What is Equilibrium of a Body on an Inclined Plane?

Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.

What are the Conditions of Limiting Equilibrium?

Acceleration of A Body on an Inclined Plane

Consider a body of mass m sliding down an inclined plane with an acceleration a [Fig.], Let the angle of inclination be θ. Since the body slides down, θ must be greater than the angle of repose. Here kinetic friction acts upwards along

the incline. Let normal force of the inclined plane be R, and coefficient of sliding friction be µ’. Hence, sliding frictional force, f = µ’R.
Also, R = mgcosθ. Component of the weight of the body along the plane is mgᐧsinθ, which is in the opposite direction of f. The equation of downward motion, of the body along this inclined plane is, mg sinθ – f = ma
or, ma = mg sinθ – µ’R = mg(sinθ – µ’cosθ)
or, a = g(sinθ – µ’cosθ)
If the inclined plane is perfectly smooth, then µ’ = 0 and a = g sinθ. Hence, the frictional force reduces the downward acceleration of the body along the plane by an amount gµ’ cosθ.

Equilibrium of a Body on an Inclined Plane

A body of weight W is kept on a plane inclined at an angle θ with the horizontal [Fig.]. If θ is greater than the angle of repose, the frictional force alone is not sufficient to keep the body at rest; an external force F is required for equilibrium. Let it be assumed that the force F makes an angle α with the inclined plane.

There can be either of the two limiting equilibrium conditions-

1. the body is about to move down along the inclined plane, or,
2. the body is about to move up along the inclined plane.

Case 1:
Equilibrium against downward motion: The force of limiting friction acts on the body in an upward direction. For equilibrium, the resultants of the components of the forces acting
1. along the incline, and
2. perpendicular to the incline, both should be zero [Fig.],

Hence, f + cosα = W sinθ …… (1)
and R + F sinα = W cosθ ……. (2)
If the coefficient of friction is μ, then f = μR and from equation (1),
μR + F cosα = W sinθ …….. (3)
Multiplying equation (2) by R and then subtracting from (3) we get,
∴ F(cosα – µsinα) = W(sinθ – μcosθ)
or, F = Wᐧ\(\frac{\sin \theta-\mu \cos \theta}{\cos \alpha-\mu \sin \alpha}\)
Again, if λ is the angle of friction then, μ = tanλ
∴ F = Wᐧ\(\frac{\sin \theta-\tan \lambda \cos \theta}{\cos \alpha-\tan \lambda \sin \alpha}\)
= Wᐧ\(\frac{\sin \theta \cos \lambda-\cos \theta \sin \lambda}{\cos \alpha \cos \lambda-\sin \lambda \sin \alpha}\)
= Wᐧ\(\frac{\sin (\theta-\lambda)}{\cos (\alpha+\lambda)}\) …… (4)
For given values of θ and λ, F will be minimum when cos(α + λ) is maximum,
i.e., when cos(α + λ) = 1 = cos 0
∴ a + λ = 0 or, α = -λ …….. (5)
∴ Minimum applied force, F_min = W sin(θ – λ)

Case 2:
Equilibrium against upward motion: If the body is about to move up along the incline, the limiting friction, f = RR acts downwards along the incline. The forces acting when the body is just about to move are shown in Fig. Conditions for limiting equilibrium are,
F cosα = µR + W sinθ
or, Fcosα – µR= W sinθ …… (6)
and Fsinα + R = Wcosθ
or, µF sinα + µR = W µcosθ ………… (7)

On adding (6) and (7), we get
F(cosα + µsinα) = W(sinθ + µcosθ)
or, F = Wᐧ\(\frac{\sin \theta+\mu \cos \theta}{\cos \alpha+\mu \sin \alpha}\)
Expressing µ in terms of the angle of friction λ, i.e., µ = tanλ,
F = Wᐧ\(\frac{\sin \theta+\tan \lambda \cos \theta}{\cos \alpha+\tan \lambda \sin \alpha}\)
= Wᐧ\(\frac{\sin \theta \cos \lambda+\sin \lambda \cos \theta}{\cos \alpha \cos \lambda+\sin \lambda \sin \alpha}\)
= Wᐧ\(\frac{\sin (\theta+\lambda)}{\cos (\alpha-\lambda)}\) ……. (8)
For given values of θ and λ for F to be minimum,
cos(a – λ) = 1 = cos 0 or, a = λ and F_min = W sin(θ + λ).

