# Rope tension equation

3 Questions about Rope tension equation

1. The picture below shows three blocks, namely A, B and C which are located on a smooth horizontal plane. If mass A = 1 kg, mass B = 2 kg and mass C = 2 kg and F = 10 N, then determine the ratio of the tension in the rope between A and B to the tension in the rope between B and C.

Known:

The mass of A (mA) = 1 kg

Mass B (mB) = 2 kg

The mass of C (mC) = 2 kg

Tensile force (F) = 10 N

Wanted: TAB : TBC

Solution:

Calculate the acceleration of the system using Newton’s Second Law formula:

ΣF = m a

F = (mA + mB + mC) a

10 = (1 + 2 + 2) a

10 = 5 a

a = 10 / 5

a = 2 m/s2

Use the rope tension formula to calculate TAB

ΣF = m a

TAB = mA a = 1 (2) = 2 Newton

Use the rope tension formula to calculate TBC

ΣF = m a

TBC = (mA + mB) a = (1 + 2) (2) = (3)(2) = 6 Newton

2. Object A with a mass of 6 kg and object B with a mass of 3 kg are connected by a rope as shown. If the coefficient of friction is 0.3 and g = 10 m/s2, determine the acceleration of the object and the tension in the ropes of each block.

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Known:

The mass of object A (mA) = 6 kg

The mass of object B (mB) = 3 kg

Coefficient of friction of block A (µk) = 0.3

Acceleration due to gravity (g) = 10 m/s2

The weight of block A (wA) = mA g = (6)(10) = 60 N

Normal force on block A (NA) = wA = 60 N

The weight of block B (wB) = mB g = (3)(10) = 30 N

Wanted: The acceleration of the system (a) and the tension in the rope (T)

Solution:

Calculate the kinetic frictional force i.e. the frictional force when block A moves:

Fk = µk NA = (0,3)(60) = 18 Newton

Calculate the acceleration of the system (a):

ΣF = m a

wB – Fk = (mA + mB) a

30 – 18 = (6 + 3) a

12 = 9 a

a = 12 / 9 = 1,3 m/s2

Calculate the tension in the string on block A (TA):

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ΣF = m a

TA – Fk = (mA) a

TA – 18 = (6)(1,3)

TA – 18 = 7,8

TA = 7,8 + 18 = 25,8 Newton

Calculate the tension in the rope on beam B (TB):

ΣF = m a

wB – TB = mB (a)

30 – TB = 3 (1,3)

30 – TB = 3,9

TB = 30 – 3,9

TB = 26,1 Newton

3. Two objects A and B with masses of 5 kg and 3 kg are connected by a frictionless pulley. The force P is applied to the pulley in an upward direction. If both blocks are initially at rest on the floor, what is the acceleration of block A, if the magnitude of P is 60 N?

Determine also the tension in the rope on blocks A and B.

Known:

Acceleration due to gravity (g) = 10 m/s2

The mass of A (mA) = 5 kg

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The weight of block A (wA) = mA g = (5)(10) = 50 Newton

Mass B (mB) = 3 kg

The weight of block B (wB) = mB g = (3)(10) = 30 Newton

Force P = 60 N

Wanted: Acceleration of the system of beams A and B (a) and the tension in the rope on beam A (TA) and beam B (TB)

Solution:

Calculate the acceleration of the system using Newton’s Second Law formula.

ΣF = m a

wA – wB = (mA + mB) a

50 – 30 = (5 + 3) a

20 = 8 a

a = 20 / 8

a = 2,5 m/s2

Use the tension force formula to calculate the tension in the rope

The tension in the rope on block A:

ΣF = m a

wA – TA = mA a

50 – TA = 5 (2,5)

50 – TA = 12,5

TA = 50 – 12,5 = 37,5 Newton

The tension in the rope on block B:

ΣF = m a

TB – wB = mB a

TB – 30 = 3 (2,5)

TB – 30 = 7,5

TB = 7,5 + 30 = 37,5 Newton