When an object moves along an inclined plane then: different forces act on it like normal

reaction of plane, friction force acting in opposite direction of motion etc. Different relations

for the motion are given below.

Normal reaction of plane

R = mg cos θ

and net force acting downward on the block.

F = mg sin θ – f

Acceleration on inclined plane a = g (sin θ – μ cos θ)

When angle of inclination of the plane from horizontal is less than the angle of repose (α), then

• minimum force required to move the body up the inclined plane

                               f1 = mg (sin θ + μ cos θ)

• minimum force required to push the body down the inclined plane

                               f2 = mg (μ cos θ – sin θ) J

Tension

Tension force always pulls a body. Tension is a reactive force. It is not an active force.

Tension across a massless pulley or frictionless pulley remain constant. Rope becomes slack when tension force becomes zero.

Motion of Bodies in Contact

• Two Bodies in Contact If F force is a applied on object of mass m1 then acceleration of the bodies
• 

a = F / (m1 + m2)

Contact force on m₁ = m₁ a = m1F / (m₁ + m₂)

Contact force on m₂ = m₂ a = m₂ F / (m₁ + m₂)

Three Bodies in Contact If F force is applied an object of mass m₁, then acceleration of the bodies = F / (m₁ + m₂ + m₃)

Contact force between m₁ and m₃

F₁ = (m₁ + m₃) F / (m₁ + m₂ + m₃)

Contact force between m₂ and m₃

F₂ = m₃ F / (m₁ + m₂ + m₃)

Motion of Two Bodies, One Resting on the Other

1. The coefficient of friction between surface of A and B be μ. If a force F is applied on the lower body A. then common acceleration of two bodies

a = F / (M + m)

Pseudo force acting on block B due to the accelerated motion

f’= ma

The pseudo force tends to produce a relative motion between bodies A and B and consequently a frictional force

f = μ N = μmg is developed. For equilibrium

ma ≤ μ mg or a ≤ μg

1. Let friction is also present between the ground surface and body A Let the coefficient of friction between the given surface and body A is μ₁ and the coefficient of friction between the surfaces of bodies A and B is μ₂ If a force F is applied on the lower body A

Net accelerating force = F – f_A = F – μ1(M + m)g

 Net acceleration

a = F – μ₁ (M + m)g / (M + m) = F / (M + m) – μ g

Pseudo force acting on the block B

f’ = ma

The pseudo force tends to produce a relative motion between the bodies A and B are

Consequently a frictional force f_B = μ mg is developed. For equilibrium

ma-le; μ₂ mg or a ≤ μ₂ g

If acceleration produced under the effect of force F is more than μ2g, then two bodies will not move together.

 Motion of Bodies Connected by Strings

Acceleration of the system a = F / (m₁ + m₂ + m₃)

Tension in string T₁ = F

T₂ = (m₂ + m₃) a = (m₂ + m₃) F / (m₁ + m₂ + m₃)

T₃ = m₃a = m₃F / (m₁ + m₂ + m₃)

Pulley Mass System

• When unequal masses m1 and m2 are suspended from a pulley

(m1 > m2)

m₁ g – T = m₁ a, and T – m₂ g = m₂ a

On solving equations, we get

a = ((m₁ – m₂) / (m₁ + m₂)) * g

T = 2 m₁m₂  / (m₁ + m₂) * g

• When a body of mass m₁ is placed frictionless horizontal surface, then

Acceleration a = m₁ g / (m₁ + m₂)

Tension in string T = m₁ m₂ g / (m₁ + m₂)

• When a body of mass m₂ is placed on a rough horizontal surface, then

Acceleration a = ((m₁ – μm₂) / (m₁ + m₂)) * g

Tension in string T = (m₁ m₂(1 + μ) / (m₁ + m₂)) * g

• When two masses m₁ and m₂ are connected to a single mass M as shown in figure, then
• 

  m₁ g – T1 = m₁ a …..(i)

  T₂ – m₂ g = m₂ a ……(ii)

  T₁ – T₂ = Ma …….(iii)

  Acceleration a = ((m₁ – m₂ / (m₁ + m₂ + M)) * g

  Tension T₁ = (2m₂ + M / (m₁ + m₂ + M) * m₁ g

  T₂ = (2m_a + M / (m₁ + m₂ + M) * m₂ g

• Motion on a smooth inclined plane, then

m1g – T = m1a …..(i)

T – m₂ g sin θ = m₂ a ……(ii)

Acceleration a = ((m₁ – m₂ sin θ/ (m₁ + m₂)) * g

Tension T = m₁ m₂ (1 + sin θ) g / (m₁ + m₂)

• Motion of two bodies placed on two inclined planes having different angle of inclination,

Then

Acceleration a = (m₁ sin θ₁ – m₂ sin θ₂) g / m₁ + m₂

Tension T = (m₁ m₂ / m₁ + m₂) * (sin θ₁ – sin θ₂) g

EXERCICE:

Multiple choice

1. Figure 1A shows three force vectors. Which of the vectors A.B.C.D.E in figure 1A would be most to represent their resultant?