Engineering mechanics irving shames pdf download

If you want to download this ebook, i provide downloads as a pdf, kindle, word, txt, ppt, rar and zip. Download pdf Engineering Irving H. Shames-Engineering Mechanics Statics and Dynamics. Shames Solution manual include answers for all chapters of textbook chapters 1 to Mechanics Free Engineering Mechanics Shames Solution - seapa. Right here, we have countless ebook engineering mechanics shames solution and Shames as PDF for free.

Solutions - Engineering Mechanics: Dynamics By. Fill epub: Try Risk Free Join our Issuu is a This irving h shames engineering mechanics download pdf, as one of the most full of zip Engineering Mechanics-Irving H. Shames Designed to provide a more In probleins o f mech;mics. For example. X are lrce vectors a s far as total dis- Lance traveled i h concerned.

Under such circunislainces the vectors are called truri. IO, we may apply the lorce anywhere alung t l i t rupc AH or inay piish at point C. The resulting motion i s lhe same in all cases. The point may he represented as the tail or head of the arrow iii thc graphical representation.

For this case. Under such circumstances. For example, if we are interested in the deformation induced by forces in the body in Fig. Clearly, force F will cause a different deformation when applied at point C than Figure 1. F i s transmissible for towing.

The force is thus a bound vector for this problem. We shall be concerned throughout this text with considerations of equivalence. Never- theless, for the student to comprehend these laws sufticiently to undertake novel and varied problems, much study will be required. We shall now discuss briefly the following laws, which are considered to be the foundation of mechanics: 1. Newtons first and second laws of motion. Newtons third law. The gravitational law of attraction. The parallelogram law.

Every particle continues in a state of rest or uniform motion in a straight line upless it is compelled to change that state by forces imposed on it. The c b g c of motion is proportional to the naturn1;ferCe impressed and is made in a direction of the straight line in which the force is impressed. Notice that the words rest, uniform motion, and change of motion appear in the statements above. For such information to he meaningful, we must have some frame of reference relative to which these states of motion can be described.

We may then ask: relative to what reference in space does every particle remain at rest or move uniformly along a straight line in the absence of any forces? Or, in the case of a force acting on the particle, relative to what reference in space is the change in motion proportional to the force? Experiment indicates that the fixed stars act as a reference for which the first and second laws of Newton are highly accurate.

Later, we will see that any other system that moves uniformly and without rotation relative to the fixed stars may be used as a reference with equal accuracy. All such references are called inertial references. The earths surface is usually employed as a refer- ence in engineering work. However, the departure i s xi small Sor m o s t situiitiiins cxccptions arc the motion iif guided missile!

We shall, therefore, usually consider the earth's s u r l x c as an inertial reference, but w i l l keep i n mind the somewhat appr iximatc iiaturu of this stcp.

Many situii- rions f a l l into this category. The study of bodies in equilibrium i s called S I U I - i c s. In addition tn the reference limitations explained above. As pointed out carlicr.

Ncar the spccd of light, they are untenable. In ciinhidci-ing the motion of high- energy elementary particles occurring i n nuclear phenomena, however, we cannot ignore relativistic effects. Finally, when we get down t o very small dis- tances. In this case, we must rexiit to quantum mechanics. Newton's Third Law. Newton stated in his third law:.

To every action rhere is always opposed an equul rcucrion, or the mutual actiuns of mu bodies upon euch other are ulwuys equal arid directed to contrary points. This i s illustrated graphically i n Fig. I I , where the action and reaction between two bodies arise Srom direct contact. Other imporrant actions i n which Newton's third law holds arc gravitationdl attractions to be discussed next and electrostatic forces between charged pat-ticks. I t should he pointed out that there are actiiiiis that dii nut fiillow this law.

Law of Gravitational Attraction. It has alrcady been piiintcd out that these i s an attraction between the earth and the bodies at its surface, such as A and B. This attraction is mutual and Newtons third law applies.

There is also an attraction between the two bodies A and B themselves, but this force because of the small size of both bodies is extremely weak. However, the mechanism for the mutual attraction between the earth and each body is the same as that for the mutual attraction between the bodies. These forces of attraction may be given by the law of gravitational attractiun:. Two anicles will be attracted toward each other along their connecting line ith a force whose magnitude is directly proponional to the product of the masses and inversely proportional to the distMce squared between the pqnicles.

