3 questions about Friction force equation

1. Block A 3 kg is placed on the table and then tied to a rope that is connected to stone B = 2 kg through a pulley as shown. The mass and friction of the pulleys are neglected. Acceleration due to gravity g = 10 m/s^{2}. Determine the acceleration of the system and the tension in the rope if:

a) smooth table

b) rough table with a coefficient of kinetic friction of 0.4

__Known:__

The mass of block A (m_{A}) = 3 kg

The mass of rock B (m_{B}) = 2 kg

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

Weight of block A (w_{A}) = m g = (3)(10) = 30 Newton

Weight of rock B (w_{B}) = m g = (2)(10) = 20 Newton

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

__Solution:__

*a) smooth table*

**Calculate the acceleration of the system using the formula for Newton’s second law:**

ΣF = m a

w_{B} = (m_{A} + m_{B}) a

20 = (3 + 2) a

20 = 5 a

a = 20 / 5 = 4 m/s^{2}

**Calculate the tension in the rope using the formula for the tension in the rope:**

**The tension in the rope on block A:**

ΣF = m_{A} a

T = m_{A} a = (3)(4) = 12 Newton

**The tension in the rope on block B:**

ΣF = m_{B} a

w_{B} – T = (2)(4)

20 – T = 8

T = 20 – 8 = 12 Newton

*b) rough table with a coefficient of kinetic friction of 0.4*

The force of the kinetic Friction:

F_{k} = µ_{k} N = (0,4)(30) = 12 Newton

**Calculate the acceleration of the system using the formula for Newton’s second law:**

ΣF = m a

w_{B} – f_{k} = (m_{A} + m_{B}) a

20 – 12 = (3 + 2) a

8 = 5 a

a = 8 / 5 = 1,6 m/s^{2}

**Calculate the tension in the rope using the formula for the tension in the rope:**

The tension in the rope on block A:

ΣF = m_{A} a

T – fk = m_{A} a

T – 12 = (3)(1,6)

T – 12 = 4,8

T = 4,8 + 12 = 16,8 Newton

The tension in the rope on block B:

ΣF = m_{B} a

w_{B} – T = (2)(1,6)

20 – T = 3,2

T = 20 – 3,2 = 16,8 Newton

2. An object with a mass of 10 kg is in a horizontal plane. The coefficient of static friction is 0.4 and the coefficient of kinetic friction is 0.35. g = 10 m/s^{2}. If an object is given a constant horizontal force of 25 N, the magnitude of the frictional force acting on the object is…

__Known:__

The mass of the object (m) = 10 kg

The coefficient of Static friction (µ_{s}) = 0.4

The coefficient of kinetic friction (µk) = 0.35

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

Horizontal force (F) = 25 N

The object’s gravity (w) = m g = (10)(10) = 100 Newton

Normal force (N) = w = 100 Newton

__Wanted:__ The amount of static friction (f_{s}) and kinetic (f_{k})

__Solution:__

The force of the static Friction::

f_{s} = µ_{s} N = (0,4)(100) = 40 Newton

The force of the Kinetic Friction:

f_{k} = µ_{k} N = (0,35)(100) = 35 Newton

The horizontal force is only 25 Newton so it can’t move objects yet.

3. The masses of blocks A and B in the figure are 10 kg and 5 kg respectively. The coefficient of friction between block A and the plane is 0.2. To prevent block A from moving, the minimum mass of block C required is…

__Known:__

The mass of block A (m_{A}) = 10 kg

The mass of block B (m_{B}) = 5 kg

Coefficient of static friction of block A (µ_{s}) = 0,2

Gravity acceleration (g) = 10 m/s^{2}

Block weight A (w_{A}) = m_{A} g = (10)(10) = 100 Newton

Block weight B (w_{B}) = m_{B} g = (5)(10) = 50 Newton

Static friction (f_{s}) = µ_{s} N = (0,2)(w_{A} + w_{C}) = (0,2)(100 + w_{C}) = 20 + 0,2 w_{C}

__Ditanya:__ The mass of block C to keep the system at rest

__Jawab:__

The system is at rest so the formula for Newton’s first law is used:

ΣF = 0

w_{B} – f_{s} = 0

50 – (20 + 0,2 w_{C}) = 0

50 – 20 – 0,2 w_{C} = 0

30 – 0,2 w_{C} = 0

30 = 0,2 w_{C}

w_{C }= 30 / 0,2 = 300 / 2 = 150 Newton

The mass of block C = 150 / 10 = 15 Kg