16.07 Dynamics Fall 2004 Version 1.1 Lecture D7-Work and Energy So far we have used Newton's second law F=ma to establish the instantaneous relation between the sum of the forces acting on a particle and the acceleration of that particle. Once the acceleration is known, the relocity(or position) is obtained by integrating the expression of the acceleration(or velocity) There are two situations in which the cumulative effects of unbalanced forces acting on a particle are of terest to us. These involve a)forces acting along the trajectory. In this case, integration of the forces with respect to the displacement leads to the principle of work and energy b) forces acting over a time interval. In this case, integration of the forces with respect to the time leads to the principle of impulse and momentum. It turns out that in many situations, these integrations can be carried beforehand to produce equations that relate the velocities velocities at the initial and final integration points. In this way, the velocity can be obtained directly, thus making it unnecessary to solve for the acceleration. We shall see that these integrated forms of the equations of motion are very useful in the practical solution of dynamics problems In this lecture, we will concentrate on situation a), and consider the space integrated form of Newtons econd law. We will defer the discussion of b), time integrated equations, to lecture D9 Mechanical work Consider a force F acting on a particle that moves along a path. Let r be the position of the particle easured relative to the origin O. The work done by the force F when the particle moves an infinitesimal amount dr is defined as dw= F. dr (1) t
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D7 - Work and Energy So far we have used Newton’s second law F = ma to establish the instantaneous relation between the sum of the forces acting on a particle and the acceleration of that particle. Once the acceleration is known, the velocity (or position) is obtained by integrating the expression of the acceleration (or velocity). There are two situations in which the cumulative effects of unbalanced forces acting on a particle are of interest to us. These involve: a) forces acting along the trajectory. In this case, integration of the forces with respect to the displacement leads to the principle of work and energy. b) forces acting over a time interval. In this case, integration of the forces with respect to the time leads to the principle of impulse and momentum. It turns out that in many situations, these integrations can be carried beforehand to produce equations that relate the velocities velocities at the initial and final integration points. In this way, the velocity can be obtained directly, thus making it unnecessary to solve for the acceleration. We shall see that these integrated forms of the equations of motion are very useful in the practical solution of dynamics problems. In this lecture, we will concentrate on situation a), and consider the space integrated form of Newton’s second law. We will defer the discussion of b), time integrated equations, to lecture D9. Mechanical Work Consider a force F acting on a particle that moves along a path. Let r be the position of the particle measured relative to the origin O. The work done by the force F when the particle moves an infinitesimal amount dr is defined as dW = F · dr . (1) 1
That is, the work done by the force F, over an infinitesimal displacement dr, is the scalar product of F and dr. It follows that the work is a scalar quantity. Using the definition of the scalar product, we have that dw= F. dr= Fds cos a. where ds is the modulus of dr, and a is the angle between F and dr. Since dr is arallel to the tangent vector to the path, et, (i.e. dr= ds et), we have that F.et=Ft. Thus, dw Ft ds which implies that only the tangential component of the force"does "work During a finite increment in which the particle moves from position ri to position r2, the total work done by F is Fdr=fe 3) Here, s1 and s2 are the path coordinates corresponding to r1 and r2 Note Units of work In the international system, SI, the unit of work is the Joule (J). We have that 1 J=1N. m. In the English system the unit of work is the ft-lb. We note that the units of work and moment are the same. It is customary to use ft-lb for work and lb-ft for moments to avoid confusion. Principle of Work and Energy We now a consider a particle moving along its path from point ri to point r2. The path coordinates at points l and 2 are si and s2, and the corresponding velocity magnitudes ui and u If we start from (3)and use Newtons second law(F=ma) to express Ft= mat, we have F Here, we have used the relationship at ds v dv, which can be easily derived from at =i and v=s(see cture D4)
That is, the work done by the force F, over an infinitesimal displacement dr, is the scalar product of F and dr. It follows that the work is a scalar quantity. Using the definition of the scalar product, we have that dW = F · dr = F ds cos α, where ds is the modulus of dr, and α is the angle between F and dr. Since dr is parallel to the tangent vector to the path, et, (i.e. dr = ds et), we have that F · et = Ft. Thus, dW = Ft ds , (2) which implies that only the tangential component of the force “does” work. During a finite increment in which the particle moves from position r1 to position r2, the total work done by F is W12 = Z r2 r1 F · dr = Z s2 s1 Ft ds . (3) Here, s1 and s2 are the path coordinates corresponding to r1 and r2. Note Units of Work In the international system, SI, the unit of work is the Joule (J). We have that 1 J = 1 N · m. In the English system the unit of work is the ft-lb. We note that the units of work and moment are the same. It is customary to use ft-lb for work and lb-ft for moments to avoid confusion. Principle of Work and Energy We now a consider a particle moving along its path from point r1 to point r2. The path coordinates at points 1 and 2 are s1 and s2, and the corresponding velocity magnitudes v1 and v2. If we start from (3) and use Newton’s second law (F = ma) to express Ft = mat, we have W12 = Z s2 s1 Ft ds = Z s2 s1 mat ds = Z v2 v1 mv dv = 1 2 mv2 2 − 1 2 mv2 1 . (4) Here, we have used the relationship at ds = v dv, which can be easily derived from at = ˙v and v = ˙s (see lecture D4). 2
