Problems of this second type are often more formidable, as the inte- inte-gration may be difficult to carry out, particularly when the force is a

mixed function of two or more motion variables. In practice, it is

fre-quently necessary to resort to approximate integration techniques,

ei-ther numerical or graphical, particularly when experimental data are

involved. The procedures for a mathematical integration of the

accelera-Article 3/3 Equation of Motion and S o l u t i o n of Problems 125

tion when it is a function of the motion variables were developed in Art.

2/2, and these same procedures apply when the force is a specified func-tion of these same parameters, since force and accelerafunc-tion differ only by the constant factor of the mass.

Constrained and Unconstrained Motion

There are two physically distinct types of motion, both described by Eq. 3/3. The first type is unconstrained motion where the particle is free of mechanical guides and followrs a path determined by its initial motion and by the forces which are applied to it from external sources. An air-plane or rocket in flight and an electron moving in a charged field are examples of unconstrained motion.

The second type is constrained motion where the path of the parti-cle is partially or totally determined by restraining guides. An ice-hockey puck is partially constrained to move in the horizontal plane by the surface of the ice. A train moving along its track and a collar sliding along a fixed shaft are examples of more fully constrained motion. Some of the forces acting on a particle during constrained motion may be ap-plied from outside sources, and others may be the reactions on the parti-cle from the constraining guides. All forces, both applied and reactive, which act on the particle must be accounted for in applying Eq. 3/3.

The choice of an appropriate coordinate system is frequently indi-cated by the number and geometry of the constraints. Thus, if a particle is free to move in space, as is the center of mass of the airplane or rocket in free flight, the particle is said to have three degrees of freedom since three independent coordinates are required to specify the position of the particle at any instant. All three of the scalar components of the equa-tion of moequa-tion would have to be integrated to obtain the space coordi-nates as a function of time.

If a particle is constrained to move along a surface, as is the hockey puck or a marble sliding on the curved surface of a bowl, only twro coor-dinates are needed to specify its position, and in this case it is said to have two degrees of freedom. If a particle is constrained to move along a fixed linear path, as is the collar sliding along a fixed shaft, its position may be specified by the coordinate measured along the shaft. In this case, the particle would have only one degree of freedom.

Free-Body Diagram

When applying any of the force-mass-acceleration equations of mo-tion, you must account correctly for all forces acting on the particle. The only forces wrhich we may neglect are those whose magnitudes are negli-gible compared with other forces acting, such as the forces of mutual at-traction between two particles compared writh their attraction to a celestial body such as the earth. The vector sum IF of Eq. 3/3 means the vector sum of all forces acting on the particle in question. Likewise, the corresponding scalar force summation in any one of the component di-rections means the sum of the components of all forces acting on the particle in that particular direction.

COAi^{C /}

1 2 6 C h a p t e r S K i n e t i c s of P a r t i c l e s

The only reliable way to account accurately and consistently for every force is to isolate the particle under consideration from all con-tacting and influencing bodies and replace the bodies removed by the forces they exert on the particle isolated. The resulting free-body dia-gram is the means by which every force, known and unknown, which acts on the particle is represented and thus accounted for. Only after this vital step has been completed should you write the appropriate equation or equations of motion.

The free-body diagram serves the same key purpose in dynamics as it does in statics. This purpose is simply to establish a thoroughly reli-able method for the correct evaluation of the resultant of all actual forces acting on the particle or body in question. In statics this resultant equals zero, whereas in dynamics it is equated to the product of mass and acceleration. When you use the vector form of the equation of mo-tion, remember that it represents several scalar equations and that every equation must be satisfied.

Careful and consistent use of the free-body method is the most im-portant single lesson to be learned in the study of engineering mechan-ics. When drawing a free-body diagram, clearly indicate the coordinate axes and their positive directions. When you write the equations of mo-tion, make sure all force summations are consistent with the choice of these positive directions. As an aid to the identification of external forces which act on the body in question, these forces are shown as heavy red vectors in the illustrations in this book. Sample Problems 3/1 through 3/5 in the next article contain five examples of free-body dia-grams. You should study these to see how the diagrams are constructed.

In solving problems, you may wonder how to get started and what sequence of steps to follow in arriving at the solution. This difficulty may be minimized by forming the habit of first recognizing some rela-tionship between the desired unknown quantity in the problem and other quantities, known and unknown. Then determine additional rela-tionships between these unknowns and other quantities, known and un-known. Finally, establish the dependence on the original data and develop the procedure for the analysis and computation. A few minutes spent organizing the plan of attack through recognition of the depen-dence of one quantity on another will he time well spent and will usually prevent groping for the answer with irrelevant calculations.

