Exercise A, Question 1

Solutionbank M1Heinemann Modular Maths for Edexcel AS and A-level

6 Connected particlesExercise A, Question 1

Question:

Two particles, of mass 3 kg and 5 kg, respectively, are attached to the ends of a light, inextensible string that passes over a smooth pulley. Both strings are vertical. Find the acceleration of the particles and the tension in the string.

Solution:

Using F = ma,

∴ Acceleration is 2.45 m s − 2

Substituting into equation [1]

∴ Tension is 36.75 N

© Harcourt Education Ltd 2005

For 5 kg particle; 5g − T = 5a [1]For 3 kg particle; T − 3g = 3aAdding; 2g = 8a

a = g1

4

5g − T = 5 × g1

4

g15

4= T

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13/02/2014file:///C:/Users/AEC/Desktop/AQA%20M1%20SolutionBank/content/header.html?un...



Question:

The diagrams below show particles that are connected by a light, inextensible string that passes over a smooth pulley. In each case find the acceleration of the particles and the tension in the string.

Solution:

(a) Using F = ma,



For 5 kg particle; 5g − T = 5a [1]For 4 kg particle; T − 4g = 4aAdding; g = 9a

a = g1

9

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(b) Using F = ma,




5g − T = 5 × g1

9

g40

9= T


a = g1

5

6g − T = 6 × g1

5

g24

5= T

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(c) Using F = ma,






a = g5

13

9g − T = 9 × g5

13

g72

13= T

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Question:

Two particles, of mass 4 kg and 7 kg, are attached to the ends of a light, inextensible string, which passes over a smooth pulley. Initially the particles are at rest and at the same level. The particles are then released.

(a) Find the acceleration of the particles.

(b) Find the time when the difference in height of the particles is 1 m.

Solution:

(a) Using F = ma,


(b) Particles are moving with constant acceleration.You known a, s, u and require t.

The equation connecting these is s = ut + at2

Substitute a = 2.67, s = , u = 0 [Particles at 1m apart. ∴ Each have moved 0.5 m]

For 7 kg particle; 7g − T = 7aFor 4 kg particle; T − 4g = 4aAdding; 3g = 11a

a = g3

11

1

2

1

2

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∴ Time is 0.612 seconds.


1

2= × 2.67 × t2

1

2

0.37453... = t2

t = 0.612

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Question:

Two particles of mass m and are connected by a light, inextensible string that passes over a smooth pulley.

Find the acceleration of the particles and the tension in the string in terms of m and g.

Solution:

Using F = ma,

Acceleration is m s − 2


Tension is mg N

3m

2

for m particle T − mg = ma [1]

for particle g − T3m

2

3m

2= a

3m

2

Addingmg

2= a

5m

2

∴ a =g

5

g

5

T − mg = mg

5

T = mg6

5

6

5

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Question:

The diagram shows a block of mass 2 kg, that slides on a rough horizontal table. It is attached by a light, inextensible string to a 3 kg mass. If the block accelerates at 4.9 m s − 2, find the coefficient of friction between the block and the table.

Solution:

Using F = ma,for 3 kg mass; 3g − T = 3a [1]for 2 kg mass (horizontally, as motion takes place horizontally) T − F = 2a [2]Resolving vertically for 2 kg mass; R = 2g [3]In limiting equilibrium, F = μR [4]It is given that a = 4.9,

[1] ⇒ 3g − T = 3 × 4.9

∴ T = 3 × 4.9

= 14.7

[2] ⇒ 14.7 − F = 2 × 4.9

∴ F = 4.9

[3] and [4] ⇒ F = μR

= μ × 2g

∴ 4.9 = μ × 2g

∴ μ = or 0.251

4

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Question:

A block, of mass 6 kg, rests on a rough, horizontal surface. The coefficient of friction between the block and the surface is 0.2. A light, inextensible string attached to the block passes over a smooth pulley. A weight, of mass 2 kg, hangs from the other end of the string, as shown in the diagram below.

