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 (m_{A}) = 1 kg

Mass B (m_{B}) = 2 kg

The mass of C (m_{C}) = 2 kg

Tensile force (F) = 10 N

__Wanted:__ T_{AB} : T_{BC}

__Solution:__

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

ΣF = m a

F = (m_{A} + m_{B} + m_{C}) a

10 = (1 + 2 + 2) a

10 = 5 a

a = 10 / 5

a = 2 m/s^{2}

Use the rope tension formula to calculate T_{AB}

ΣF = m a

T_{AB }= m_{A} a = 1 (2) = 2 Newton

Use the rope tension formula to calculate T_{BC}

ΣF = m a

T_{BC} = (m_{A} + m_{B}) 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/s^{2}, determine the acceleration of the object and the tension in the ropes of each block.

__Known____:__

The mass of object A (m_{A}) = 6 kg

The mass of object B (m_{B}) = 3 kg

Coefficient of friction of block A (µ_{k}) = 0.3

Acceleration due to gravity (g) = 10 m/s^{2}

The weight of block A (w_{A}) = m_{A} g = (6)(10) = 60 N

Normal force on block A (N_{A}) = w_{A} = 60 N

The weight of block B (w_{B}) = m_{B} 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:

F_{k} = µ_{k} N_{A }= (0,3)(60) = 18 Newton

**Calculate the acceleration of the system (a):**

ΣF = m a

wB – F_{k} = (m_{A} + m_{B}) a

30 – 18 = (6 + 3) a

12 = 9 a

a = 12 / 9 = 1,3 m/s^{2}

**Calculate the tension in the string on block A (T _{A}):**

ΣF = m a

T_{A} – F_{k} = (m_{A}) a

T_{A} – 18 = (6)(1,3)

T_{A} – 18 = 7,8

T_{A} = 7,8 + 18 = 25,8 Newton

**Calculate the tension in the rope on beam B (T _{B}):**

ΣF = m a

w_{B} – T_{B }= m_{B} (a)

30 – T_{B }= 3 (1,3)

30 – T_{B }= 3,9

T_{B }= 30 – 3,9

T_{B }= 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/s^{2}

The mass of A (m_{A}) = 5 kg

The weight of block A (w_{A}) = m_{A} g = (5)(10) = 50 Newton

Mass B (m_{B}) = 3 kg

The weight of block B (w_{B}) = m_{B} 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 (T_{A}) and beam B (T_{B})

__Solution:__

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

ΣF = m a

w_{A} – w_{B }= (m_{A }+ m_{B}) a

50 – 30 = (5 + 3) a

20 = 8 a

a = 20 / 8

a = 2,5 m/s^{2}

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

The tension in the rope on block A:

ΣF = m a

w_{A} – T_{A} = m_{A} a

50 – T_{A }= 5 (2,5)

50 – T_{A }= 12,5

T_{A} = 50 – 12,5 = 37,5 Newton

The tension in the rope on block B:

ΣF = m a

T_{B} – w_{B} = m_{B} a

T_{B} – 30 = 3 (2,5)

T_{B} – 30 = 7,5

T_{B} = 7,5 + 30 = 37,5 Newton