In the actions involv- ing the earth and the bodies discussed above, we may consider each body as a particle, with its entire mass concentrated at its center of gravity. Parallelogram Law. Stevinius 1 was the first to demonstrate that forces could be combined by representing them by arrows to some suitable scale, and then forming a parallelogram in which the diagonal repre- sents the sum of the two forces. As we pointed out, all vectors must combine in this manner.

These hasic, and from them secondary, dimensions may be related by dimensionally homoge- ncous equations which, with suitable idcalirations, can represent certain actions i n nature. The baric laws of mechanics were thus introduccd. Since the equations of these laws relate vector quantities, we shall introduce a use- ful and highly dercriplive set of vector operations in Chapter 2 in order to learn to handle these laws effectively and to gain more insight into mechanics in general.

These operations are generally c:illcd. Check-Out for Sections with 'i 1. What are two kinds of limitations on Newtonian mechanics'? What are the two phenomena wherein mass plays a key role? If a pound force is defined by the extension of a standard spring, define the pound mass and the slug.

Express mass density dimensionally. How many scale units of mass density mass per unit volume in the SI units are equivalent to I scale unit in the American system using a slugs, ft, sec and b Ibm, ft, sec? The proportionality constant pis called the coefficient of viscosity. What is its dimensional representation? Define a vector and a scalar. What is meant by line of action o f a vector?

What is a di. What is an inertial reference? Vector Algebra The basic algebraic operations for the handling of scalar quantities are those famil- iar ones studied in grade school, so familiar that you now wonder even that you had to be introduced to them. For vector quantities, these methods may be cumbersome since the directional aspects must be taken into account.

Therefore, an algebra has evolved that clearly and concisely allows for certain vely useful manipulations of vectors. It is not merely for elegance or sophisti- cation that we employ vector algebra. Indeed, we can achieve greater insight into the subject matter-particularly into dynamics-by employing the more powerful and descriptive methods introduced in this chapter.

Note that the magnitude of a quantity is its absolute value. The mathe- matical symbol for indicating the magnitude of a quantity is a set of vertical lines enclosing the quantity. The instmcter may wish tu assign the reading of these seclians along with the aforemen- tioned questions. The definitinn 71 the product o i vectoi A h y wiliir iii,written simply iis m A , i s given i n the following IiiiinncI:.

The magnitude it. A tI n-ci lhc final iirrou' ciiii then he iiiterpretcd in teiniir o i i t s length by cinployinf tlic chosen scale f k t o r. Addition hy pmillclograiii iau. Example 2. Onc force has a magnitude of 10 Ib act- ing in the positive x direction, whereas the other has a magnitude of 5 Ib acting at an angle of " with a sense directed away from the origin. A- li 0 IOlb Figure 2.

Find F and a using trigonometry. Thus, using triangle OBA,. The direction of the vector may he described by giving the angle and the sense. The angle is determined by employing the law of sines for triangle OBA.? Thc total unstretched length of the rubber band i s 5 in. The top view 01 the slingshot i s shown in Fig.

The change in overall length of the ruhher hand A L from i t s unstretchcd length i s. The teiisiiin i n the entire extended rubber hand i s then II. Simplc dingshut. We ciin use the law Figure 2.

A more direct calculation can be used by considering two right triangles within the chosen triangle. Parsllclogram orcrs. It must be emphatically pointed out that thc additiim if vectorsA and R only involves the vectors themselves and iiot thcir lines of actions or thcir posi- tions along their respective lincs 0faction.

That is. The sum F then closes the triangle and is OB. Also, in Fig. The vectors are connected in Fig. The sum of the vectors then is the dashed vector that closes the polygon. Nevertheless, it is seen that the sum is the same vector as in Fig. Clearly, the order of laying off the vectors is not significant.

Figure 2. Subtraction of vectors. A simple physical interpretation of the above vector sum can he formed for vectors each of which represents a movement of a certain distance and , direction i. Then, traveling along the system of given vectors you start from one point the tail of the first vector and end at another point the head of the last vector.

The vector sum that closes the polygon is equivalent to the system of given vectors, in that it takes you from. The extension of this proce- dure to any number of vectors is obvious. This process may also be used in the polygon construction. Addition and we proceed a s shown in Fig. Again, the order of the process is not sig- subtraction using polygon nificant, as can be seen in Fig. Add ii N force pointing in the positive r direction to a magnitude of force 8 and the direction of forcc C?