Find the tension in the string and the acceleration of the block. [A]

Solution:

Using F = mafor 2 kg mass; 2g − T = 2a [1]for 6 kg mass (horizontally, as motion takes place horizontally) T − F = 6a [2]

i.e.: Acceleration is 0.98 m s − 2

Resolving vertically for 6 kg mass; R = 6g [3]In limiting equilibrium F = μR

= 0.2R ( since μ = 0.2 ) [4]

[3] & [4] ⇒ F = 0.2R = 0.2 × 6g

= 1.2g

[2] ⇒ T − 1.2g = 6a [5]

3 × [1] ⇒ 6g − 3T = 6a

∴ T − 1.2g = 6g − 3T

4T = 7.2gT = 17.6 NAdding [1] and [5] 0.8g = 8a

∴ a = 0.1g

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Question:

The diagram shows a mass A of 5 kg initially at rest on a horizontal table. A resistance force of 10 N acts against the motion of A which is connected to mass B of 3 kg by a light, inextensible string which passes over a smooth pulley. The system is released from rest.

(a) Calculate the acceleration of A.

(b) Calculate the tension in the string.After a short time, B reaches the floor.

(c) Calculate the acceleration of A now.

Solution:

(a) Using F = ma

Acceleration is 2.43 m s − 2

for 3 kg mass; 3g − T = 3afor 5 kg mass; T − 10 = 5a [1]Adding 3g − 10 = 8aa = 2.425

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(b)

Tension is 22.1 N

(c) After 3kg reaches the floor, tension in string becomes zero 5 kg mass starts decelerating.

∴ Acceleration is − 2 m s − 2

( − because the mass is decelerating)


From [1] T − 10 = 5 × 2.425T = 22.125

Using F = ma (horizontally) for 5 kg mass10 = 5a

∴ a = 2

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Question:

A breakdown truck tows a car of mass 1200 kg. Assume that the rope is horizontal and that the truck moves on a horizontal surface. Calculate the tension in the tow rope if the car is:

(a) accelerating at 0.5 m s − 2 and experiencing a resistance force of 500 N;

(b) travelling at constant speed but experiencing a resistance force of 400 N.

Solution:

(a) Using F = ma, horizontally for the car,

∴ Tension is 1100 N

(b) Using F = ma, horizontally for the car,

∴ Tension is 400 N

T − 500 = 1200a= 600 ( since a = 0.5 )

T − 400 = 1200a= 0 since a = 0 [If travelling at constant speed, acceleration is zero]

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Question:

Two bodies A and B of mass 4 kg and 2 kg, respectively, are attached by a light, inextensible string passing over a smooth pulley. A rests on a table and B hangs over the side. Resistance forces on A amount to 8 N.The system is released from rest. Calculate the acceleration of the system and the tension in the string. Find the speed of B when it has fallen 2 m.

Solution:

Using F = mafor 2 kg body; 2g − T = 2a [1]for 4 kg body (horizontally, as motion takes place horizontally); T − 8 = 4a [2]


Tension is 15.7 NTo find the speed when B has fallen 2 m;you know s, a, u and require v.

Velocity is 2.78 m s − 1

Adding [1] and [2] 2g − 8 = 6aa = 1.933...

Substituting into [1] 2g − T = 2 × 1.93T = 2g − 3.86

∴ Use v2 = u2 + 2as

∴ v2 = 0 + 2 × 1.93 × 2

v2 = 7.72

v = \ 7.72 = 2.778...

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Question:

Two particles, of masses 4 kg and 6 kg, are connected by a light, inextensible string that passes over a smooth, light pulley. The two particles are released from rest, with the string taut, as shown in the diagram.

(a) Show that the acceleration of each particle is 1.96 m s − 2.

(b) Calculate the tension in the string. [A]

Solution:

(a) Using F = ma,

∴ Acceleration of each particle is 1.96 m s − 2

for 6 kg particle; 6g − T = 6afor 4 kg particle; T − 4g = 4a [1]Adding 2g = 10a

∴ a =g

5

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(b)

or 47.04 N


From [1] , T = 4g + 4a

=24g

5

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Question:

Two particles, of masses 0.3 kg and 0.4 kg, are connected by a light, inextensible string which hangs over a smooth fixed peg, as shown in the diagram. The system is released from rest.

(a) (i) Show that, in the subsequent motion, the acceleration of the particles is of magnitude 1.4 m s − 2.(ii) Find the tension in the string during this motion.

(b) Find the distance travelled by each particle during the first 2 seconds of the motion. [A]

Solution:

(a) (i) Using F = ma


for 0.4 kg particle; 0.4g − T = 0.4afor 0.3 kg particle; T − 0.3g = 0.3a [1]Adding 0.1g = 0.7a

a = = 1.4g

7

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(ii)

Tension is 3.36 N

(b)

∴ Distance travelled by each particle is 2.8 m.