3 / 4 R E C T I L I N E A R M O T I O N

We now apply the concepts discussed in Aits. 3/2 and 3/3 to prob-lems in particle motion, starting with rectilinear motion in this article and treating curvilinear motion in Art. 3/5. In both articles, we will ana-lyze the motions of bodies which can be treated as particles. This simpli-fication is possible as long as we are interested only in the motion of the mass center of the body. In this case we may treat the forces as concur-rent through the mass center. We will account for the action of noncon-current forces on the motions of bodies when we discuss the kinetics of rigid bodies in Chapter 6.

Article 3/4 Rectilinear Motion 127

If we choose the ^-direction, for example, as the direction of the rec-tilinear motion of a particle of mass m, the acceleration in the y- and 2-directions will be zero and the scalar components of Eq. 3/3 become

TFX — max

W y = 0 (3/4) I FZ = 0

For cases where we are not free to choose a coordinate direction along the motion, we would have in the general case all three compo-nent equations

TFX — max

Wy = may (3/5)

IFz - maz

where the acceleration and resultant force are given by

a = ari + ay j + a,k a = Ja2 + a2 + a/

IF = IFvi + I Fvj + I F J t

|IF| " v i l Fxf + ( I F / +

ilFf-This vievu of a car-collision test suggests that very large accelerations and accompanying large forces occur throughout the system of the two cars.

The crash dummies are also subjected to large forces, primarily by the shoulder-harness/seat-belt restraints.

128 Chapter S K i n e t i c s of P a r t i c l e s

S a m p l e P r o b l e m 3/1

A 75-kg man stands on a spring scale in an elevator. During the first 3 sec-onds of motion from rest, the tension T in the hoisting cahle is 8300 N. Find the reading if of the scale in newtons during this interval and the upward velocity v of the elevator at the end of the 3 seconds. The total mass of the elevator, man, and scale is 750 kg.

Solution. The force registered by the scale and the velocity both depend on the acceleration of the elevator, which is constant during the interval for which the forces are constant. From the free-body diagram of the elevator, scale, and man taken together, the acceleration is found to be

[EF^ = mav] 8300 - 7360 = 750a^y ay = 1.257 m/s2

The scale reads the downward force exerted on it by the man's feet. The equal and opposite reaction if to this action is shown on the free-body diagram of the man alone together with his weight , and the equation of motion for him gives

The velocity reached at the end of the 3 seconds is

[Ali =

J

a dt] v - 0 =

J

1.257 dt v = 3.77 m/s Ans.

J !

Helpful Hint

(T) If the scale were calibrated in kilo-grams it would read 830/9.81 = 84.6 kg which, of course, is not his tine mass since the measurement was made in a noninertial (accelerat-ing) system. Suggestion: Rework this problem in U.S. customary units.

S a m p l e P r o b l e m 3/2

A small inspection car with a mass of 200 kg runs along the fixed overhead cable and is controlled by the attached cable at A. Determine the acceleration of the car when the control cable is horizontal and under a tension T 2.4 kN.

Also find the total force P exerted by the supporting cable on the wheels.

Solution. The free-body diagram of the car and wrheels taken together and treated as a particle discloses the 2.4-kN tension T, the weight W mg = 200(9.81) = 1962 N, and the force P exerted on the wheel assembly by the cable.

The Car is in equilibrium in the y-direction since there is no acceleration in this direction. Thus,

[ I F , = 0] P - 2.4(^3) - 1 . 9 6 2 ( j f ) = 0 In the x-direction the equation of motion gives [LFt = maj 2400(TI) - 1 9 6 2 ( f ; ) = 200c

(Î) By choosing oui- coordinate axes along and normal to the direction of the acceleration, we are able to solve the two equations independently.

Would this be so if x and y were cho-sen as horizontal and vertical?

Article 3/4 R e c t i l i n e a r Motion 129

S a m p l e P r o b l e m 3 / 3

The 250-lb concrete block A is released from rest in the position shown and pulls the 400-lb log up the 30° ramp. If the coefficient of kinetic friction between the log and the ramp is 0.5, determine the velocity of the block as it hits the ground at B.

Solution. The motions of the log and the block A are clearly dependent. Al-though by now it should be evident that the acceleration of the log up the incline is half the downward acceleration of A, we may prove it formally. The constant total length of the cable is L = 2s^L- + s^A + constant, where the constant accounts for the cable portions wrapped around the pulleys. Differentiating twice with re-spect to time gives 0 = 2 sc + s^, or

0 = 2 ac + aA

We assume here that the masses of the pulleys are negligible and that they turn with negligible friction. With these assumptions the free-body diagram of the pulley C discloses force and moment equilibrium. Thus, the tension in the cable attached to the log is twice that applied to the block. Note that the acceler-ations of the log and the center of pulley C are identical.

