We look at the reaction
between the man and the
floor of the lift at constant
velocity (acceleration = 0)
N
gR
822
90
=
=
R
90g N
(b) We now look at the
reaction between the man
and the floor of the lift
during acceleration.
Resolving upward
and using F = ma
NR
R
gR
945
82263
7.09090
=
+=
×=−
2ms7.0
−
R
90g N
(a) For this part, consider the mass
of the life and the man inside it as
one mass of 390kg.
Then by F = ma
NT
T
gT
4095
3822273
7.0390390
=
+=
×=−
2ms7.0
−
T
390g N
(b) Finally, we look at the
reaction between the man and the
floor of the lift during
deceleration.
Resolving upward
and using F = ma
NR
R
gR
777
82245
5.09090
=
+−=
−×=−
2ms5.0
−
R
90g N
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Motion on inclined plane When solving problems involving motion on an incline plane, you need to resolve forces parallel and
perpendicular to the plane.
Example 6
An object of mass 8 kg is being towed by a light string up a slope incline at 20o to the horizontal. The string
is inclined at 30o to the slope. There is a resistance force of 40 N. The object is acceleration up the slope at
2ms8.0
− .
(a) Find the tension in the string
(b) Find the normal reaction exerted by the slope on the object.
As usual, draw a good diagram showing all the forces acting on the object.
To find the answers to (a) and (b),we form two equations using the resolved forces. One equation comes
from the forces parallel to the plane (given by the i direction in the diagram). The other equation comes
from the forces perpendicular to the plane (given by the j direction in the diagram).
As usual, draw a good diagram showing all
the forces acting on the object.
Tcos30o
Tsin30o
8gsin20o
20o
30o i
j
40 N 8g N
R T
As usual, draw a good diagram showing
all the forces acting on the object.
We now resolve the force T and the
force 8g N parallel and perpendicular to
the plane as shown in the next box.
20o
30o i
j
40 N 8g N
R T
8gcos20o
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Connected Particles Connected particles problems will involve to particles (cars, lorries, vans, caravans, masses, in fact any two
masses modeled as particles) connected by a light inextensible string, cable, rope etc… You will need to draw a clear diagram of the whole system. Each particle will give you one equation of motion which can be
solved to find tension, acceleration, masses and so on depending on the information you have been given in
the question. Here is an example.
Example 7
The diagram shows a car mass 1100 kg pulling a trailer of mass 900 kg.
The action of the engine causes a force of magnitude 3000N to act on
the car. Resistance forces of magnitude 300N and 450N act on the car
and trailer respectively.
Find
(a) the acceleration of the car and trailer
(b)force that the car exerts on the trailer.
It is important to draw a diagram and add all the forces acting on the system of car and trailer. To find the
common acceleration you can combine the masses of the particles (car and trailer) into a single mass and all
the resistance forces as a single force.
You can combine this diagram into this one.
driving force
T
R2 R1
1100g 900g
300N 450N
(a) Parallel to the plane (in the i direction)
NT
gT
gT
o
oo
oo
5.8430cos
21.73
4.64020sin830cos
8.084020sin830cos
==
++=
×=−−
(b) Parallel to the plane (in the i direction)
NR
R
gR
gRT
oo
oo
42.31
25.4267.73
30sin5.8420cos8
20cos830sin
=
−=
×−=
=+
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
(a) Using Newton’s Second Law maF = , we can write the equation:
213.1125.1
2000
225020007503000
−
≈==⇒=− msaa
Now that you know the common acceleration, you can use one of the particles (car or trailer) to find the force
that the car exerts on the trailer (this force is just the tension in the tow bar). You only need the diagram of
the car and all the forces acting on it in the direction of motion.
(b)
You will come across problems involving pulleys and connected particles. We use a similar strategy by
forming two equations of motion each involving only one of the particles. Here is an example
Example 8
A block of mass 4kg rests on a horizontal table. It is attached by means of a light, inextensible string to a
particle of mass 9kg. There is a resistance force of 20N opposing the motion of the block. Find:
(a) the acceleration of the system,
(b) the tension in the string,
(c) the resultant force acting on the pulley.
Always draw a diagram, showing all the forces acting on the particles and the
string. To find the acceleration, the tension in the string and the force acting
on the pulley, you have to solve a pair of simultaneous equations: one
equation from each particle.