From [1] T = 0.3g + 0.3 × 1.4= 3.36

When t = 2 ,

using s = ut + at2 ( we do not know v )1

2

s = × 1.4 × 221

2

= 2.8

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Question:

A box, of mass 6 kg, is at rest on a rough horizontal table. A light, inextensible string is attached to the box. This string passes over a smooth, light pulley and the other end is attached to a sphere of mass 4 kg, as shown in the diagram. Assume that the string is horizontal between the box and the pulley. The coefficient of friction between the box and the table is 0.5.

The sphere is released from rest, with the string taut, and the box slides along the table.

(a) Calculate the magnitude of the friction force acting on the box.

(b) Show that the acceleration of the system is 0.98 m s − 2.

(c) Calculate the tension in the string. [A]

Solution:

(a) At the box;

Friction is 29.4 N

resolve vertically R = 6gusing F = μR ,F = 0.5 × 6g

= 3g

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(b)

(c)

∴ Tension is 35.3 N.


Using F = ma at 4 kg sphere; 4g − T = 4aat 6 kg box (in the direction of movement); T − F = 6a [1]Adding 4g − F = 10a10a = 4g − 29.4a = 0.98


From [1] , T = F + 6a= 29.4 + 6 × 0.98= 35.28

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Question:

Two particles are connected by a light, inextensible string, which passes over a smooth, light pulley, as shown in the diagram. The particle A, of mass 0.5 kg, is in contact with a rough horizontal surface, and the particle B,

of mass 0.2 kg, hangs freely. The coefficient of friction between A and the surface is .

The system is released from rest with the string taut and A moves towards the pulley.

(a) Show that the frictional force between A and the surface is of magnitude 1.4 N.

(b) Find the acceleration of the particles.

(c) Find the tension in the string.

(d) Find the time taken for the particles to travel 0.625 metres, given that A has not then reached the pulley. [A]

Solution:

2

7

Page 1 of 2


Frictional force is 1.4 N

(b) Using F = ma at

(c)

(d)

∴ Time is 1.25 sec.


At A , resolve vertically R = 0.5gUsing F = μR ,

F = × 0.5g = 1.42

7

0.2 kg particle; 0.2g − T = 0.2a0.5 kg particle (in the direction of movement); T − F = 0.5a [1]Adding 0.2g − F = 0.7a0.7a = 0.2g − 1.4

a = 0.8 m s − 2

From [1] , T = F + 0.5 × 0.8

= 1.8 ⇒ Tension is 1.8 N

Using s = ut + at2 ,1

2

0.625 = × 0.8t21

2

t2 = 1.5625

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Question:

A block, of mass 6 kg, is held at rest on a rough horizontal table. The block is attached, by a light string that passes over a light, smooth pulley, to a sphere of mass 4 kg, that hangs freely, as shown in the diagram.

(a) The block is released and travels 60 cm in 2 seconds. Show that the acceleration of the block is 0.3 ms − 2.

(b) Find the magnitude of the tension in the string.

(c) Find the magnitude of the friction force that acts on the block.

(d) Find the coefficient of friction between the block and the table. Give your answer correct to two significant figures. [A]

Solution:

(a) Using s = ut + at2, 0.6 = × a × 22


(b) Using F = ma for 4 kg sphere;

1

2

1

2

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∴ Tension is 38 N

(c) Using F = ma for 6 kg block, in the direction of movement,

∴ Magnitude of frictional force is 36.2 N

(d) Resolving vertically at the block, R = 6g

∴ Coefficient of friction is 0.62 (to 2 s.f.).


4g − T = 4a4g − 4 × 0.3 = T

T − F = 6a

∴ F = T − 6a

= 38 − 6 × 0.3

F = μR ⇒ μ =F

R

=36.2

6g

= 0.6156

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Question:

Two particles, P and Q, are connected by a light, inextensible string which passes over a smooth, fixed peg, as shown in the diagram.

The particle P is of mass 2m and Q is of mass m. The particle P is in contact with a horizontal surface, which may be modelled as either rough or smooth.

(a) In the first case, the surface is modelled as rough and the particle Q hangs at rest. The coefficient of friction between P and the surface is μ.(i) Find the tension in the string.(ii) Find the range of possible values of μ.

(b) In the second case, the surface is modelled as smooth. The system is released from rest with the string taut and the particle P moves towards the peg.Find the tension in the string. [A]

Solution:

(a) (i) No movement, hence particle m is at rest.Resolve vertically mg − T = 0Tension is mg.