Special cases: When F is applied along the plane, i.e., if α = 0, then the force required to prevent the body from moving downwards is
F = Wᐧ\(\frac{\sin (\theta-\lambda)}{\cos \lambda}\) …(9)
Again, the force required to bring the body on the verge of moving upwards along the inclined plane is
F = \(\frac{W \sin (\theta+\lambda)}{\cos \lambda}\) …… (10)

Numerical Examples

Example 1.
A body is kept on a horizontal rough plane. The plane is then gradually raised to an inclination of 30° with the horizontal and the body starts to slide down. The body descends 12 m along the inclined plane in the next 4 s. Find the coefficients of static and kinetic friction between the surfaces in contact.
Solution:
In this case, angle of repose = 30°.
∴ Coefficient of static friction, µ = tan30° = \(\frac{1}{\sqrt{3}}\) = 0.577
As the body begins to slide down, kinetic friction starts acting against the motion. Downward acceleration,
a = g( sinθ – µ’costθ)
[µ’ = coefficient of kinetic friction]
= 9.8(\(\frac{1}{2}\) – µ’\(\frac{\sqrt{3}}{2}\)) = 9.8(0.5 – 0.866 µ’) m ᐧ s^-2
As the body travels 12 m in 4 s from rest, from the equation, s = \(\frac{1}{2}\) at², we get,
12 = \(\frac{1}{2}\) a ᐧ (4)² or, a = \(\frac{3}{2}\) = 1.5 m ᐧ s^-2
∴ 9.8(0.5 – 0.866µ’) = 1.5
or, 4.9 – 8.5µ’= 1.5 or, µ’ = \(\frac{3.4}{8.5}\) = 0.4

Example 2.
To initiate an upward motion of a body along an inclined plane, the minimum force required is twice the force required to keep the body at rest on the same incline. If the coefficient of friction is µ, prove that the inclination of the plane is θ = tan^-1(3µ).
Solution:
Let the mass of the body be m, and the minimum force required to keep the body at rest on the inclined plane be F₁. In this case, the force of limiting friction f acts in the direction of the applied force.
∴ F₁ + f = mg sinθ

or, F₁ + µmg cosθ = mg sinθ or, F₁ = mg(sinθ – µcosθ)
If F₂ is the minimum force needed to set the body in an upward motion, then limiting friction acts downwards.
∴ F₂ = mg sinθ + f = mg sinθ + µmg cosθ
= mg(sinθ + µ cosθ)
As F₂ = 2F₁ (given),
mg( sinθ + µ cosθ) = 2 mg( sinθ – µ cosθ)
or, sinθ = 3µ cosθ
or, tanθ = 3µ or, θ = tan^-1(3µ) (Proved).

Example 3.
A body of mass 5 × 10³ kg is projected upwards along a plane Inclined at an angle of 30° with the horizontal. If the time required by the body to move up the incline is half the time required for it to slide down, find the coefficient of friction between the surface and the body.
Solution:
Let the mass of the body be m, and its upward acceleration be a₁. Let the force of friction be f.
∴ ma₁ = mg sinθ + µmg cosθ
or, a₁ = g(sinθ + µ cosθ)

If a₂ is the acceleration during the downward motion, then,
ma₂ = mgsinθ – f = mgsinθ – µmgcosθ
or, a₂ = g(sinθ – µcosθ)

If the displacement along the inclined plane is s, and the time taken to move up and to come down are t₁ and t₂ respectively, then,

Example 4.
The upper half of an inclined surface is perfectly smooth but the lower half is rough. A body starts sliding down the plane and stops immediately on reaching the bottom. The inclination of the plane is 30° with the horizontal. Show that the frictional resistance of the rough part of the surface is equal to the weight of the body.
Solution:
Let the length of the inclined plane be 2l, and the vertical height from the ground be h. Then we get,
sin30° = \(\frac{h}{2 l}\) or, \(\frac{1}{2}\) = \(\frac{h}{2 l}\) or, h = l.
Potential energy of the body of mass m at maximum height on the plane = mgh = mgl.

The body comes to rest at the bottom. Hence, all its potential energy at the top is spent in doing work against the force of friction, which acts along the length of the lower half of the surface. If f is the frictional force, then
f × l = mgl
or, f = mg = weight of the body (Proved).