So the mass 4 kg
4 kg
R a
a
T
T
T T
9g N
20 N
4g N
3000N T
300N
750N
2000g
R
3000N
a
a Using Newton’s Second Law maF = , we can
write the equation:
NT
aT
5.1462)125.1(11003003000
11003003000
=−−=
=−−
9 kg
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
)1.....(420 aT =−
Notice that R and 4g do not contribute to this equation.
So the mass 9 kg
)2.....(99 aTg =−
(a) By adding equations (1) and (2) we get 225.5
13
2.682.681320913
−
==⇒=⇒−= msaaga
(b) Substitute for 225.5
−
= msa in equation (1) to get NTT 412120)25.5(420 =+=⇒=−
c) The force acting on the pulley is shown in the diagram on the right. The force T is also
shown on this diagram. We have to find the resultant of the force T. By Pythagoras
Theorem
Resultant force = N5898.573362414122
≈==+
A final example on connected particle
Example 9
The diagram shows two blocks connected by a string passing over a
smooth pulley. Find the acceleration of each mass and the tension
in the string if there is no friction present.
Add the forces contributing to motion and direction of acceleration.
For the 6kg mass: )1.....(620sin6 agT =−
For the 7kg mass: )2.....(770sin7 aTg =−
Add equations (1) and (2) to get
241.3
13
35.44
11.2046.6413
20sin670sin713
−
==
−=
−=
msa
a
gga
Use equation (1) to get T.
NT
T
agT
57.40
46.2011.20
620sin6
=
+=
+=
T
7gsin70o
6gsin20o
T
70o 20
o
7 kg
T
T
T
a
a
6 kg
70o 20
o
6 kg 7 kg
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The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
Rheolir FMSP partneriaeth â hariannir gan
Cymru gan
Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
TAFLEN ADOLYGU – M1 (CBAC)
Dynameg gronyn, Deddfau mudiant Newton
Grymoedd yn weithredol ar fàs yn cynhyrchu mudiant Grymoedd yn baralel i gyfeiriad y mudiant Mae’n bwysig cofio y gallwn fodeli fudiant màs a weithredir arno gan rym gan Ail Ddeddf mudiant Newton,
h.y. F=ma, lle F yw’r grym, m yw màs y gwrthrych ac a yw’r cyflymiad.
Dangosir y diagramau canlynol yn glir beth yw’r F sydd yn Ail Ddeddf Newton.
Dychmygwch un grym yn unig yn weithredol ar y car yma, sy’n gwneud iddo gyflymu. Felly'r hafaliad
mudiant am y car yma yw P = ma, gan fod y grym P yng nghyfeiriad y mudiant.
P = ma
Cyn yr arholiad dylech wybod
• Sut i linio diagramau grymoedd fel pwysau, ffrithiant, adwaith normal, tensiwn a gwthiad yn
gweithredu ar fàs.
• Defnydd o Ail Ddeddf mudiant Newton (F = ma) i ysgrifennu hafaliadau mudiant ar fàs symudol.
• Datrys problemau’n ymwneud â lifftiau.
• Cydrannu grymoedd yn baralel a pherpendicwlar i blân ar oledd.
• Datrys problemau yn ymwneud â blân ar oledd.
• Llunio’n glir grymoedd sy’n gweithredu ar ronynnau wedi’u cysylltu, lle mae un gronyn yn
hongian yn rhydd a’r llall yn hongian yn rhydd neu
ar fwrdd llorweddol neu ar blân ar oledd.
• Darganfod y cyflymiad pan mae gronynnau wedi’u cysylltu gan linyn anestynadwy.
• Darganfod y tensiwn yn y llinyn sy’n cysylltu dau
ronyn.
Y prif syniadau yw
CB
AC
2il Deddf mudiant Newton M1
Grymoedd cyson yn weithredol ar
wrthrychau sydd â màs M1
Problemau ar lifftiau M1
Mudiant ar blân ar oledd M1
Gronynau wedi’u cysylltu M1
P
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Nawr, dychmygwch fod yna wrthiant yn weithredol ar y car, a bod y car yn teithio o’r chwith i’r dde. Y tro
yma'r F yn Ail Ddeddf Newton yw P- R a’r hafaliad rydym angen yw P – R=ma.