(ii) At P, resolve horizontally

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F = T = mgResolve vertically R = 2mg

From [1]; ∴ Tension is mg .


F ≤ μR (friction need not be limiting)

∴ mg ≤ μ.2mg

i.e. μ ≥1

2

Using F = mafor Q ; mg − T = mafor P ; T = 2ma [1]Adding mg = 3ma

a = g1

3

2

3

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Question:

A builder uses a rope that passes over a pulley to raise and lower concrete blocks while building a house. The diagram shows the initial positions of two loads of the blocks.

Each load consists of eight blocks, each of mass 5 kg. Initially the lower load is resting on the ground. When two blocks are accidentally removed from the lower load the system begins to move.

(a) Making any necessary modelling assumptions:(i) show that the acceleration of the loads is 1.4 m s − 2,(ii) find the tension in the rope.

(b) State two assumptions made in (a).

(c) Find the speed of the two loads when they are at the same level. [A]

Solution:

(a) (i) Using F = ma for


40 kg mass; 40g − T = 40a30 kg mass; T − 30g = 30a [1]Adding 10g = 70a

a = × g1

7

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(ii) From [1] T = 30g + 30 × 1.4Tension is 336 N

(b) Assumptions are: smooth pulley light, inextensible string.

(c) When at the same level, each load has moved 2.5 m

∴ speeds are 2.65 m s − 1 .


Using v2 = u2 + 2as

v2 = 0 + 2 × 1.4 × 2.5

= 7v = 2.6457

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Question:

Two children are holding the ends of a light, inextensible rope, which passes over a light, smooth pulley. Initially Tom, who has a mass of 40 kg, is standing at ground level and Simon, who has a mass of 60 kg, is on the edge of a fixed platform 2 metres above ground level. Model the two boys as particles, one initially at ground level, and the other initially at a height of 2 metres. The rope is taut.

Simon steps off the platform and as he falls vertically, Tom rises vertically.

(a) Assume that the rope remains taut while the boys are moving.(i) Show that the acceleration of each boy is 1.96 m s − 2.(ii) Find the tension in the rope.

(b) Find the total distance that Tom travels upwards. [A]

Solution:

(a) (i) Using F = ma for


Simon; 60g − T = 60aTom; T − 40g = 40a [1]Adding 20g = 100a

a =g

5

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(ii)

∴ Tension in the rope is 470 N

(b) When Simon hits the floor, to find the speed of Tom use v2 = u2 + 2as,

Tom now moves subject to gravitational force

∴ Tom travels 2 m + 0.4 m = 2.4 m .


From [1] T = 40g + 40 × 1.96= 470.4

v2 = 2 × 1.96 × 2

∴ v = 2.8

Using v2 = u2 + 2as with a = − g

0 = 2.82 − 2gs

s =2.82

2g

= 0.4

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Question:

The diagram shows a car pulling a trailer in a straight line on a horizontal stretch of road.

The mass of the car is 1250 kg and the total resistance force acting on the car is 500 N.The mass of the trailer is 250 kg and the total resistance force acting on the trailer is 100 N.

(a) During part of the journey, a constant braking force is applied to the car, causing the car and trailer to decelerate at a constant rate of 0.5 m s − 2.(i) By considering the forces on the trailer, find the magnitude of the force in the towbar between the car and the trailer.(ii) Find the magnitude of the braking force applied to the car.

(b) Later in the journey, the car and trailer travel with constant speed.State the magnitude of the tension in the towbar. [A]

Solution:

(a) (i) Using F = ma horizontally

∴ Magnitude of the force in the towbar is 25 NForces acting on trailer;

(ii) Let B be the braking force applied to the car.

Using F = ma horizontally

− T − 100 = − 250 × 0.5T = 25

− B − 500 + T = − 1250 × 0.5B = 150

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∴ Magnitude of the braking force is 150 N

(b) When travelling at constant speed, the forces acting on the trailer are as shown.T = 100 (no acceleration)∴ Tension is 100 N


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Question:

A particle, of mass M, is on a rough horizontal table. A light, inextensible string is attached to this particle and passes over a smooth pulley. Attached to the other end of the string, so that it hangs vertically, is another particle of mass m. The coefficient of friction between the particle and the table is μ. The particles are initially at rest.

(a) Show that the acceleration of the particles is .

(b) Find the tension in the string.

(c) Find the time that it takes the particle on the table to move 0.5 m and its speed at this time.