The free-body diagram of the log shows the friction force N for motion up the plane. Equilibrium of the log in the y-direction gives

[ZFV = 0] N - 400 cos 30° = 0 N = 340 lb

and its equation of motion in the x-direction gives

[ZF^z = m a j 0.5(346) - 2 T + 400 sin 30° = a^c

For the block in the positive downward direction, we have

l+l IF^ma] 250-T=^aA

Solving the three equations in ac, Q^, and T gives us

a^A = 5.83 ft/sec² a^c = - 2 . 9 2 ft/sec² T 205 lb

For the 20-ft drop with constant acceleration, the block acquires a velocity

[v2 = 2ax] vA = ¿2 (5.831(20) 15.27 ft/sec Ares.

2501b 1 +

Helpful Hints

(T) The coordinates used in expressing the final kinematic constraint rela-tionship must be consistent with those used for the kinetic equations of motion.

(2) We can verify that the log wdl in-deed move up the ramp by calculat-ing the force in the cable necessary to initiate motion from the equilib-rium condition. This force is 2T 0.5JV + 400 sin 30" = 373 lb or T 186.5 lb, which is less than the 250-lb weight of block A. Hence, the log will move up.

(3) Note the serious error in assuming that T 250 lb, in which case, block A would not accelerate.

(4) Because the forces on this system re-main constant, the resulting acceler-ations also remain constant.

130 Chapter S K i n e t i c s of P a r t i c l e s

S a m p l e P r o b l e m 3 / 4

The design model for a new ship has a mass of 10 kg and is tested in an exper-imental towing tank to determine its resistance to motion through the water at various speeds. The test results are plotted on the accompanying graph, and the resistance if may be closely approximated by the dashed parabolic curve shown. If the model is released when it has a speed of 2 m/s, determine the time t required for it to reduce its speed to 1 m/s and the corresponding travel distance x.

Solution. We approximate the re si stance-velocity relation by if kv2 and find k by substituting if = 8 N and u = 2 m/s into the equation, which gives k = 8/22 2 N • s2!m². Thus, if 2c2,

The only horizontal force on the model is if, so that

© [ZFX = maz] - i f = max or -2v2 = 10 ^ I iL1: dt We separate the variables and integrate to obtain

Jo^J2 V² \V 2}

Thus, when V = VQ/2 1 m/s, the time is t 5 ( j - *) = 2.5 s.

The distance traveled during the 2.5 seconds is obtained by integrating v dxidt. Thus, v = 10/(5 + 2t) so that

(2) Suggestion: Express the distance x after release in terms of the velocity v and see if you agree with the re-sulting relation x 5 In (v0/u).

S a m p l e P r o b l e m 3 / 5

The collar of mass in slides up the vertical shaft under the action of a force F of constant magnitude but variable direction. If 0 • kt where k is a constant and if the collar starts from rest with ft 0, determine the magnitude F of the force which will result in the collar coming to rest as if reaches jr/2. The coeffi-cient of kinetic friction between the collai- and shaft is t-i/,.

Solution. After drawing the free-body diagram, we apply the equation of mo-tion in the y-direcmo-tion to get

where equilibrium in the horizontal direction requires N F sin 0. Substituting 0 = kt and integrating first between general limits give

Helpful Hints

Article 3/4 P r o b l e m s 131

PROBLEMS

Introductory Problems

3/1 During a brake test, the rear-engine car is stopped from an initial speed of 100 km/h in a distance of 50 m. If it is known that all four wheels contribute equally to the braking force, determine the braking force F at each wheel. Assume a constant deceleration for the 1500-kg car.

Aits. F = 2890 N

50 m

= 100 km/h = 0

Problem 3/1

3 / 2 The 50-kg crate is stationary when the force P is ap-plied. Determine the resulting acceleration of the crate if (a) P 0, (b) P 150 N, and (c) P = 300 N

3 / 4 The 300-Mg jet airliner has three engines, each of wliich produces a nearly constant thrust o f 2 4 0 kN dur-ing the takeoff roll. Determine the length s of runway required if the takeoff speed is 220 km/h. Compute s first for an uphill t akeoff direction from A to B and sec-ond for a downhill takeoff from B to A on the slightly inclined runway. Neglect air and rolling resistance.

Horizontal

Problem 3/4

3 / 5 The 10-Mg truck hauls the 20-Mg trailer. If the unit starts from rest on a level road with a tractive force of 20 kN between the driving wheels of the truck and the road, compute the tension T in the horizontal drawbar and the acceleration a of the rig.