For the Eim- N forcc at an nnple 45" to the. Do this first graphically. Fieure P. Whal 2. At the outset, i t goes rluc east for 5 km, then due north for 7 kin. See Fig. Figure P. A homing pigcon i s released at point A and is observed. I t flier I O krn due south, then gocs duc east for IS km. Next i t goes southcast for I O k m and finally gocs due south 5 km to reach i t s destination H.

What is the Figure P. If the difference between forces B and A in Fig. A man pulls with force Won a rope through a simple fric- force D having a magnitude of 25 N, what i s the magnitude of B tionless pulley to raise a weight W. What total force is exerted on and the direction o f D? What is the sum ofthe forces transmitted by the structural rods to the pin at A? Suppose in Problem 2. If we do not change the force transmitted by the horizontal 2.

Add the three vectors using the parallelogram law twice. What is the total force? Give the magnitude of 2. Using the parallelogram law, find the tensile force in the sum and the angle if forms with f h e x axis. We will do this problem differ- ently in Example S. In the preceding problem, what should the angle 6 be so 2. A mass M is supported by cables I and 2. The tension that the sum of the forces from cable DE and cable EA is colinear in cable I is N, whereas the tension in 2 is such as to main- with the boom CE?

You will leam very shortly that the weight of M must be equal 2. Three forces act on the block. The N and the N and opposite to the vector sum of the supporting forces for forms act, respectively, on the upper and lower faces of the block, equilibrium. Give the magnitude of the sum nf these forces using the parallelogram law twice. Two foothall player, are pushing a hlocking dummy. What i s the total furce cxertcd on the dummy hy the players'! Dn prohlcin 2. Twu soccer player5 approach a stntir,nary hall I O ft away t r i m the gnal.

Does the offense score asuming lhiit the g d i c i s asleep '! Thc rollers on thc side u l the hlock tlo not 70 Ih crmtrihute to thc vertical support nf the hlock. The wire5 cnnnect til the gcrrmerric center of the hlock C. Form three independent equations for any given a involving the unknowns f ;.

Stipulated directions 2. Two-dimensional original vector. We shall discuss three- dimensional resolution involving three noncoplanar component vectors later in the section. The two-dimensional resolution citn he accomplished by graphical construction 0 1 the parallelogram. Thc two vectors C , and C , formed in this way are the compomnt vectors.

Wc olten replace a vector b y its components siticc the cnniponents are alway:, cquivalcnt i n rigid-body nicchanics to the original vector. When this i s done. I I wherein a sailboat is going from marker A to marker I3 5, meters apart. Clearly the displacement vectofl pARis equivalent to the vector sum of displacement vectors pAc plus pcR in that the same starting points A , and the same destination points B,are involved in each case.

Thus, vectors pAc and pcR are two-dimensional components of vector pas Accordingly, we can show a parallelogram for those vectors for which triangle ABC forms half of the parallelogram see Fig. We leave it for you to jus- tify the various angles indicated in the diagram. Now we first use the law if sines. Sailboat tacking. And BC - s,oon sin y sinn. Enlarged parallelogram.

It is also readily possible to find three components not in the. This is the aforementioned three-dimensional resolution. Consider the specification of three orthopnal directions' for the resolution of C positioned in the first quadrant, as is shown in Fig. The resolution may be accomplished in two steps. Resolve C along the z direction, and along the. Orthogonal or the orthogonal directions are used most often in engineering practice.

Next take vector C,, and r c s o h e it along axes. Hence C ,. Now le1 us consider the right triangle.

This sh iuld spui- you lo obscrvc care i n your notation Sotnctimcs only o w of the sciiliir iirthiigiiiiiil coniporients o f a ngular sciilar coniponciit i h desired. Rcctaogular component OF C. However, it is only the single component C, that we often use, disregarding other rectangular components. It is always the case that the triangle formed by the vector and its scalar rec- tangular component is a right triangle. In establishing C, we therefore speak of dropping a perpendicular from C to s or of projecting along s.

The scalar rectangular component Cycould also be the result of a fwo- dimensionul orthogonal resolution wherein the other component is in the plane of C and Cyand is normal to C,. It is important to remember, however,. As a final consideration, let us examine vectors A and B , which, along with directions, form a plane as is shown in Fig.

The sum of the vectors A and B is found by the parallelogram law to be C. We shall now show that the projection ofC along s is the same as the sum of the projections of the two-dimensional components of A and B , taken along s.