P- R = ma
Nesaf, dychmygwch fod dau rym yn gwrthwynebu gweithrediad P fel y dangosir isod. Y tro yma'r F yn Ail
Ddeddf Newton yw P - R - F a’r hafaliad angenrheidiol yw P- R - F=ma.
P- R - F=ma.
Enghraifft 1
Mae tynfad (tug) yn tynnu bâd màs 10000 kg gyda chyflymiad 0.2ms-2
. Gwrthiant y dŵr i fudiant y bâd yw
900 N. O wybod bod y rhaff tynnu yn llorweddol, cyfrifwch y tensiwn yn y rhaff.
Tynnwch ddiagram bob tro i ddangos y grymoedd i gyd sy’n weithredol ar y bâd. Modelwch y bâd
gan ddiagram syml fel y dangosir isod.
Cyfeiriad y mudiant
Defnyddio Ail Ddeddf mudiant Newton maF = .
Cawn 2.010000900 ×=−T , sy’n rhoi NT 29002000900 =+=
Ar gyfer maes llafur CBAC, mae’r grymoedd yn gyson yn yr adran hon. Felly, mae’r cyflymiad hefyd yn
gyson. Gellir gofyn cwestiynau lle bydd angen yr hafaliadau mudiant a ddefnyddir yn Adran 1. Dyma
enghraifft:
Enghraifft 2
Weithredir gwrthiant cyson 2500N ar gar, màs 900 kg, sy’n symud ar heol lorweddol fel bod ei fuanedd yn
arafu o 25 ms-1
i 15 ms-1
. Darganfyddwch, ar gyfer y cyfnod lle mae’r car yn arafu,
(a) yr amser a gymerir
(b) y pellter a deithir.
I ateb y cwestiwn, mae angen darganfod yr arafiad cyson er mwyn defnyddio’r hafaliadau mudiant ar gyfer
cyflymiad cyson i ddatrys y broblem. Lluniwch ddiagram fel arfer
Cyfeiriad y mudiant Felly F = - 2500N
P R
P R
F
T 900
a = 0.2 ms-2
2500
a
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Defnyddio Ail Ddeddf Newton, 2278.0
9000
250090002500
−
−=−=⇒=− msaa .
Nawr,
?
?
278.0
15
25
2
1
1
=
=
=
=
=
−
−
−
t
s
msa
msv
msu
Pan mae grymoedd yn baralel i gyfeiriad y mudiant, mae’n eithaf rhwydd ffurfio’r hafaliad mudiant, fel y
dangosir uchod.
Pan mae’r grymoedd ar ongl i gyfeiriad y mudiant bydd rhaid eu cydrannu’n baralel a pherpendicwlar i’r
cyfeiriad hwn.
Dim ond y grymoedd sy’n gweithredu yng nghyfeiriad y mudiant sy’n cyfrannu i’r hafaliad mudiant.
Nid yw’r grymoedd sy’n berpendicwlar i gyfeiriad y mudiant yn cyfrannu i’r hafaliad mudiant, heblaw y defnyddir ffrithiant.
Grymoedd ar ongl i gyfeiriad y mudiant Ystyriwch y sefyllfa ganlynol:
Gweithredir un grym ar wrthrych, màs m fel y dangosir ar y chwith. Ar y dde, mae’r grymoedd wedi’u
cydrannu’n llorweddol a fertigol.
Cyfeiriad y mudiant
Felly’r hafaliad mudiant am y màs hwn yw maPo
=35cos , gan dybio ei fod yn symud i’r dde gyda
chyflymiad a.
Os oes sawl grym yn weithredol ar onglau gwahanol, bydd angen cydrannu pob un yn baralel a
pherpendicwlar i gyfeiriad y mudiant. Mae’r hafaliad mudiant yn dilyn wrth ddefnyddio Ail Ddeddf mudiant
Newton.
(a) Defnyddio atuv +=
t278.02515 −+=
st 97.35278.0
10
278.0
2515=
−
−
=
−
−
=
(b) Defnyddio tvu
s ×
+=
2
ms 4.71997.352097.352
1525=×=×
+=
a
35o
P
35o
P
Psin35o
Pcos35o
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Ystyriwch y sefyllfa ganlynol:
Cyfeiriad y mudiant
Felly’r hafaliad mudiant am y màs yma yw maPQ oo
=− 35cos50cos , gan dybio ei fod yn symud i’r dde
gyda chyflymiad a.