Ans. T = 13.33 kN, a = 0.067 m/s2

Problem 3/5

Problem 3/2

3 / 3 At a certain instant, the 80-lb crate has a velocity of 30 ft/sec up the 20" incline. Calculate the time t re-quired for the crate to come to rest and the corre-sponding distance d traveled. Also, determine the distance d' traveled when the crate speed has been re-duced to 15 ft/sec.

Ans. t = 1.615 sec, d 24.2 ft, d' 18.17 ft

3 / 6 A skier starts from rest on the 40° slope at time t 0 and is clocked at t 2.58 s as he passes a speed check-point 20 m down the slope. Determine the coefficient of kinetic friction between the snow and the skis. Ne-glect wind resistance.

P r o b l e m 3/6 Ms

ßk

= 0.40

= 0.25

P r o b l e m 3/3

1 3 2 Chapter S K i n e t i c s of P a r t i c l e s

3 / 7 Calculate the vertical acceleration a of the 100-lb cylinder for each of the two cases illustrated. Neglect friction and the mass of the pulleys.

Ans. (a) a = 6.44 ft/sec2

{6} a = 16.10 ft/see12

(a) (b)

Problem 3/7

3 / 8 The 170-lb man in the bosun's chair exerts a pull of 60 lb on the rope for a short interval. Find his acceler-ation. Neglect the mass of the chair, rope, and pulleys.

Problem 3/8

3 / 9 A man pulls himself up the 15° incline by the method shown. If the combined weight of the man and cart is 250 lb, determine the acceleration of the cart if the man exerts a pull of 60 lb on the rope. Neglect all fric-tion and the mass of the rope, pulleys, and wheels.

Ans, a = 14.85 ft/sec²

3 / 1 0

Problem 3/9

The 750,000-lb jetliner A has four engines, each of which produces a nearly constant thrust of 40,000 lb during the takeoff roll, A small commuter aircraft B taxis toward the end of the runway at a constant speed VJJ 15 mi/hr. Determine the velocity and ac-celeration which A appeal's to have relative to an ob-server in B 10 seconds after A begins its takeoff roll.

Neglect air and rolling resistance.

S A

y I t 1

30°

3/11

Problem 3/10

A car is descending the hill of slope 0j with the brakes slightly applied so that the speed u is con-stant. The slope decreases abruptly to 02 at point A.

If the driver does not change the braking force, deter-mine the accelerat ion a of the car after it passes point A. Evaluate your expression for flj 6° and <)² = 2°.

Ans. a = g(sin b² — sin 0{), a = -0.0696^

A r t i c l e 3/4 P r o b l e m s 1 3 3

v = constant

Problem 3/11

3/12 The block-and-tackle system is released from rest with all cables taut. Neglect the mass and friction of all pulleys and determine the acceleration of each cylinder and the tensions T\ and T-A in the two cables.

Problem 3/12

3/13 Determine the tension P in the cable which will give the 100-lb block a steady acceleration of 5 ft/see up the incline.

Ans. P 43.8 lb

fjk = 0.25

3/14 A toy train has magnetic couplers whose maximum attractive force is 0.2 lb between adjacent cars. What is the maximum force P with which a child can pull the locomotive and not break the train apart at a coupler? If P is slightly exceeded, which coupler fails? Neglect the mass and friction associated with all wheels.

2 oz 5 oz

1 oz 1 oz l o z 1 B ~ L H

Problem 3/14

Representative Problems

3/15 A train consists of a 400,000-lb locomotive and one hundred 200,000-lb hopper cars. If the locomotive exerts a friction force of 40,000 lb on the rails in starting the train from rest, compute the forces in couplers 1 and 100. Assume no slack in the couplers and neglect friction.

Ans. Ti = 39,200 lb, T1(W = 392 lb

^ r u s ^ û s

. j m B I S ,

100 99 98 3 2 1 Problem 3/1S

3/16 A .small package is deposited by the conveyor belt onto the 30° ramp at A with a velocity of 0.8 m/s.

Calculate the distance s on the level surface BC at which the package comes to rest. The coefficient of kinetic friction for the package and supporting sur-face from A to C is 0.3.

Problem 3/16

Problem 3/13

1 3 4 Chapter 3 K i n e t i c s of P a r t i c l e s

3 / 1 7 The steel ball is suspended from the accelerating frame by the two cords A and B. Determine the ac-celeration a of the frame which will cause the ten-sion in A to be twice that in B.

In document Engineering Mechanics Dynamics 6th Edition (Page 133-166)