The vector a is called a unit vector. The unit vector is also at times denoted ash. You will write it as 6. It has no dimensions. We formu- late this vector as follows:. This will be determined entirely by its use. However, we can represent the vector D , shown in Fig. Unit vector a.

It thus acts as a free vector. Next, if a given vector is represented using a lowercase letter, such as the vector r , then we oflen make use of the circuni- flex mark to indicate the. As will be seen Iaitcr. IOb as the corresponding scalar equations. The purpose of this rectangular parallelepiped and diagonal is to allow for the easy determination of the ori- entation of the line of action and hence the orientation of a vector.

AB in the diagram is such a diagonal used for the determination of the line of action of vector F. Numbers for this purpose are shown along the sides of the rectan- gular parallelepiped without units. Any set of numbers can be used as long as the ratios of these numbers remain the ones required for the proper determi- nation of the orientation of the vector. That determination proceeds by first replacing the displacement vector p,,, from comer A to comer B by a set of three vector displacements going from A to B along the sides of the rectangu- lar parallelepiped.

We thereby can replace the vector p,, by the sum of its rectangular components. Thus, for the case shown in Fig. Rectangular parallelepiped used for specifying the direction of a vector. XImagineyou are "walking" from A to B hut restricting your movements to he along the coordinate directions. This movement i s equivalent to going directly from A to B in that the Same endmints result. In the two-dimensional case. This i s shown i n Fig. Here we can say. Other ways to use the reclangular parallelepiped.

Right triangle used fur sppecifying the direction ot it vector m two dirnensionq. There are times when the rectangular parallelpiped i s not shown explic- itly.

The simplest procedure i s IO mcntally move from the beginning point o f the diagonal to the final point always moving along coordinatc directions, or, i n other words, always mov- ing along the sides 0 1 the hypothetical rectangular parallelepiped.

This would take us from initial point A to final point IT. The corresponding displacement veclor would then he. Note that D is at the center of the outer edge of the crate; C is 1. What are the forces Fl, F2, and Fi transmitted by the cables? We will soon l e a n formally what our common sense tells us, namely that the vector sum of force F,, force 5.

We first express these three forces in terms of rectangular components. Ii -. A crate is supported by three forces. Resolve thc h force into a set of components along the slot shown and in the vertical direction. Two tughants are maneuvering an w e a n liiw The desired iota1 inrcc i s 3, Ih at an angle r r f I S " a h cliown i n the diagram. A lamer needc to build a fence from the corner of his ham to the corner of hic chicken house 10 m away in the NE dircction.

However, he wants to enclose ils much of the harnyard as possi- hie. Thus, he nins thc fcncc cmt, from thc corner of his ham to thc prnpeny line and then NNE to thc corncr of his chickcn housc. How long i s the fencc? If the component along AH i s N. A simple truss to be studied later in detail supports two forces.

If the forces in the members are colinear with the mem- hcrs, what arc the forces in thc mcmbcrs? Him; The lorccs in thc mcmherq must have a vector w m q u a l and opposite to the vcctoi sum of E; and F,. The entire Eystem i s coplanar. Two men are trying to pull a crate which will not move y until a lb total force is applied in any one direction.

Man A can pull only at 45' to the desired direction of crate motion, whereas man B can pull only at 60" to the desired motion. What force must each man exert to start the box moving as shown? The orthogonal components of a force are: x comvonent 10 Ib in oositive x direction y component 20 Ib in positive y direction z component 30 Ib in negative z direction a What is the magnitude of the force itself? What are the rectangular components of the lb force? What IS the sum of the three forces?

The 2,N force IS , in the y z plane. What is the vector sum of these forces? The N force is to be resolved into components along the AC and AB directions in the xy plane measured by the angles a and p.

What is y? How long mist rncmhel-c OA. HC, and 'E hi. What is thc orthogonal total f h x cwnponcnt in tlic. I direction 01 the ioice tiansmittcd to pin A of a roo1 t n h i h i tlic four rncrnher,'! What is the total cirmpi,nent in the? Express the N force in terms of the unit vectors i, j , 2. Express the unit vectors i,j , and k in terms of unit vectors and k.

What is the unit vector in the direction of the N force'! These are unit vectors for cvlindricul coordinates. The force lies along diagonal AB. Express the 1,lb force going through the origin and through point 2, 4, 4 in terms of thc unit vectors i, j , k and E? See the footnote o n p. In effect, two vectors, force and displacement, are employed to give a scalar, work.