Enghraifft 3
Darganfyddwch werth y grymoedd R a T yn y
sefyllfa canlynol.
Cydranwch y grym T yn baralel a pherpendicwlar i gyfeiriad y mudiant.
Felly, Tcos30o – 1 = 10 x 4, sy’n rhoi
NNT 473.4730cos
41≈==
Hefyd, R + Tsin30o = 98 (dim mudiant
fertigol)
NR 7435.7430sin3.4798 ≈=−=
a = 4ms-2
35o 50
o
Q P
35o 50
o
Q P
Qcos50o Pcos35
o
a Qsin50o Psin35
o
Enghraifft 4
Darganfyddwch y grym T a’r ongl θ yn y sefyllfa
canlynol, o wybod bod a = 3 ms-2
.
Unwaith eto, cydrannu’r grym T yn baralel a
pherpendicwlar i gyfeiriad y mudiant.
Lluniwch dau hafaliad, sydd angen eu datrys yn
gydamserol.
3152cos ×=−θT ……(1)
147100sin =+θT ..….(2)
47cos =θT ………… ..(3)
47sin =θT ……………(4)
)3()4( ÷ yn rhoi o
451tan =⇒= θθ
Amnewid am θ yn (3) i gael NT 5.6645cos
47==
Tcos 30o
R Tsin30o
1N
98N
Tcosθ
Tsinθ 100N
2N
147 N
30o
T m = 10 kg
1N
98 N
R 100 N
θ
T m = 15 kg
2N
a 147 N
www.furthermaths.org.uk
The FMSP Wales in partnership with funded by Is managed by mewn ac fe’i
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Problemau yn ymwneud â lifft Gall problemau’n ymwneud â lifft gael eu datrys gan ddefnyddio’r un technegau a ddefnyddir uchod.
Cofiwch lunio hafaliad mudiant yn defnyddio F = ma. Dangosir enghraifft hyn yn glir
Enghraifft 5
Rhoddir mudiant màs 300 kg, sy’n codi o ddisymudedd, mewn tri rhan.
• Yn gyntaf mae’n cyflymu ar raddfa 0.7 ms-2
tan iddo gyrraedd cyflymder penodol.
• Mae’n cadw’r cyflymder yma am gyfnod.
• Yn olaf, mae’n arafu gydag arafiad 0.5 ms-2
.
(a) Darganfyddwch y tensiwn yng nghebl y lifft tra ei fod y cyflymu ar i fyny.
(b) Darganfyddwch adwaith llawr y lifft ar ddyn màs 90kg yn ystod pob un o’r rhannau uchod.
Gweithiwch ar un rhan ar y tro, lluniwch fraslun.
Nodwch fod yr adwaith rhwng y dyn a’r lifft fwyaf yn ystod y cyflymu.
Wrth arafu, mae’r adwaith rhwng y dyn a’r lifft lleiaf.
Ar fuanedd cyson, pan fydd y cyflymiad yn sero, mae’r adwaith rhwng y gwerth mwyaf a lleiaf.
(a) Ar gyfer y rhan yma, ystyriwch
màs y lifft a’r dyn tu fewn fel un màs
390kg.
Defnyddio F = ma
NT
T
gT
4095
3822273
7.0390390
=
+=
×=−
2ms7.0
−
(b) Edrychwn nawr ar yr
adwaith rhwng y dyn a
llawr y lifft yn ystod y
cyflymiad.
Defnyddio F = ma
i fynny
NR
R
gR
945
82263
7.09090
=
+=
×=−
2ms7.0
−
(b) Ystyriwn yr adwaith
rhwng y dyn a llawr y lifft
pan mae’r cyflymder yn
gyson (cyflymder = 0)
N
gR
822
90
=
=
(b)Yn olaf, ystyriwn yr adwaith
rhwng y dyn a llawr y lifft yn
ystod yr arafu
Defnyddio F = ma
i fynny
NR
R
gR
777
82245
5.09090
=
+−=
−×=−
2ms5.0
−
R
90g N
R
90g N
T
390g N
R
90g N
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Mudiant ar blân ar oledd Wrth ddatrys problemau yn ymwneud â mudiant ar blân ar oledd, bydd angen cydrannu’r grymoedd yn
baralel a pherpendicwlar i’r plân.
