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Problem Set : Given the vectors P=-5i+2j-2k, Q = 1k+4j-3k, and S = i-2j+2k, compute the scalar products P.Q, P.S, and Q.S.

1.

Determine the volume of the parallelepiped of Forces when P= 2i-7j+5k, Q=6i+2j-k,S=4i-j-4k

2.

Cube of side a=3m is acted upon by a force P =20kN as shown determine themoment of P (a) about the edge AB, (b) About the diagonal AG of the cube? [ Ans:(a)

42.42 KNm, (b)-24.5 KNm ]

3.

Consider the volleyball net shown. Determine the angle formed by guy wire AB and AC (Hint: Find the location of B and C with respect to A. after apply the equation fromreading material Angle formed by two given vectors,[ Ans:43.6degree ]

4.

Engineering Mechanics / Unit 1 / Module 3 Momentabout an axis and couples

ct/Unit Name/Module Name file:///C:/Users/user/Desktop/EM/EM_1.4/EM_1.4_HTML/index.html

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Practice the problem given in Example 4 and additional problems.5.

Additional Problems:

If there are forces F=20N and having angle 45 degrees acting in North East directionand force F = -20N and having an angle 45 degrees in south west direction, Thenwill they form couple? If they forms the couple what is the magnitude of moment of couple?

1.

If there are three vectors P = 2i+3j+6k, Q=3i+4j+k, R = 2i+2j+rk, find the value of thecomponent ‘r ’ if they are coplanar? (Hint: Mixed triple product) [ Ans:r= -10 ]

2.

Examples

1. Given the vectors P=-4i+8j-3k, Q = 9i-j-7k, and S = 5i-6j+2k, compute the scalar products P.Q, P.S, and Q.S.

Solution:

P ⋅Q = (−4i +8j −3k) ⋅ (9i − j −7k)

= (−4)(9)+(8)(−1)+(−3)(−7)

= −23

or P ⋅Q = −23

P ⋅S = (−4i +8j −3k) ⋅ (5i −6j +2k )

= (−4)(5) + (8)(−6) + (−3)(2)


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= −74

or P ⋅S = −74

Q⋅S = (9i − j −7k) ⋅ (5i −6j +2k )

=(9)(5)+(−1)(−6)+(−7)(2)

= 37

or Q ⋅S = 37

2. Determine the volume of the pallaelopiped of Forces when P= 3i-4j+k,Q=-7i+6j-8k, S=9i-2j-3k, all these forces have dimensions in inches?

Solution:

Volume of the parallelepiped is obtained by mixed triple product

3. A cube of side a is acted upon by a force P as shown determine the moment of P(a) about A, (b) About the diagonal AG of the cube?

Solution:a) Let ’s take Moment about A


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For that we have to choose the coordinate system x, y, and z axes asshown in figure and corresponding unit vectors are i, j, k. For calculatingmoment we want perpendicular distance of the force P from the point A. Thisperpendicular distance shows by r F/A = AF drawn from A to the point of application F of P.

r F/A = ai-aj = a(i-j)

The moment of P about A is cross product of perpendicular distance (r F/A) andforce P.

MA = r F/A

xP =

(b) Moment about diagonal AG: The moment of P about AG is obtained byprojecting M A (moment about A) on AB.

For that let ’s take unit vector λ along AG as show in figure with blue color,


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Moment about diagonal AG is projection of MA (moment about A) on thediagonal AG for that we have to multiply unit vector of diagonal AG with MA.

4. A square foundation mat supports the four columns shown. Determine themagnitude and point of application of the result of four loads?

Solution:


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we first reduce the given system of force to a force-couple system at origin O of the coordinate system. This force-couple system consists of a force R and a

couple vector MOR defined as follows.

R = ∑F M OR = ∑(rxF )

Here force R is sum of individual force acting on column and that forces have

been transfer to point O. MOR is sum of moments which generate by individual

force acting on column and perpendicular distance from O . clear idea can becarried out by figure given below.