In other physical problems, vectors are associated in this same manner so as to result in a scalar quantity. A vector operation that represents such operations con- cisely is the scalar product or dot product , which, for the vectorA and B in Fig. I 1 where a is the smaller angle between the two vectors. Note that the dot prod- uct may involve vectors of different dimensional representation, and may be positive or negative, depending on whether the smaller included angle a is less than or greater than 90".

The appro- priate sign must, of course, be assigned positive if the projected component of vector A and vector B point in the same direction; negative, if nol.

The work concept for a force F acting on a particle moving along a path described by s can now be given as. As with addition and subtraction of vectors, the dot product operation involves only the vectors themselves and not their rcspective lines of action.

Accordingly, for a dot product of two vectors. Remember in so doing we must not alter the magnitudcs and directions of the vectors. Let us next consider the scalar product of mA and n u. From the definition, clearly thc dot product is commutnrivr. How- ever, in Section 2. Thus, the dot product is distributive. The scalar product between unit vectors will now be carried out. I I is 90". On the other hand. We can thus conclude that the dot product of equal orthogonal unit vectors for a given reference is unity and that of un- equal orthogonal unit vectors is zero.

Thus, wc see that B scalar producl of two vectors is the sum of the ordinary products of the respective components. The dot product may be of immediate use in expressing the scalar rec- tangular component of a vector along a given direction as discussed in Sec- tion 2. If you refer back to Fig.

Unit vector idirected from 0. Kvdio transmission tiiwers. Given the vectors 2. What is cos A, E? A block A is constrained to move along a 20' incline in 2. Given the vectors the yz plane. How far does the block have to move if the force F is to do I O ft-lb of work? A sailboat is tacking into a knot wind.

The boat has a velocity component along its axis of 6 kn but because of side slip X and water currents, it has a speed a speed of. What are the x and y components of the wind velocity and the boat velocity? What is the angle between the wind d o c - ity and the sailboat velocity'? An electrostatic field E exerts a force on a charged pani- cle of qE, where q is the charge of the particle.

The vector points away from the origin. Hint: Whenever simply a component is asked for, it is virtually Figure P. What is the anglc bctwccn the I ,N force and the axis 2. The force is i n thc diagonal plane GCDE.

A radio tiiwci is held by guy wires. Xk, what shuuld A hc'! What is the angle hetween F arid i? Find the dot prtiduct ul the vectors represented hy thz diagonal5 from A to I" arid trmn 1 to G. Whal is thc angle hetween them? What is the angle bctweeii the 1.

One such interaction is the moment of a force, which involves a special product of the force and a position vector to he studied in Chapter 3. To set up a convenient operation for these situations, the vecfor cross product has been established.

For the two vectors having possibly different dimensions shown in Fig. The angle a is the smaller of the two angles between the vectors, thus making sin a always positive. The vector C has an orientation normal to the plane of the vectors A and B. The sense, furthermore, corresponds to the advance of a right-hand screw rotated about C as an axis while turning from A to B through a-that is, from the first stated vector to the second stated vector through the smaller angle between them.

The reader can easily ver- ify this. The description of vector C is now complete, since the magnitude and direction are fully established. The line of action of C is not determined by the cross product; it depends on the use of the vector C. Again we remind you that the cross product, like the other vector alge- braic operations, does not involve lines of action, so in taking a cross product we can move the vectors so as to come together at their tails as in Fig.

As in the previous case, the coefficients of the vectors will multiply as ordinary scalars. This may he deduced from the nature of the definition. However, the commutative law breaks down for this product. P We can readily show that the cross product, like the dot product. To do this, consider in Fig. It will be left to the student to justil'y the given formulation for each of the vectors in Fig.

Since the prism is a closed surface, the net projected area in any direclion must be zero, and this. Prim using A , B, and C. Area vzctorc tor prism f i c m. Noting that the second and third expressions cancel each olher.

Here, the product of equal vector5 is zero hecause a and. The product i X j is unity i n magnitude. If the :a x i s has been erected in a seiise con-. Istt-hand triad. Diffcrent kinds of rcfcrcnccs.

For the products along the dashed diagonals, we must remember in this method tu multiply by -I. It must be cautioned that this method of evaluating a determinant is correct only for 3 x 3 determinants.

If the height of the pyramid i s I t. S00k ft? And once again. Substituting from Eqs. We see from this example that a plane surface can be represented as a vector, and if that plane surface i s part of a closed surface, by convention the area vector is in the direction of the outward normal.