Enghraifft 6
Tynnir gwrthrych màs 8 kg gan linyn ysgafn i fyny llethr wedi’i oleddu ar ongl 20o i’r llorwedd. Mae’r
llinyn ar ongl 30o i’r llethr. Mae yna wrthiant 40 N i’r gwrthrych. Cyflymiad y gwrthrych i fynnu’r llethr yw
2ms8.0
− .
(a) Darganfyddwch y tensiwn yn y llinyn.
(b) Darganfyddwch yr adwaith normal rhwng y llethr a’r gwrthrych.
Fel arfer, lluniwch ddiagram da yn dangos y grymoedd i gyd sy’n weithredol ar y gwrthrych.
I ddod o hyd i’r atebion ar gyfer (a) a (b), luniwn ddau hafaliad gan ddefnyddio’r grymoedd sydd wedi’i
gydrannu. Ddaw un hafaliad o’r grymoedd sy’n baralel i’r plân (rhoddir gan gyfeiriad yr i yn y diagram).
Ddaw’r hafaliad arall o’r grymoedd sy’n berpendicwlar i’r plân (rhoddir gan gyfeiriad yr j yn y diagram).
Fel arfer, lluniwch ddiagram da yn dangos y grymoedd i gyd sy’n weithredol ar y
gwrthrych.
Cydrannwn y grym T a’r grym 8g N yn baralel
a pherpendicwlar i’r plân fel y ddangosir yn y
bocs nesaf..
i
20o
30o
j
40 N 8g N
R T
Tcos30o
Tsin30o
8gsin20o
20o
30o i
j
40 N 8g N
R T
8gcos20o
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Gronynnau wedi’u cysylltu Bydd y problemau ar ronynnau (ceir, loriau, faniau, carafanau, unrhyw dau fàs wedi’u modeli fel gronynnau)
sydd wedi’u cysylltu â llinyn ysgafn anestynadwy, cebl, raff, a.y.b. . . . Bydd angen lunio diagram clir ar
gyfer y system gyfan. Rhoddir bob gronyn un hafaliad yr un, a gallwn ddatrys y rhain er mwyn darganfod y
tensiwn, y cyflymiad, y màs a.y.b.; yn dibynnu ar y wybodaeth rhoddir yn y cwestiwn. Dyma enghraifft
Enghraifft 7
Dangosir gar, màs 1100 kg, yn y diagram yn tynnu ôl-gerbyd, màs 900
kg. Mae effaith yr injan yn rhoi grym, maint 3000N ar y car. Mae
grymoedd gwrthiant, maint 300N a 450N yn weithredol ar y car a’r ôl-
gerbyd yn ôl eu trefn.
Darganfyddwch
(a) cyflymiad y car a’r ôl-gerbyd
(b) y grym mae’r car yn rhoi ar yr ôl-gerbyd.
Mae’n bwysig llunio diagram ac adio’r grymoedd i gyd sy’n weithredol ar system y car a’r ôl-gerbyd. I
ddarganfod y cyflymiad cyffredin gallwch gyfuno màs y ddau ronyn (car a’r ôl-gerbyd) i mewn i un màs, a’r
holl wrthiant i mewn i un grym.
(a) Paralel i’r plân (cyfeiriad i)
NT
gT
gT
o
oo
oo
5.8430cos
21.73
4.64020sin830cos
8.084020sin830cos
==
++=
×=−−
(b) Perpendicwlar i’r plân (cyfeiriad j)
NR
R
gR
gRT
oo
oo
42.31
25.4267.73
30sin5.8420cos8
20cos830sin
=
−=
×−=
=+
grym gyrru T
R2 R1
1100g 900g
300N 450N
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Gallwn gyfuno rhain i’r diagram yma
(a) Defnyddio Ail Ddeddf mudiant Newton maF = , cawn yr hafaliad yma:
213.1125.1
2000
225020007503000
−
≈==⇒=− msaa
Gan ein bod yn gwybod y cyflymiad cyffredinol, gallwn ddefnyddio un o’r gronynnau (car neu ôl-gerbyd) i
ddarganfod y grym mae’r car yn rhoi i’r ôl-gerbyd (hwn yw’r tensiwn yn y bar tynnu). Defnyddiwn lun y car a’r grymoedd sy’n weithredol arno yng nghyfeiriad y mudiant.
