Coefficient of Static Friction $μ_s = {F_s}/{R}$ …For impending motion Coefficient of Kinetic Friction $μ_k = {F_k}/{R}$ For actual motion R -> Magnitude of normal reaction Angle of Friction φ = $tan^{-1} μ $ For impending motion the angle between normal reaction and resultant reaction is called as angle of static friction. For actual motion when P = $F_k$ the angle between normal reaction and resultant reaction is called angle of kinetic friction. Angle of Respose Angle of inclined plane with horizontal at which body is just on verge of sliding (impending motion) and so $F_s$ = $μ_s$ R Angle ...
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There are as such no formulas in this chapter.
To find tension at extreme points of cable/at support: T = $R_A$ = $√{{A_x^2}+{A_y^2}}$ tan θ = ${A_y}/{A_x}$ For finding other tensions use Lami’s theorem at intermediate joints To find sag at any point Use geometry for example, ${A_y}/{A_x} = {\text"Verticle distance/sag"}/{\text"Horizontal distance"}$ Important tips 1. Observe number of unknown forces. If no. of unknown forces is ‘n’ then consider (n-3) parts of cable for finding the unknown quantities. 2. While selecting no. of parts care should be taken that parts must be starting from “same support” upto the points where sags of points are known. 3. Although the tensions ...
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Perfect Truss (Stable Truss) Condition: n = 2j – R j = Number of joints n = Number of members R = No. of reaction components (external) Imperfect Truss (unstable) n ≠ 2j – R Deffecient Truss n < 2j - R Redundant Truss n > 2j – R Table to remember: $\table Sr.No. ,Member ,Magnitude,Nature$
Rectangular Components of a Force in Space $F_x$ = F cos$θ_x$ , $F_y$ = cos$θ_y$ , $F_z$ = cos$θ_z$ $\ov{F}$ = $F_x$i + $F_y$j + $F_z$k l = cos$θ_x$ = ${F_x}/{F}$ Similarly for others. $l^2 + m^2 + n^2 = 1$ Unit Vector $\ov{F}$ = F $\ov{e}$ where, $\ov{e}$ = unit vector in the direction of force F = (cos$θ_x$)i + (cos$θ_y$)j + (cos$θ_z$)k Unit Vector when Force is Specified by Two Points $\ov{F}$ = F $[{(x_2 – x_1)i+(y_2 – y_1)j+(z_2 – z_1)k}/{√{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2}}]$ Components of Force when Orientation of Planes ...
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Types of Beams (1)Simply supported beam, (2)Overhanging beam (3)Cantilever beam, (4)A beam with hinge and roller supports (5)A fixed beam, (6)A continuous beam (7)Prop Cantilever Types of loads on the beam (1)Point load or Concentrated load Always represented in Newton or kN (2)Uniformly Distributed Load (UDL) Always represented in N/m or kN/m Converted in point load by taking product of intensity of udl and spreading distance. This acts on centre of udl (3)Uniformly Varying Load (UVL) $\table Shape,\text"Point load",\text"Point of action (From left & Right)";\text"Right triangle", 1/2 wl,{2l}/3 \text" &" l/3;\text"Trapezoidal",w_1l \text" & " (w_2 – w_1)l/2,l/2 \text"&" 2/3 l;\text"Pyramidal ...
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Conditions of Equilibrium (a) For co-planar concurrent forces ∑$F_x$ = 0 and ∑$F_y$ = 0 Thereforce, R = 0 (b) For co-planar non-concurrent forces ∑$F_x$ = 0 and ∑$F_y$ = 0 and ∑M = 0 Types of supports and Corresponding Reactions: $\table Support\/Connection,Specification,No. of unknowns;Rollers,\text"Known reaction which is perp. to plane of roller",One;\text"Smooth Surface",\text"Reaction is perp. to the surface",One;\text"Rough Surface",\text"Two reactions components with unknown directions",Two;\text"Smoth pin or Hinge",\text"Two reaction components with unknown directions",Two;\text"Flexible cord,rope or cable of negligible weight",\text"One axial force acting away from body",One;\text"Fixed support",\text"Two reaction components and one moment with all components unknown in directions",Three;\text"A smooth pin in ...
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Section I – Impulse – Momentum Theorem Linear Momentum: Linear momentum = mass * velocity = m * v Momentum is vector quantity and its SI unit is kg.m/sec. Angular Momentum Angular momentum = I . w Angular momentum is vector quantity and its SI unit is kg . $m^2$ rad/s Linear Impulse If force is constant, Impulse = F Δt Impulse is a vector quantity and its SI unit is N.sec Angular Impulse If torque is constant, Angular Impulse = T Δt It is a vector quantity and its SI unit is N.m.sec Impulsive – Momentum Theorem for Particles ...
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The S.I. unit of work done is N.m or Joule W.D. is a scalar quantity has only magnitude but no direction. Work-Energy Principle for a Particle $U_{1→2} = KE_2 – KE_1$ where, $U_{1→2}$ = Algebric sum of work done by all forces from position 1 → 2 $KE_1$ = KE of system at position 1 $KE_2$ = KE of system at position 2 Work Done Calculations for Different Forces: W.D. by External Force: (constant) W.D. = F * S (Positive if work is done in the direction of force. W.D. by Variable Force: (When F v/s S curve is given) ...
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Steps for application: 1. Draw F.B.D of particle 2. Apply D’Alembert’s force m x $a_N$ = ${m v^2 }/{ρ}$ in opposite direction of $a_N$ (i.e. away from centre of curvature) 3. Apply Σ$F_N$ = 0 Some important concepts: (a) When particle is moving along a horizontal circle, mark y axis and normal axis in F.B.D. (b) When particle is moving along a vertical circle, mark normal and tangential axis in F.B.D. (c) When a particle attached by a string and whirled in verticle circle $T_{top} = {mv^2}/{r} – mg$ $T_{bottom} = {mv^2}/{r} + mg$ (d) When an aeroplane flying along ...
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