The position vector of points of application of various forces R determined whencomputations are arranged in tabular form.

r (ft) F (kip) rxF (kip-ft)

0

10 i

10 i+5k

4i+10 k

-40 j

-12 j

-8 j

-20 j

0

-120 k

40 i-80 k

200 i-80k R = -80 j M OR = 240 i-280 k

Since the force R and couple vector MOR are mutually perpendicular, the force-

couple system obtained can be reduced further to a single force R . The newpoint of application of R will be selected in the plane of the mat and in such

away the moment of R about O will be equal to M OR . Denoting by r the position

vector of the desired point of application and by x and z its coordinates we write

rxR = M OR


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(xi+zk )x(-80 j ) = 240 i-280 k

-80x k + 80z i = 240 i – 280 k

From which it follows that

-80x = -280 80z = 240

x= 3.50ft z= 3ft

we can conclude that the result of given system of force is R=80kpi ↓ atx=3.5ft and z=3ft

Faculty Notes

1. Scalar product of two vectors

The scalar product of two vectors P and Q is defined as the product of the

magnitude of P and Q and of the cosine of the angle θ formed by P and Q.Thescalar product of P and Q is denoted by P.Q . We write therefore

Note that the expression just defined is not a vector but a scalar, which explainsthe name scalar product; because of the notations used, P.Q is also referred to asthe dot product of the vectors P and Q. It follows from its very definition that the

scalar product of two vectors is commutative, that is, thatP.Q = Q.P

To prove the scalar product is also distributive, we must prove the relation

P.(Q 1 + Q 2) = P.Q 1 + P.Q 2

We can, without any loss of generality; assume that P is directed along they-axis. Denoting by Q the sum of Q1 and Q2 and by the θy the angle Q forms withy-axis, we express the left-hand member as follows


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P. (Q 1 + Q 2) = P.Q = PQ cos θy = PQ y

Where Q y is the y component of Q . We can, in a similar way, express the right-handed member of as P.Q 1 + P.Q 2 = P(Q 1)y + P(Q 2)y

Since Q is the sum of Q1 and Q2, its y component must be equal to the sum of y components of Q1 and Q2. Thus, the expression obtained in the two expressions isequal, and the relation has been proved.

As far as the third property the associative property is concerned, we note thatthis property cannot apply to scalar products. Indeed, (P.Q).S has no meaning, sinceP.Q is not a vector but a scalar.

The scalar product of two vectors P and Q can be expressed in terms of rectangular components. Resolving P and Q into components, we first write

P.Q = (P xi + P y j + P zk).(Q xi + Q y j + Q zk)

Making use of the distributive property, we express P.Q as the sum of the scalar

products, such as P xi . Q xi and P xi.Q y j. However, from the definition of the scalar product it follows that the scalar product of the unit vectors are either zero or one.

i.i = 1 j.j = 1 k.k = 1 i.j = 0 j.k = 0 k.i = 0

Thus, the expression obtained for P.Q reduced to

P.Q = P xQx + P yQy + P zQ z

In particular case when P and Q are equal, we note that

P.P = P x2 + P y

2 + P z2 = P 2

Applications :

1.Angle formed by two given vectors: let two vectors be given in terms of their components be

P = P xi + P y j + P zk

Q = Q xi + Q y j + Q zk

To determine the angle formed by the two vectors, we equate the expressions

obtained by and for their scalar product and write

PQ cos θ = P xQ x + P yQy + P zQ z

Solving for cos θ , we have

2 . Projection of a vector on a given axis:


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Consider a vector P forming an angle θ with an axis, or directed line, OL (fig. 1).The projection of P on the axis OL is defined as the scalar

P OL = P cos θ

We note that projection P OL is equal in absolute value to the length of thesegment OA; it will be positive if OA has the same sense as the axis OL are at aright angle, the projection of P on OL is zero.

Consider now a vector Q (fig. 2) directed along OL and of the same sense as OL.The scalar product of P and Q can be expressed as P.Q = PQ cos θ = P OL Q

From which it follows that

In the particular case when the vector selected along OL is the unit vector λ, wewrite

P OL = P. λ


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Resolving the P and λ into rectangular components and recalling from thecomponents of λ along the coordinate axes are respectively equal to the directioncosines of OL, we express the projection of P on OL as P OL = P x cos θx + P y cos θy+P z cos θz

Where θx, θy, θz denote the angles that the axis OL forms with the coordinate axesas shown in figure 3.

3. Mixed Triple Product of Three Vectors:

We define the mixed triple product of the three vectors S, P and Q as thescalar expression

S.(P x Q)

Obtained by forming the scalar product of S with the vector product of P and Q asshown in fig. 4

A simple geometrical interpretation can be given for the mixed triple product of S,P and Q. We know that the vector PxQ is perpendicular to the plane containing Pand Q and its magnitude is equal to the area of the parallelogram which has sides Pand Q. On other hand, equation indicates that the scalr product of S and PxQ canbe obtained by multiplying the magnitude of PxQ ( that is, the area of theparallelogram defined by P and Q) by the projection of S on the vector PxQ (that is,by the projection of S on the normal to the plane containing the parallelogram).