(b)
Fe ddewch o hyd i broblemau’n ymwneud â phwlïau a gronynnau wedi’u cysylltu. Defnyddiwn strategaeth
debyg i ffurfio dau hafaliad mudiant yn cynnwys un gronyn yn unig. Dyma enghraifft
Enghraifft 8
Mae blocyn, màs 4kg yn gorwedd ar fwrdd llorweddol. Mae wedi’i gysylltu gan linyn ysgafn anestynadwy i
ronyn, màs 9kg. Mae yna wrthiant, 20N yn gwrthwynebu mudiant y bloc. Darganfyddwch:
(a) cyflymiad y system,
(b) y tensiwn yn y llinyn,
(c) y grym cydeffaith sy’n gweithredu ar y pwli.
Tynnwch ddiagram bob amser, i ddangos yr holl rymoedd sy’n gweithredu ar
y gronyn a’r llinyn. I ddarganfod y cyflymiad, y tensiwn yn y llinyn a’r
grym sy’n gweithredu ar y pwli, rhaid datrys pâr o hafaliadau cydamserol: un
hafaliad ar gyfer pob gronyn.
750N
2000g
R
3000N
a
a
Defnyddio Ail Ddeddf Newton maF = , cawn:
NT
aT
5.1462)125.1(11003003000
11003003000
=−−=
=−−
3000N T
300N
4 kg
R a
a
T
T
T T
9g N
20 N
4g N
9 kg
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Disclaimer: Every effort has gone into ensuring the accuracy of this document. However the Further Maths Support Programme Wales can accept no responsibility for its content matching WJEC specifications exactly. Ymwadiad: Mae pob ymdrech wedi mynd mewn i sicrhau cywirdeb y ddogfen hon. Er hyn, ni all y Rhaglen Cymorth Mathemateg Bellach Cymru dderbyn unrhyw gyfrifoldeb am ei chynnwys yn cyfateb yn union gyda manylebau CBAC.
Ar gyfer y màs 4 kg
)1.....(420 aT =−
Nodwch nad yw R nac 4g yn cyfrannu i’r hafaliad.
Ar gyfer y màs 9 kg
)2.....(99 aTg =−
(a) Gan adio hafaliadau (1) a (2) cawn 225.5
13
2.682.681320913
−
==⇒=⇒−= msaaga
(b) Amnewid am 225.5 −
= msa yn hafaliad (1) cawn NTT 412120)25.5(420 =+=⇒=−
(c) Dangosir y grym sy’n gweithredu ar y pwli yn y diagram ar y dde. Dangosir y grym T
hefyd yn y diagram. Rydym angen y grym cydeffaith ar gyfer y ddau T. Defnyddio
Theorem Pythagoras
Y grym cydeffaith = N5898.573362414122
≈==+
Enghraifft olaf ar ronynnau wedi’u cysylltu
Enghraifft 9
Mae’r diagram yn dangos dau floc wedi’u cysylltu gan linyn sy’n
pasio dros bwli llyfn. Darganfyddwch gyflymiad y ddau fàs a’r tensiwn yn y llinyn o wybod nad oes dim ffrithiant yn bresennol.
Adiwch y grymoedd sy’n cyfrannu i’r mudiant a chyfeiriad y
cyflymiad.
Ar gyfer y màs 6kg: )1.....(620sin6 agT =−
Ar gyfer y màs 7kg: )2.....(770sin7 aTg =−
Adio hafaliadau (1) a (2) cawn
241.3
13
35.44
11.2046.6413
20sin670sin713
−
==
−=
−=
msa
a
gga
Defnyddio hafaliad (1) i gael T.
NT
T
agT
57.40
46.2011.20
620sin6
=
+=
+=
T
T
T
70o 20
o
7 kg 6 kg
T
7gsin70o
6gsin20o
T a
a
70o 20o
6 kg 7 kg