Solution:

g ( m − μM )

m + M

Resolve vertically for particle M ; R = MgUsing F = μR for this particle, F = μMgUsing F = ma for particle M (horizontally); T − μMg = Mafor particle m (vertically); mg − T = ma [1]Adding mg − μMg = ma + Ma

∴ a =g ( m − μM )

m + M

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(b)

(c)


From [1] T = mg − ma

= mg − mg ( m − μM )

m + M

=Mmg + mMμg

m + M

=mMg ( 1 + μ )

m + M

Using s = ut + at2 to find the time taken to move 0.5 m,1

2

0.5 = × t21

2

g ( m − μM )

m + M

∴ t = \m + M

g ( m − μM )

To find the speed, using v2 = u2 + 2as

v2 = 2 × × 0.5g ( m − μM )

m + M

v = \g ( m − μM )

m + M

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Question:

The diagram shows two particles connected by a light, inelastic rope. One has a mass of 4 kg and is on a smooth slope and the other has a mass of 6 kg and hangs freely.

(a) Find the acceleration of the particles if α = 30 ° .

(b) Find the tension in the rope if α = 10 ° .

Solution:

and for 4 kg particle (in the direction of movement, i.e. along the slope with α = 30 )


Using F = ma for 6 kg particle 6g − T = 6a

T − 4g sin 30 ° = 4a

Adding 6g − 4g sin 30 ° = 10a4g = 10a

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(b)



If α = 10 ° , the use of F = ma gives 6g − T = 6a [1]T − 4g sin 10 ° = 4aAdding 6g − 4g sin 10 ° = 10a

∴ a = 5.199

[1] ⇒ 6g − T = 6 × 5.199

T = 6g − 6 × 5.199= 58.8 − 31.195= 27.605

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Question:

A car of mass 1200 kg is to be pulled 30 m up a slope inclined at 5° to the horizontal. A rope is attached to the car and passes over a smooth pulley and hangs vertically in an old mine shaft. A concrete weight of mass 200 kg is attached to the other end of the rope and released from rest. Assume that there is no resistance to the motion of the car.

(a) Find the acceleration of the car.

(b) How long does the car take to travel the 30 m?

Solution:


(b) For time to travel 30 m,

Using F = ma for 200 kg concrete weight; 200g − T = 200afor 1200 kg car (in the direction of movement, i.e. up the slope); T − 1200g sin 5 ° = 1200aAdding 200g − 1200g sin 5 ° = 1400aa = 0.66789

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∴ Time is 9.48 s.


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Question:

At an old mine trolleys move on tracks up a slope inclined at 10° to the horizontal. They are pulled up by a rope that passes over a pulley and then hangs down the mine shaft. A weight of mass 250 kg is attached to this end of the rope. The mass of an empty truck is 800 kg. The truck is subject to a resistance force of 100 N as it moves.

(a) Find the acceleration of an empty truck as it moves up the slope.

(b) At the top of the slope the truck is loaded with 1000 kg of coal. Find the acceleration of the truck as it moves down the slope.

Solution:

∴ Acceleration of empty truck is 0.942 m s − 1 .

(b)

Using F = ma for 250 kg weight; 250g − T = 250afor 800 kg trolley (along the slope); T − 800g sin 10 ° − 100 = 800aAdding 250g − 800g sin 10 ° − 100 = 1050a988.59 = 1050aa = 0.94151

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∴ Acceleration is 0.250 m s − 2 .


Using F = ma for 250 kg weight; T − 250g = 250afor 1800 trolley (along the slope); 1800g sin 10 ° − 100 − T = 1800aAdding 1800g sin 10 ° − 100 − 250g = 2050a513.15 = 2050aa = 0.2503

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Question:

Two particles are connected by a light string that passes over a smooth, light pulley as shown in the diagram.The 4 kg particle is on a smooth, fixed slope, which is at an angle of 60° to the horizontal. The 3 kg particle hangs with the string vertical.The particles are released from rest at the position shown.

(a) Show that the acceleration of the particles is approximately 0.65 m s − 2.

(b) By considering the 3 kg particle, determine the tension in the string. [A]

Solution:

(a)

∴ Acceleration of the particles is approximately 0.65 m s − 2

Using F = ma for3 kg particle; T − 3g = 3a [1]4 kg particle (along the slope); 4g sin 60 ° − T = 4aAdding 4g sin 60 ° − 3g = 7a7a = 4.548

∴ a = 0.6497

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(b) From [1]



T = 3g + 3a= 3g + 3 × 0.6497= 31.349

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Exercise A, Question 1