Defining the kinetic energy, as 1 re have that V12=T2-71 or T1+w12=T2 (5) The above relationship is known as the principle of work and energg, and states that the mechanical work done on a particle is equal to the change in the kinetic energy of the particle Example Block sliding down an incline a block is released from rest at the top of a ramp. The coefficient of kinetic friction between the surface of the ramp and the block is u. We want to determine the velocity of the block as a function of the distance traveled on the ramp, s N 2 nng The forces on the block are: the weight, mg, the normal force, N, and, the friction force, uN. We have that f an and Ft= mat. Since Fn =N-mg cos a and an =0, we have n= mg cos a. Thus Ft= mg sina-HN= mg sin a- umg cos a, which is constant. If we apply the principle of work and energy between the position (1), when the block is at rest at the top of the ramp, and the position(), when the block has travelled a distance s, we have T1=0, T2=(mu)/2, and the work done by Ft is simply W12= fts. Thus T1+W12=T2 mg(sin- ucos a)=mu From which we obtain, for the velocit 2g(sin a-u cos a)s We make two observations: first, the normal force, N, does no work since it is, at all times, perpendicular to the path, and second, we have obtained the velocity of the block directly without having to carry out ny integrations. Note that an alternative, longer approach would have been to directly use F= ma, and integrate the corresponding expression for the acceleration. The use of T to denote the kinetic energy, instead of K, is customary in dynamics textbooks
Defining the kinetic energy1 , as T = 1 2 mv2 , we have that, W12 = T2 − T1 or T1 + W12 = T2 . (5) The above relationship is known as the principle of work and energy, and states that the mechanical work done on a particle is equal to the change in the kinetic energy of the particle. Example Block sliding down an incline A block is released from rest at the top of a ramp. The coefficient of kinetic friction between the surface of the ramp and the block is µ. We want to determine the velocity of the block as a function of the distance traveled on the ramp, s. The forces on the block are: the weight, mg, the normal force, N, and, the friction force, µN. We have that Fn = man and Ft = mat. Since Fn = N − mg cos α and an = 0, we have N = mg cos α. Thus, Ft = mg sinα − µN = mg sin α − µmg cos α, which is constant. If we apply the principle of work and energy between the position (1), when the block is at rest at the top of the ramp, and the position (2), when the block has travelled a distance s, we have T1 = 0, T2 = (mv2 )/2, and the work done by Ft is simply W12 = Fts. Thus, T1 + W12 = T2 , or, mg(sin α − µ cos α) = 1 2 mv2 . From which we obtain, for the velocity, v = p 2g(sinα − µ cos α)s . We make two observations: first, the normal force, N, does no work since it is, at all times, perpendicular to the path, and second, we have obtained the velocity of the block directly without having to carry out any integrations. Note that an alternative, longer approach would have been to directly use F = ma, and integrate the corresponding expression for the acceleration. 1The use of T to denote the kinetic energy, instead of K, is customary in dynamics textbooks 3
Alternative expressions for dw We have seen in expression(2)that a convenient set of coordinates to express dw are the tangential- normal- binormal coordinates. Alternative expressions can be derived for other coordinate systems. For instance, we can express dw= F.dr in dw= Fr dr Fy dy+F2 da cylindrical(polar) coordinates, dw Fr dr+ Fo rde+ F2 dz or spherical coordinates, dw= Fr dr Fe rcos d8 Fo rdo As an illustration, let's calculate the work done by a constant force, such as that due to gravity. The force on a particle of mass m is given by F= -mgk. When the particle moves from position r1=11i+913+a1k to position r2=12i+y2]+22k, work is done, and the work may be written as IE.dr=/ ower In many situations it is useful to consider the rate at which a device can deliver work. The work per unit ower, P. Thus, dm The unit of power in the SI system is the Watt(W). We have that 1 W=1J/s In the English system the unit of power is the ft-1b/s. A common unit of power is also the horse power(hp), which is equivalent to 550 ft-lb/s, or 746 W Note Efficiency The ratio of the power delivered out of a system. the power delivered in to the system, Pin, is called he efficiency, e, of the system. This definition assumes that the energy into and out of the system flows continuously and is not retained within the system. The efficiency of any real machine is always less than unity since there is mechanical energy dissipated as heat due to friction forces
Note Alternative expressions for dW We have seen in expression (2) that a convenient set of coordinates to express dW are the tangential- normalbinormal coordinates. Alternative expressions can be derived for other coordinate systems. For instance, we can express dW = F · dr in: cartesian coordinates, dW = Fx dx + Fy dy + Fz dz , cylindrical (polar) coordinates, dW = Fr dr + Fθ rdθ + Fz dz , or spherical coordinates, dW = Fr dr + Fθ r cos φ dθ + Fφ rdφ . As an illustration, let’s calculate the work done by a constant force, such as that due to gravity. The force on a particle of mass m is given by F = −mgk. When the particle moves from position r1 = x1i+y1j +z1k to position r2 = x2i + y2j + z2k, work is done, and the work may be written as W12 = Z r2 r1 F · dr = Z z2 z1 −mg dz = −mg(z2 − z1) . Power In many situations it is useful to consider the rate at which a device can deliver work. The work per unit time is called the power, P. Thus, P = dW dt = F · dr dt = F · v . The unit of power in the SI system is the Watt (W). We have that 1 W = 1 J /s. In the English system the unit of power is the ft-lb/s. A common unit of power is also the horse power (hp), which is equivalent to 550 ft-lb/s, or 746 W. Note Efficiency The ratio of the power delivered out of a system, Pout, to the power delivered in to the system, Pin, is called the efficiency, e, of the system. e = Pout Pin . This definition assumes that the energy into and out of the system flows continuously and is not retained within the system. The efficiency of any real machine is always less than unity since there is always some mechanical energy dissipated as heat due to friction forces. 4
ADDITIONAL READING J. L. Meriam and L G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 3/6
ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 3/6 5
