Statics and Dynamics Formulas

In this section, students will find most important Statics and Dynamics Formula Sheets, Tables and lists with some example probelms. It helps students to enhance their probelm-solving skills and get experience in solving the problems easily and fastly. So, refer to the below list of Statics and Dynamics Formulas finish your homework, assignments and exam preparation.

A. Statics

Statics is that branch of mechanics which deals with the forces acting on a body so that the body is at rest.

1. Resultant of Two Coplanar Forces

If P & Q are two forces and a is the angle between them, then the resultant of these two forces is
R = \(\sqrt{P^{2}+Q^{2}+2 P Q \cos \alpha}\) and tan θ = \(\frac{Q \sin \alpha}{P+Q \cos \alpha}\)
where θ is the angle between resultant R and force P.

• Maximum R = P + Q, when α = 0
• Minimum R = |P – Q|, when α = π.
• If α = π/2, then R = \(\sqrt{P^{2}+Q^{2}}\), tan θ = \(\frac{Q}{\mathrm{P}}\)

2. Component of Force

Component of Force F in two direction is
p = \(\frac{\mathrm{F} \sin \alpha}{\sin (\alpha+\beta)}\), and Q = \(\frac{\mathrm{F} \sin \beta}{\sin (\alpha+\beta)}\)
where α is the angle between P & F and p is the angle between Q & F.

3. Lami’s Theorem

If three forces acting at a point are in equilibrium, then each is proportional to the sine of angle between the other two i.e.
\(\frac{P}{\sin \alpha}=\frac{Q}{\sin \beta}=\frac{R}{\sin \gamma}\)
where α, β, γ are respectively the angles between Q, R ; R, P & P, Q.

4. Triangle Law of Forces

If three coplanar forces acting at a point be represented in magnitude and direction by the sides of a triangle, taken in order, they are in equilibrium.

5. Trigonometrical Theorem

Let P be a point on the base BC of the ΔABC such that BP : CP = m: n If ∠ BAP = α and ∠ CAP = β, and ∠ APC = γ, then

• (m + n) cot θ = m cot α – n cot β
• (m + n) cot θ = n cot B – m cot C

6. Friction

Friction is a retarding force which prevent one body from sliding on another.
(i) Limiting Friction:
When the body is just on the point of moving the force of friction F is limiting and it bears constant ratio to the normal reaction R. The constant ratio p is called the coefficient of friction
i.e. \(\frac{F}{R}\) = µ ⇒ F = µR

(ii) Angle of Friction:
The angle which the resultant force makes with the direction of the normal reaction is called the angle of friction.

S cos λ = R; S sin λ = µR
∴ S = \(\sqrt{1+\mu^{2}}\) and tan λ = µ = coefficient of friction.

(iii) Least Force on the Horizontal plane:
Least force required to j move the body with weight W on the rough horizontal plane is W sin λ.

(iv) Least Force on the Inclined plane:
Let α be the inclination of rough inclined plane to the horizontal and λ, be the angle of friction. Then

• If α = λ, then the body is in limiting equilibrium and is just on the point of moving downwards.
• If α < λ, then the least force required to move the body j down the plane is W sin (λ, – α)
• If α = λ, α > λ or α < λ, then the least force required to move the body up the plane is W sin (α + λ).
• If α > λ, then the body will move down the plane under the action of its weight and normal reactions.

B. Dynamics

Dynamics is that branch of mechanics which deals with the action of forces on bodies in motion.

7. Velocity

The velocity of moving point is the rate of its displacement. It is a vector quantity.
v = \(\frac{d x}{d t}\)

8. Acceleration

It is the rate of change of velocity. It is also a vector quantity.
f = \(\frac{d v}{d t}=\frac{d^{2} x}{d t^{2}}\)

9. Retardation

The retardation of moving point is the rate of decrease of its velocity. Thus if f is the acceleration of a particle then its
retardation = -f = \(-\frac{d v}{d t}=-\frac{d^{2} x}{d t^{2}}\)

10. Rectilinear motion

If u = initial velocity, v = final velocity, f = uniform acceleration, t = time, then

• v = u + ft
• s = ut + \(\frac{1}{2}\) ft²
• v² – u² = 2fs
• Distance travelled in n^th seconds = S_nth = u + \(\frac{1}{2}\) (2n – 1) f

11. Motion Under Gravity

(i) Downward motion: Here f = g

• v = u + gt
• s = ut + \(\frac{1}{2}\) gt²
• v² = u² + 2gs

(ii) Upward motion: Here f = – g

• v = u – gt
• s = ut – \(\frac{1}{2}\) gt²
• v² = u² – 2gs

(iii) When a particle is thrown vertically upward then the maximum height reached by the particle = \(\frac{u^{2}}{2 g}\) and time taken to reach this height = \(\frac{u}{g}\)

(iv) Time of flight:
The total time taken by the particle in going up and then coming back to the point of projection = \(\frac{2u}{g}\).

