Equivalent Force System and Free Body Diagram MCQ Quiz in मराठी - Objective Question with Answer for Equivalent Force System and Free Body Diagram - मोफत PDF डाउनलोड करा

The correct answer is option 3):(Lami's theorem)

Concept:

• LAMI’S Theorem states, “If three coplanar forces acting at a point be in equilibrium, then each force is proportional to the sine of the angle between the other two.”
• Consider three forces F1, F2, F3 acting on a particle or rigid body making angles α, β, and γ with each other.

\(\rm\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}\)

Additional Information

• Lambert's law of absorption states that equal parts in the same absorbing medium absorb equal fractions of the light that enters them
• The equilibrium law states that the concentrations of the products multiplied together, divided by the concentration of the reactants multiplied together, equal an equilibrium constant (K)
• The superposition theorem states that a circuit with multiple voltage and current sources is equal to the sum of simplified circuits using just one of the sources.

Given:

Mass of wardrobe (m) = 100 kg, height = 4 m, width = 2 m, depth = 1 m

The wardrobe will be on the verge of tip if the normal reaction passes through the point Q as shown in the free body diagram:

Taking moments about Q, ∑M_Q = 0

⇒ W × 1 = P × 2

\(\Rightarrow {\rm{P}} = \frac{{100 \times 9.81}}{2} = 490.5{\rm~{N}}\)

∑H = 0

⇒ F_F = P = 490.5 N

∑V = 0 ⇒ R_N = mg = 981 N

Friction Force = μR_N

\({\rm{\mu }} = \frac{{490.5}}{{981}} = 0.5\)

v² = u² + 2as

\({19.86^2} = 0 + 2 \times gsin\theta \times 19.86\sqrt 2\)

7.021 = 9.8 × sin θ

sin θ = 0.716

Explanation:

As the lamina is symmetrical about the y-y axis, bisecting the lamina, therefore its center of gravity lies on this axis.

Let O be the reference point. From the geometry of the lamina. We find the semi-vertical angle of the lamina.,\(\alpha = 30 ^\circ =\frac{\pi}{6}~rad\)

We know that distance between the reference point O and center of gravity of the lamina,

\(\bar y=\frac{2r}{3} \frac{sin \alpha}{\alpha }\)

\(\bar y=\frac{2 \times 220}{3} \times \frac{sin ~30}{\frac{\pi}{6}}\)

\(\bar y=\frac{440}{3} \times \frac{0.5}{\frac{\pi}{6}}\)

\(\bar y=140~mm\)

Lami's Theorem:

If a body is in equilibrium under the effect of three forces, then each force is proportional to the sine of the angle between the other two.

\(\frac {T_{QR}}{\sin (180\ -\ \alpha)} = \frac {{W}}{\sin (\beta \ +\ \alpha)} = \frac {T_{PR}}{\sin(180\ -\ \beta)} \)

sin(180° - θ) = sin θ 

Calculation: 

Given:

P = Q = 100 N, ∠POR = ∠QOR = 150, R = ?

By applying Lami's theorem, we get

\(\frac{P}{Sin(150)}=\frac{Q}{sin(150)}=\frac{R}{sin(60)}\)

​\(\frac{P}{Sin(180-150)}=\frac{Q}{sin(180-150)}=\frac{R}{sin(60)}\)​​

\(\frac{P}{Sin(30)}=\frac{Q}{sin(30)}=\frac{R}{sin(60)}\)​

\(R =\frac{P}{sin~30}\times sin~ 60\)

\(R = \frac{100}{\frac{1}{2}}\times \frac{\sqrt{3}}{2}=100\sqrt{3}~N\)

Explanation:

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 FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.

Then from Lami's theorem,

\(\frac{{{F_A}}}{{sin\alpha }} = \frac{{{F_B}}}{{sin\beta }} = \frac{{{F_C}}}{{sin\gamma }}\)

Explanation:

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 FA, FB, FC acting on a particle or rigid body making angles α, β and γ with each other.

Therefore,

 \(\frac{{{F_A}}}{{sin\alpha }} = \frac{{{F_B}}}{{sin\beta }} = \frac{{{F_C}}}{{sin\gamma }}\)

Important Points

Triangle Law of Forces: 

It states that if two forces acting simultaneously on a particle, be represented in magnitude and direction by the two sides of a triangle, taken in order; their results may be represented in magnitude and direction by the third side of the triangle, taken in the opposite order.”

Law of the parallelogram of force: 

It states that if two forces, acting at the point of a body, be represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram contained in between the two adjacent forces.

Concepts:

The Force system is classified according to the orientation of the line of action forces acting on that body.

Coplanar forces and Non-coplanar forces

In coplanar forces, the line of action of all forces is lying on the same plane. If the lines of action of all the forces are lying on a different plane, then they are called non-coplanar forces.

Concurrent and Non-concurrent types of force system

If the line of action of all forces passes through a single point then the force system is called a concurrent force system. Similarly, if the line of action of all forces does not pass through a single point, then they are called non-concurrent forces.

A parallel system of forces

In a parallel system of forces, the line of actions of all forces is parallel to each other. The parallel system is again classified into like and unlike. If the forces are in the same direction then it is like a parallel system and if the forces are in opposite direction then it is unlike a parallel system.

General Force: A system of which is non-concurrent and non-parallel is called a general force system.

In the given figure all the forces are not meeting at a point and also, it can be seen that line of action of forces is not parallel to each other, so the force system is General.

The different types of force systems can be summarized in the following schematic diagram.

Concept:

In free body diagram we replace the constraints by equivalent reaction forces.

Concept:

Methods of finding resultant of forces

A) Law of a parallelogram of forces

Where P and Q are two forces acting on a body with angle θ between them

α = Angle made by resultant with respect to force P

α is the direction of resultant R

\({\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}} \)

\({\rm{\alpha }} = {\tan ^{ - 1}}\left[ {\frac{{{\rm{Q}}\sin {\rm{\theta }}}}{{{\rm{P}} + {\rm{Q}}\cos {\rm{\theta }}}}} \right]{\rm{\;\;}}\)

B) Method of resolution of forces

ΣF_X = summation of forces acting in X-direction

ΣF_Y = Summation of forces acting in Y-direction

\({\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}} \)

\({\rm{\theta }} = {\tan ^{ - 1}}\left| {\frac{{{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}}}{{{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}}}} \right|\)

Calculation:

Let the force P = 9 N and Q = 12 N and θ = 90°

By law of parallelogram of forces

\({\rm{R}} = \sqrt {{{\rm{P}}^2} + {{\rm{Q}}^2} + 2{\rm{PQ}}\cos {\rm{\theta }}} \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2} + 2 \times 9 \times 12 \times \cos 90^\circ } \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2}} \)

R = 15 N

Alternate method:

By the method of resolution

ΣF_X = 9 N, ΣF_Y = 12 N

\({\rm{R}} = \sqrt {{{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{X}}}} \right)}^2} + {{\left( {{\rm{\Sigma }}{{\rm{F}}_{\rm{Y}}}} \right)}^2}} \)

\({\rm{R}} = \sqrt {{9^2} + {{12}^2}} \)

R = 15 N