A simple geometric meaning can be associated with this operation. E , and C as an arbitrary set of concurrent vec- tors. We have set up an xyz reference such that the A and B vectors are in the xy plime. Further, a parallelogram abcd in the xy p k n e is shown in the dia- gram. A and R in p planc. Using thi. It will he left iis an cxercise Prohleni 2. I n liiler chapters. Thc vector triple product i s a vector quantity and w i l l appear quitc o l t c n i n sttidier oidynamics.

It w i l l be left h r you lo demonstrate that. The normal n to the infinite plane must have three equal direction cosines. Hence, noting Eq. When ire simply identify- ing quantities i n il di. A correct representation o l this Iiirce i n a vector equation would he I.

Thus, in Fig. If the coordinates of vertex E of the inclined pyramid are x A. What is the magnitude of the resulting vector'! If vectors A and B in the xy plane have a dot product of SO units, and if the magnitudes of these vectors are I O units and R units, respectively.

Y Explain. What is the cross product of the displacement vector from A to B times the displacement vector from C to D'! I Figure P. In Prvblem 2. Compute the determinant I. Compare the result with the computation of A x B C by using the dot-product and crowproduct operations.

In Example 2. Give the results in kilo- Figure P. Fixure P. Check-Oulfor Sections with t 2. What sign iiwst i t hauc'? If so, dcscribc the rcsull. What are the distances that he must travel from Dallas to Topeka and from Topeka ti Chicago'! How long is E the road? H i m : See the compass-settings diagram, Fig. Dallas Figure P. Sum all forces acting on the block. Plane A is parallel to the. We will later m d y the special properties of two par- allel forces called a c o u p k that are opposile in direction atid 2.

What is thc cross product between the 1,N force and in magnitude' the diiplaccrnent vector p,,,"'? Four member, of a space frame are loaded as shown. What are the orthogunill scalar components of the forces on the ball 2. The r and I compunsnti of the force F arc known tu be joint at O? What i q thc f k c F and what are rectangular p;irallelepiped. What are the force components on the electron'?

The charge of the electron is l. A skeet shooter is aiming his gun at point A. What is the L. A 2,15,z m 2. For the line segment A X , determine cosines m and n. Using the scalar triple product, find the area projected onto the d a n e N from the surface ABC.

A lh crate is held up by three forces. What Suppose that an electron moves through a uniform magnetic field should forces F , and F, he for this condition'? Con- sider first the path of motion of a particle shown dashed in Fig. As indi- cated in Chapter I, the di,splacemenr vecfor p is a directed line segment connecting any two points on the path of motion, such as points I and 2 in Fig.

The displacement vector thus represents the shortest movement of the panicle to get from one position on the path of motion to another. The Figure 3. Displacement vector p purpose of the rectangular parallelepiped shown in the diagram is to convey hetween points 1 and 2.

We can readily express. The directed line segment r from the origin of a coordinate system to a. The notations R and p are also used for position vectors. You can conclude from Chapter 2 that the. The scalar components of a position vector are simply the coordi- Figure 3. Position vector. To express r in Cartesian components, we then have. Figure 3. Relation hetween a displacement vector and position Yectors.

Example 3. YK and XYZ. What arc the coor- dinates. From Fig. For Simple Cases. The moment of a force about a point 0 see Fig. And the direction of this vector is perpendicular to the plane of the point and the force, with a sense determined from the familiar right-hand-screw rule.

Case B. For Complex Cases. Another apprwach is to employ a position vcctcir r from point 0 to nnypoinf P along the line ol'action of force F as shown Figure 3.

The nioment M of F about point 0 will he shown to he given as1. For the purpose of forming the cross product, the vectors in Fig. Thus, we get the same magnitude of M as with the elementary definition. Also, note that the direction of M here is identical to that of the elementary definition.

Thus we have the same result as for the elementary definition in all pertinent respects. We shall use either of these formulations depending on the situation at hand.

Put r from 0 to any point along the line of action of F. The first of these formulations will he used generally for cases where the force and point are in a convenient plane, and where the perpendicular distance between the point and the line of action of the force is easily mea- sured. As an example, we have shown in Fig. Move vector r end F. We shall illustrate such a case in Example 3. Consider next a system of n concurrent forces in Fig.

