Test: Equilibrium of Forces - JEE MCQ

Concept:
Making free body diagram of each block:

Balancing forces in positive x direction

∴ F = μR_N1 + μR_N2 = μ(R_N1 + R_N2)

where, R_N1 = Normal reaction force on block R and  RN₂ = Normal reaction force on block S

Calculation:

Given:

μ = 0.4

R_N1 = 100g = 100 × 9.81 = 981 N

R_N2 = (100 + 150)g = 250 × 9.81 = 2452.5 N

Minimum force (F) is

F = μ(R_N1 + R_N2)

F = 0.4 × (981 + 2452.5)  = 1373.4 N = 1.37 kN

When two force making an angle θ, the resultant (R) of the two force is given by:

Since the two forces are equal;

We know that;
1 + cos θ = 2 cos²(θ/2)

R = 2F cos(θ/2)

Free-Body Diagram: These are the diagrams used to show the relative magnitude and direction of all external forces acting upon an object in a given situation. A free-body diagram is a special example of vector diagrams.

Some common rules for making a free-body diagram:

• The size of the arrow in a free-body diagram reflects the magnitude of the force.
• The direction of the arrow shows the direction that the force is acting.
• Each force arrow in the diagram is labelled to indicate the exact type of force.
• It is generally customary in a free-body diagram to represent the object by a box and to draw the force arrow from the centre of the box outward in the direction that the force is acting.

Example:

If the sum of all forces acting on the body is zero, then it is not necessary that the body will be in equilibrium. For the explanation, let us take an example.

Let us assume that two equal and opposite forces having their line of action at a certain distance apart act on a body. Then the body in this case will not have translational motion but have rotation due to torque produced. Hence not in equilibrium. If two equal and opposite force act a point or concurrent, then the torque produced will be zero. The body will not have translational and rotational motion and will be in equilibrium.

Concept:
Conditions for the system to be in equlibrium
ΣF_x = 0, ΣF_y = 0, ΣM = 0
Calculation:
Given:
m = 2 kg, Assume g = 10 m/s²

Lager will be the moment, smaller will be the force required to lift the rod. Hence, Applying Moment about 0 cm point we get.

w × 50 = F × 100

m × g × 50 = F × 100

2 × 10 × 50 = F × 100

F = 10 N

Equilibrium of planar forces:

If the forces acting on the free-body are non-concurrent then the equivalent resultant will be a single force acting at a common point and a moment about the same point. The effect of such a force system will be to translate the body as well as to rotate it. Hence, for equilibrium to exist, both the force and moment must be null vectors.

i.e ΣF = 0, ΣM = 0

When the forces acting on a body lie in a plane (X - Y plane) but are non-concurrent, the body will have rotational motion perpendicular to the plane in addition to the translational motion along the plane, Hence necessary and sufficient condition for static equilibrium are,

ΣF_x = 0, ΣF_y = 0, ΣM_z = 0

Concept:

For three force system, the static equilibrium is a state in which the net force and net torque acted upon the system is zero.

Calculation:

Given:

Coplanar forces equal to 7P, 8P, and 5P acting on a particle are in equilibrium.

Let the vectors a, b, and c be equal to 7P, 8P and 5P respectively. Therefore we have to find the angle between 8P and 5P i.e. between b and c.

Squaring both sides

64P² + 25P² + 80P² cos θ = 49P²

80cos θ = 40

cos θ = -1/2

θ = π - cos^-1(-1/2)

θ = 2π/3 = 120°.

Concept:

Lami's theorem:

It states that if three forces acting at a point are in equilibrium, each force is proportional to the sine of the angle between the other two forces. Consider three forces F_A, F_B, F_C acting on a particle or rigid body making angles α, β and γ with each other.

Calculation:
Given:

We have angle BAC = 180° - (30° + 60°)
angle BAC = 90°
By Lami’s Theorem,

 
T_AC = 400 N

Law of Parallelogram of forces: This law is used to determine the resultant of two coplanar forces acting at a point.

• It states that “If two forces acting at a point are represented in magnitude and direction by two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram which passes through that common point.”

Let two forces F₁ and F₂, acting at the point O be represented, in magnitude and direction, by the directed line OA and OB inclined at an angle θ with each other.

Then if the parallelogram OACB be completed, the resultant force R will be represented by the diagonal OC.

CALCULATION:

Given F₁ = P, F₂ = √2P, θ = 135∘ 

Then the resultant force is given by 

Concept:

Lami's Theorem: It is an equation that relates the magnitude of the three co-planner, concurrent and non-collinear forces that keeps a body in equilibrium. It states that each force is proportional to the sine of the angle between the other two forces.

Calculation:

T₁ = 500 × sin 120° and  T₂ = 500 sin 150°
T₁ = 433 N and T₂ = 250 N