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The mixed triple product is thus equal, in absolute value, to the volume of theparallelepiped having the vectors S, P, and Q for sides (fig. 5). We note that the signof the mixed triple product will be positive if S, P and Q form a right handed triadand negative if they form the left-handed triad (that is, S.(P x Q) will be negative if the rotation which brings P into line with Q is observed as clockwise from the tip of S). The mixed triple product will be zero if S, P and Q are coplanar.

Since the parallelepiped defined in the preceding paragraph is independent of theorder in which the three vectors are taken, the six mixed triple products which canbe formed with S, P and Q will all have the absolute value, although not the samesign. It is easily shown that

S.(P x Q) = P.(Q x S) = Q.(S x P) = -S.(Q x P) = -P.(S x Q) = -Q.(P x S)

Arranging in a circle and in counterclockwise order the letters representing thethree vectors, we observe that the sign of the mixed triple product remainsunchanged if the vectors are permuted in such a way that they are still read incounterclockwise order. Such a permutation is said to be a circular permutation. Italso follows from equation and from the commutative property of scalar products thatthe mixed triple product of S, P and Q can be defined equally well as

S.(P x Q) or (S x P).Q.


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The mixed triple product of the vectors S,P and Q can be expressed in terms of the rectangular components of these vectors. Denoting PxQ by V and using formulato express the scalar product of S and V, we write

S.(P x Q) = S.V = S xVx + S yVy + S zVz

Substituting from the relations for the components of V, we obtain

S.(P x Q) = S x(P yQ z – P zQy) + S y(P zQx – P xQz) + S z(P xQy – P yQx)

The expression can be written in a more compact form if we observe that itrepresents the expression of a determinant:

By applying the rules governing the permutation of rows in a determinant, wecould easily verify the relation which was derived earlier from geometricalconsiderations.

4. Moment of a Force about A Given Axis:

Now that we further increased our knowledge of vector algebra, we canintroduce a new concept, the concept of moment of a force about an axis. Consider again a force F acting on a rigid body and the moment MO of that force about O. Let

OL be an axis through O; we define the moment M OL of F about OL, as theprojection OC of the moment MO onto the axis OL. Denoting λ the unit vector alongOL, the projection of a vector on a given axis and for the moment MO of a force F , wewrite

MOL = λ.MO = λ.(r x F)

This shows that the moment M OL of F about the axis OL is the scalar obtainedby forming the mixed triple product of λ, r, and F. Expressing M OL in the form of adeterminant, we write

Where λx, λy, λz = direction cosines of axis OL x, y, z = coordinates of point of application of F Fx, F y, F z = components of force F


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The physical significance of the moment M OL of a force F about a fixed axis OLbecomes more apparent if we resolve F into two rectangular components F1 and F2,with F1 parallel to OL and F2 lying in a plane P perpendicular to OL. Resolving r similarly into two components r 1 and r 2 and substituting for F and r into we write

MOL = λ.[(r 1+r 2)x(F 1+F 2)]= λ.(r 1 x F 1) + λ.(r 1 x F 2) + λ.(r 2 x F 1) + λ.(r 2 x F 2)

Noting that the entire mixed triple product except the last one is equal to zero,since they involve vectors which are coplanar when drawn form a common origin, wehave

MOL = λ.(r 2 x F 2)

The vector product of (r 2xF 2) is perpendicular to the plane P and represents themoment of the component F2 of F about the point Q where OL intersects P.

Therefore, the scalar M OL, which will be positive if r 2xF 2 and OL have same senseand negative otherwise, measures the tendency of F2 to make the rigid body rotateabout fixed axis OL. Since the other component F1 of F does not tend to make thebody rotate about OL, we conclude that the moment M OL of F about OL measuresthe tendency of the force F to impart to the rigid body a motion of rotation about thefixed axis OL

It follows from the definition of the moment of a force about an axis that themoment of F about a coordinate axis is equal to the components of MO along thataxis. Substituting successively each of the unit vector i, j, and k for λ , we observethat the expression thus obtained for the moment of F about the coordinate axes arerespectively equal to the expression obtained for the component of the moment MOof F about O.

Mx = yF z – zF y

My = zF x – xFz

Mz = xF y – yF x

We observe that just as the components F x, F y and F z of a force F acting on a


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rigid body measure respectively, the tendency of F to move rigid body in the x, y andz directions, the moment M x, My and M z of F about the coordinate axes measure thetendency of F to impart to the rigid body a motion of rotation about the x, y and zaxes, respectively.

5. Moment of a Couple :

Two forces F and – F having the same magnitude, parallel line of action andopposite sense are said to form a couple as shown in fig. 9. Clearly, the sum of thecomponents of the two forces in any direction is zero. The sum of the moments of the two forces about a given point, however, not zero. While the two forces will nottranslate the body on which they act, they will tend to make it rotate.