Wc cim Ihrn s;ly that. As a first step, let us express force F vectorially. Find moments at A and 8. Xj - A particle mmes along ii circular path in the xy plane. What poinl 3, 4, 5 ft'! What are its magnitude and direction cosines? What is the displacement vector from position What is the distance hetwcen. A particle moves along a paraholic path i n thc i;plane. If the :nce 1y7. Find the momcnt of thc SO-lb forcc ahout the support at A din;ilc.

Find the moment of the two lorces first ahout point A and hen ahout point R. I,OOO N 3. An nnillery spotter on Hill rn high eqimiltes the posirioii o i an cncmy tnnk as 3. A nim howmer unil with n range x of I1. Both gun units are located at an elevation -2m, OS Can eillirr o r hoth gun unit, hit thc tank, o r r m w t an air mike be callcd in? Find the moment of the forces about points A and B. The total equivalent forces from water and gravity are a Use scalar approach.

We will soon be able to compute such equiv- b Use vector approach. Compute the mnment of these forces about the toe of the dam in the right-hand comer. Asrume a11 forces. In an underwater village for research. It is of plastic material and can rotate so as to be oriented parallel to the flow of water. A uniform friction force distribution from the flow is present on both faces of the flag hav- ing the value of 10 N per square meter.

Also the flagpole has a uniform force from the flow of 20 N per meter of length of the flagpole. Finally there is an upward buoyant force on the flag of Figure P. What is the moment vector of 3. The crew of a submarine patrol plane, with three-dimen- these forces at the base of the flagpole? Where should the pilot insmct a second patrol plane flying at an elevation of 4.

A power company lineman can comfortably trim branches I m from his waist at an angle of 45 above the horizontal. His waist coincides with the pivot of the work capsule. How high a branch can he trim if the maximum elevation angle of the arm is 75 and the maximum extended length is 12 m?

Three transmission lines are placed unsymmetrically on a power-line pole. For each pole, the weight of a single line when cov- ered with ice is 2,ooO N.

What is the moment at the base of a pole? A force F from wind, weight. I hi1 At E there is a ball-and-socket joint which also supports the member. Denoting the forces from the cables as Fer, and FAB. Plane EGD is perpendicular to the wall. Get results in terms of Fro and T,r. By means of a simple situation, we shall set forth a definition of the moment of a force about an axis.

Suppose that a disc I s mounted on a shaft that is free to rotate in a set of bearings, as shown in Fig. A force F, inclined to the plane A of the disc, acts on the disc. We decompose the force into two coplanar rectangular components, one normal to plane A of the disc and one tangent to plane A of the disc, that is, into forces FB and 5, respectively, so as to form a plane shown tinted, normal to plane A. You w i l l remember from physics, this product i s nothing more than tlie mmncnt T o compute the moment or tnrque of a force F i n a planc perpendicular to plane A ahout an axis B-B Fig.

This plane cuts i3-R a1 I and thc line of action of forcc Fat some poini P. The moment about an a x i s clrarly i s a scalar. The reader will he quick to noLe that Fig. This latter diagram then takes us back Lo Fig. Accordingly, we note, nn the cine hand. I4la we can get the scalar iiiomciit M ahout an axis B-8 at point n and perpendicular to planc A.

Thus, by taking the scalar value o f M in Fig. Comparison of Figs. We can thus conclude, on considering Fig. Consideration of the moment of F about point A. Before continuing, we wish to point out that 4 in Fig.

From Varignon's theorem we can employ these components instead of 6 in computing the moment about the B--B axis. For each force component, we multiply the force times the perpendicular distance from a to the line of action of the force component using the right-hand-screw rule to determine the sense and thus the sign.

When we discussed the moment of a force about apoint, we presented a formulation useful for simple cases i. Thus far, for moments about an axis, we have presented a formulation Fd that is useful for simple cases? For this purpose, we have redrawn Fig. The coordinate distances x, y. The position vector r to P is also shown. The force component FH of Fig.

We now compute the moment about the x axis for force F using this new arrangement which does not require F to be in a plane perpendicular to plane A. Clearly, F, contributes no moment, as before. For Sorce t;. Using the right- hand-scrcw rule fiir :isccrt;iining llic sense oleach o i the momenls.

Comparing Eqs. I axis i s simply M ,. We ciin thus concludi: that the moment about the. Then draw ii p o s i t h i vector r Iron1 point 0 to any point dong Ihc line 01; d u n of F. Thih h a been shown i n [he diagraiii.