12. Parallelogram Law of Velocities

If v₁ and v₂ are the two velocities and α is the angle between them
then the resultant velocity v² = v₁² + v₂² + 2 v₁ v₂ cos α and tan θ = \(\frac{v_{2} \sin \alpha}{v_{1}+v_{2} \cos \alpha}\); where θ is the angle between resultant velocity v and velocity v₁.

13. Projectile

(i) The path of a projectile is a parabola whose equation is
y = x tan α – \(\frac{g x^{2}}{2 u^{2} \cos ^{2} \alpha}\)
and its vertex is \(\left(\frac{\mathrm{u}^{2} \sin \alpha \cos \alpha}{\mathrm{g}}, \frac{\mathrm{u}^{2} \sin ^{2} \alpha}{2 \mathrm{g}}\right)\) and
focus \(\left(\frac{u^{2} \sin 2 \alpha}{2 g},-\frac{u^{2} \cos 2 \alpha}{2 g}\right)\)

(ii) Time to reach the greatest height = (u sin α)/g

(iii) Time of flight, T = (2u sin α)/g

(iv) Greatest height, H = (u² sin² α)/2g

(v) Horizontal range, R = (u² sin 2α)/g

(vi) Maximum horizontal range = u²/g. Angle of projection for maximum horizontal range = π/4

(vii) If a particle is projected with velocity u at an angle α to the horizontal then it will move at right angles to its direction of projection after the time u/(g sin α).

(viii) There are two angles of projection i.e. α and 90° – α for the same horizontal range.

(ix) If there are two times t₁ & t₂ for a given height h then, t₁t₂ = 2h/g

14. Projection on an Inclined plane

A particle is projected with velocity u at angle α to the horizontal from a point on a plane inclined at an angle β to the horizontal, then (i) Time of flight
(i) Time of flight
Up the plane = \(\frac{2 u \sin (\alpha-\beta)}{g \cos \beta}\), Down the plane = \(\frac{2 u \sin (\alpha+\beta)}{g \cos \beta}\)

(ii) Range
Up the plane = \(\frac{2 u^{2} \cos \alpha \sin (\alpha-\beta)}{g \cos ^{2} \beta}\),
Down the plane = \(\frac{2 u^{2} \cos \alpha \sin (\alpha+\beta)}{g \cos ^{2} \beta}\)

(iii) Maximum Range
Up the plane = \(\frac{u^{2}}{g(1+\sin \beta)}\), Angle α = \(\frac{\pi}{4}+\frac{\beta}{2}\),
Down the plane = \(\frac{u^{2}}{g(1-\sin \beta)}\), Angle α = \(\frac{\pi}{2}-\frac{\beta}{2}\)

15. Apparent weigh

In a lift when in motion = \(\left\{\begin{array}{ll}\mathrm{m}(\mathrm{g}+\mathrm{f}), & \text { when lift moves upwards } \\\mathrm{m}(\mathrm{g}-\mathrm{f}), & \text { when lift moves downwards }\end{array}\right.\)

16. Newton’s Law of Collision

Let u₁ and u₂ be the velocities of two bodies before collision and v₁ and v₂ that after the collision, along the line of impact or line of collision.
\(\frac { v_{2} – v_{1} }{ u_{2} – u_{1} } \) = – e ; Where e = coefficient of restitution. I
e = 1 for elastic collision
e < 1 for inelastic collision
e = 0 for perfectly inelastic collision.
(i) In the direct elastic collision both momentum and kinetic energy remain conserved.
(ii) In the direct inelastic collision only momentum is conserved.
(iii) Rebounds of a particle on smooth plane:
If a smooth ball falls from a height h upon a fixed horizontal plane and if e is the coefficient of restitution, then whole time before the rebounding ends \(\sqrt { \frac { 2h }{ g } } \)\(\frac { 1+e }{ 1-e } \) and the total distance covered before finishing rebounding = \(\frac{1+e^{2}}{1-e^{2}}\)h

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