Denoting r A and r B, respectively, the position vectors of the point of application of F and – F , we find the sum of the moments of the two forces about O is (fig. 10)

r A x F + r B x (-F) = (r A – r B) x F

setting r A-r B = r , where r is the vector joining the point of application of the twoforces, we conclude that the sum of the moment of F and – F about O is representedby vector

M= r x F

The vector M is called Moment of Couple, it is a vector perpendicular to the planecontaining the two forces, and its magnitude is

M = rF sin θ = Fd


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Where d is the perpendicular distance between the line of action of F and – F . Thesense of M is defined by the right-handed rule. (fig. 11)

Since the vector r is independent of the choice of the origin O of the coordinateaxes, we note that the same result would have been obtained if the moments of Fand – F had been computed about a different point O ’. Thus, the moment M of acouple is a free vector which can be applied at any point.

From the definition of the moment of a couple, it also follows that the two couples,one consisting of the forces F

1and – F

1, the other of the forces F

2 and – F

2, will have

equal moment if F 1d 1 = F 2d2And if the the two couples lie in parallel planes or in the same plane and have the

same sense

6. Addition of Couples:

Consider two intersecting planes P 1and P 2 and two couple actingrespectively in P 1 and P 2. We can,without any loss of generality, assumethat the couple in P 1 consist of two forcesF 1 and – F1 perpendicular to the line of intersection of the two planes and actingrespectively at A and B. Similarly, weassume that the couple in P 2 consists of two forces F 2 and – F 2 perpendicular to


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AB and acting respectively at A and B. It is clear that theresultant R of F1 and F2 andresultant – R of – F 1 and – F2form a couple. Denoting by r the vector joining B to A andrecalling the definition of themoment of a couple, weexpress the moment M of theresulting couple as follows:

M = r x R = r x (F 1+F 2)

And by Varignon ’s theoremM = r x F 1 + r x F 2

But the first term in the expression obtained represents the moment M1 of thecouple in P 1, and the second term represents the moment M2 of the couple P 2. Wehave

M = M1 + M2

And we conclude that the sum of two couples of moment M1 and M2 is a coupleof moment M equal to the vector sum of M1 and M2

7. Resolution of a given Force into a Force at O and a couple :

Consider a force F acting on a rigid body at a point A (fig 14) defined by the

position vector r . Suppose that for some reason we would rather have the force act atpoint O. while we can move F along its line of action (principle of transmissibility), wecannot move it to a point O which does not lie on the original line of action withoutmodifying the action of F on the rigid body.

We can, however, attach two forces at point O, one equal to F and the other equalto – F , without modifying the action of the original force on the rigid body. As a resultof this transformation, a force F is now applied at O; the other two forces form acouple of moment MO = rxF . Thus, any force F acting on a rigid body can be movedto an arbitrary point O provided that a couple is added whose moment is equal to themoment of F about O. the couple tends to impart to the rigid body the same rotary

motion about O that the force F tended to produce before it was transferred to O.The couple is represented by a couple vector MO perpendicular to the planecontaining r and F . since MO is a free vector, it may be applied anywhere; for convenience, however the couple vector is usually attached at O, together with F ,and the combination obtained is referred to as a force-couple system.


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8. Reduction of a system of forces to one force and one couple:

Consider a system of forces F1, F 2, F 3 ….. acting on a rigid body (fig 15) atthe point A 1, A2, A3, .. . . defined by the position vectors r 1, r 2, r 3, etc. As seen in thepreceding section, F1 can be moved from A1 to a given point O if a couple of moment M1 equal to the moment r 1xF 1 of F 1 about O is added to the original systemof forces. Repeating the procedure with F2, F 3, . . . we obtain the system shown infig which consists of the original forces, now acting at O, and the added couplevectors. Since the forces are now concurrent, they can be added vectorially and

replaced by their resultant R . Similarly, the couple vector M1, M 2, M 3, . . . can beadded vectorially and replaced by a single couple vector MOR . Any system of forces,however complex, can thus be reduced to an equivalent force-couple system actingat a given point O. We should note that while each of the couple M1, M 2, M 3, . . . infig. is perpendicular to its corresponding force, the resultant force F and the resultantcouple vector MOR will not, in general, be perpendicular to each other. Theequivalent force-couple system is defined by the equation

R = ΣF MOR = ΣMO = Σ (r x F)

Which express that the force R is obtained by adding all the forces of the system,while the moment of the resultant couple vector MOR , called the moment of resultantof the system, is obtained by adding the moment about O of all the force of thesystem.

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