Friction force equation

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/s2. Determine the acceleration of the system and the tension in the rope if:

a) smooth tableFriction force equation 1

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

Known:

The mass of block A (mA) = 3 kg

The mass of rock B (mB) = 2 kg

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

Weight of block A (wA) = m g = (3)(10) = 30 Newton

Weight of rock B (wB) = 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

wB = (mA + mB) a

20 = (3 + 2) a

20 = 5 a

a = 20 / 5 = 4 m/s2

See also  Kirchhoff's rules – problems and solutions

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

The tension in the rope on block A:

ΣF = mA a

T = mA a = (3)(4) = 12 Newton

The tension in the rope on block B:

ΣF = mB a

wB – T = (2)(4)

20 – T = 8

T = 20 – 8 = 12 Newton

b) rough table with a coefficient of kinetic friction of 0.4Friction force equation 2

The force of the kinetic Friction:

Fk = µk N = (0,4)(30) = 12 Newton

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

ΣF = m a

wB – fk = (mA + mB) a

20 – 12 = (3 + 2) a

8 = 5 a

a = 8 / 5 = 1,6 m/s2

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

The tension in the rope on block A:

ΣF = mA a

T – fk = mA 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 = mB a

wB – T = (2)(1,6)

20 – T = 3,2

T = 20 – 3,2 = 16,8 Newton

See also  Determine resultant of two vectors using cosines equation

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/s2. If an object is given a constant horizontal force of 25 N, the magnitude of the frictional force acting on the object is…

Known:Friction force equation 3

The mass of the object (m) = 10 kg

The coefficient of Static friction (µs) = 0.4

The coefficient of kinetic friction (µk) = 0.35

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

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 (fs) and kinetic (fk)

Solution:

The force of the static Friction::

fs = µs N = (0,4)(100) = 40 Newton

The force of the Kinetic Friction:

fk = µk N = (0,35)(100) = 35 Newton

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

See also  Boyle's law Charles's law Gay-Lussac's law – Problems and Solutions

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:Friction force equation 4

The mass of block A (mA) = 10 kg

The mass of block B (mB) = 5 kg

Coefficient of static friction of block A (µs) = 0,2

Gravity acceleration (g) = 10 m/s2

Block weight A (wA) = mA g = (10)(10) = 100 Newton

Block weight B (wB) = mB g = (5)(10) = 50 Newton

Static friction (fs) = µs N = (0,2)(wA + wC) = (0,2)(100 + wC) = 20 + 0,2 wC

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

wB – fs = 0

50 – (20 + 0,2 wC) = 0

50 – 20 – 0,2 wC = 0

30 – 0,2 wC = 0

30 = 0,2 wC

wC = 30 / 0,2 = 300 / 2 = 150 Newton

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

Print Friendly, PDF & Email