1.
The Work of a Force•A force F does work on a particle only when the particle undergoes a displacement inthe direction of the force.•Consider the force acting on the particle•If the particle moves along the path s from position r to new position r’, displacementdr = r’ – r•Magnitude of dr is represented by ds, differential segment along the path•If the angle between tails of dr and F is θ, work dU done by F is a scalar quantitydU = F ds cos θdU = F·dr•Resultant interpreted in two ways Product of F and the component of displacement in the direction of the force ds cos θ Product of ds and component of force in the direction of the displacement F cos θ Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
2.
•If 0° < θ < 90°, the force component and the displacement has the same sense so thatthe work is positive•If 90° < θ < 180°, the force component and the displacement has the opposite sense sothat the work is negative• dU = 0 if the force is perpendicular to the displacement since cos 90° = 0 or if the forceis applied at a fixed point where displacement = 0•Basic unit for work in SI units is joule (J)•This unit combines the units for force and displacement•1 joule of work is done when a force of 1 newton moves 1 meter along its line of action1J = 1N.m•Moment of a force has this same combination of units, however, the concepts ofmoment and work are in no way related•A moment is s vector quantity, whereas work is a scalarWork of a Variable Force.•If the particle undergoes a finite displace along its path from r1 to r2 or s1 to s2, thedetermined by integration.•If F is expressed as a function of positio Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
3.
•If the working component of the force, F cos θ, is plotted versus s, the integral in thisequation can be interpreted as the area under the curve from position s1 to position s2Work of a Constant Force Moving Along a Straight Line.•If the force Fc has a constant magnitude and acts at a constant angle θ from its straightline path, then the components of Fc in the direction of displacement is Fc cos θ Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
4.
•The work done by Fc when the particle is displaced from s1 to s2 is determinedThe work of Fc represents the area of the rectangleWork of a Weight.•Consider a particle which moves up along the path s from s1 to position s2.•At an intermediate point, the displacement dr = dxi +dyj + dzk. Since W = -Wj•Work done is equal to the magnitude of the particle’s weight times its verticaldisplacement.•If W is downward and ∆y is upward, work is negative•If the particle is displaced downward (-∆y), the work of the weight is positive. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
5.
Work of a Spring Force.•The magnitude of force developed in a linear elastic spring when the spring isdisplaced a distance s from its unstretched position is Fs = ks.•If the spring is elongated or compressed from a position s1 to s2, the W.D on spring byFs is positive, since force and displacement are in the same direction. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
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•If a particle is attached to a spring, then the force Fs exerted on the particle is oppositeto that exerted on the spring.•The force will do negative work on the particle when the particle is moving so as tofurther elongate (or compress) the spring. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
7.
ExampleThe 10-kg block rest on a smooth incline. If the spring is originally stretched 0.5 m,determine the total work done by all forces acting on the block when a horizontal forceP = 400 N pushes the block up the plane s = 2 m. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
8.
Principle of Work and Energy•Consider a particle P, which a the instant considered located on the path as measuredfrom an inertial coordinate system•For the particle in the tangential direction, ∑Ft = mat•For principle of work and energy for the particle, Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
9.
•Term on the LHS is the sum of work done by all the forces acting on the particle as theparticle moves from point 1 to point 2•Term on the RHS defines the particle’s final and initial kinetic energy•Both terms are always positive scalars•The particle’s initial kinetic energy plus the work done by all the forces acting on theparticle as it moves from initial to its final position is equal to the particle’s final kineticenergy•For example, if a particle’s initial speed is known and the work of all the forces actingon the particle can be determined, the above eqn provides a direct means of obtainingthe final speed v2 of the particle after it undergoes a specified displacementPROCEDURE FOR ANALYSISWork (Free-Body Diagram)•Establish the initial coordinate system and draw a FBD of the particle to account for allthe forces that do work on the particle as it moves along its pathPrinciple of Work and Energy•Apply the principle of work and energy Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
10.
•The kinetic energy at the initial and final points is always positive since it involves thespeed squared•A force does work when it moves through a displacement in the direction of the force•Work is always positive when the force component is in the same direction as itsdisplacement, otherwise, it is negative•Forces that are functions of displacement must be integrated to obtain the work•Graphically, the work is equal to the area under the force-displacement curve•The work of a weight is the product of the weight magnitude and the verticaldisplacement•It is positive when the weight moves downwards•The work of the spring is in the form of where k is the spring stiffness and s is thestretch or compression of the spring•Principle of work and energy can be extended to include a system of n particlesisolated within an enclose region of space•An arbitrary ith particle, having a mass mi, is subjected to a resultant force Fi and aresultant internal force fi , which each of the other particles exerts on the ith particle Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
11.
• Principle of work and energy for the ith particle• Since both work and force are scalars, the results may be added together algebraically• We can write this equation symbolically• Equation states that the system’s initial energy plus the work done by all the external and internal forces acting on the particles of the system is equal to the system’s final kinetic energy Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
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• Internal forces on adjacent articles will occur in equal but opposite collinear pairs, the total work done by each of these forces will not cancel out since paths over which corresponding particles travel will be differentTwo important exception:1.if the particles are contained within the boundary of a translating rigid body, theinternal forces will undergo the same displacement and therefore the internal work willbe zero2.Adjacent particles exert equal but opposite internal forces that have componentswhich undergo the same displacement and therefore the work of these forces cancelsWork of Friction Caused by Sliding.•Consider a case when a body is sliding over the surface of another body in thepresence of friction•Consider a block translating a distance s over a rough surface• If the applied force P just balances the resultant frictional force μkN then due to equilibrium a constant velocity v is maintained Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
13.
• Sliding motion will generate heat, a form of energy which seems not to be accounted for in the work energy equation• Model the block so that the surfaces of contact are deformable (nonrigid)• Rough portions at the bottom of the block act as teeth and when the block slides these teeth deform slightly and either break off or vibrate due to interlocking effects and pull away from teeth at the contracting surface• As a result, frictional forces that act on the block at these points are displaced slightly, due to localized deformations, and then they are replaced by other frictional forces as other points of contact are made• At any instant, the resultant F of these frictional forces remain essentially constant, μkN, however, due to many localized deformations, the actual displacement s’ of μkN is not the same displacement s as the applied force P• s’ < s and the external work done by the resultant force will be μkNs’ and not μkNs the remaining work μkN(s –s’) manifests itself as the increase in internal energy which in fact, causes the block temperature to increase Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
14.
ExampleThe 17.5-kN automobile is traveling down the 10° inclined road at a speed of 6 m/s. ifthe driver jams on the brakes, causing his wheels to lock, determine how far s his tiresskid on the road. The coefficient of the kinetic friction between the wheels and the roadis μk = 0.5 Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
15.
ExampleFor a short time the crane lifts the 2.50-Mg beam with a force of F = (28 + 3s2) kN.Determine the speed of the beam when it has risen s = 3 m. How much time does ittake to attain this height starting from rest. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
16.
ExampleThe platform P is tied down so that the 0.4-m long cords keep a 1-m long springcompressed 0.6-m when nothing is on the platform. If a 2-kg platform is placed on theplatform and released from rest after the platform is pushed down 0.1-m, determine themax height h the block rises in the air, measure from the ground. Disediakan oleh SHAIFUL ZAMRI, JKM, POLIMAS
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