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Resultant of Coplanar Forces

When a number of coplanar forces are acting on a rigid* body, then these forces can be replaced by a single force which has the same effect on the rigid body as that of all the forces acting together, then this single force is known as the resultant of several forces. Hence a single force which can replace a number of forces acting on a rigid body, without causing any change in the external effects on the body, is known as the resultant force.

The resultant of coplanar forces may be determined by the following two methods :

1. Graphical method

2. Analytical method.

The resultant of the following coplanar forces will be determined by the above two methods:

(i) Resultant of collinear coplanar forces

(ii) Resultant of concurrent coplanar forces.

 

1.10.1. Resultant of Collinear Coplanar Forces. As defined in Art. 1.9.1, collinear coplanar forces are those forces which act in the same plane and have a common line of action. The resultant of these forces are obtained by analytical method or graphical method.

 

1. Analytical method. The resultant is obtained by adding all the forces if they are acting in the same direction. If any one of the forces is acting in the opposite direction, then resultant is obtained by subtracting that force.

Fig. 1.36 shows three collinear coplanar forcesF1, F2 andF3 acting on a rigid body in the same direction. Their resultant R will be sum of these forces.

R = F 1 + F2 + F3

 

If any one of these forces (say force F2 ) is acting in the opposite direction, as shown in

Fig. 1.37, then their resultant will be given by

R =F1 -F2 +F3

 

2. Graphical method. Some suitable scale is chosen and vectors are drawn to the chosen scale. These vectors are added/ a or subtracted to find the resultant. The resultant of the three collinear forces F1‘ F2 and F3 acting in the same direction will  be obtained by adding all the vectors. In Fig. 1.38, the force F1 = ab to some scale, force F2 =be and force Fa= ed. Then the length ad represents the magnitude of the resultant on the scale chosen.

The resultant of the forces F1‘ F2 and Fa acting on a body shown in Fig. 1.37 will be obtained by subtracting the vector F2. This resultant is shown in Fig. 1.39, in which the force F1 = ab a to some suitable scale. This force is acting from a to b. The force 1 F2 is taken equal to be on the same scale in opposite direction. This force is acting from b to c. The force Fa is taken equal to cd. This force is acting from c to d. The resultant force is represented in magnitude by ad on the chosen scale.

Problem 1.21. Three collinear horizontal forces of magnitude 200 N, 100 N and 300 N are acting on a rigid body. Determine the resultant of the forces analytically and graphically when

(i) all the forces are acting in the same direction,

(ii) the force 100 N acts in the opposite direction.

 Sol. Given : F1 = 200 N, F2 = 100 N and Fa= 300 N

(a) Analytical method

(i) When all the forces are acting in the same direction, then resultant is given by equation (1.18) as

R = F 1 + F2 + F3 = 200 + 100 + 300 = 600 N. Ans.

(ii) When the force 100 N acts in the opposite direction, then resultant is given by equation (1.19) as

R=F1 -F2 +F3 =200-100+300=400N. Ans.

 

(b) Graphical method

Select a suitable scale. Suppose 100 N = 1 cm. Then to this scale, we have

F1 = 200 / 100 = 2 cm,

F2 = 100 / 100 = 1 cm ,

F3 = 300 / 100 = 3 cm

(i) When all the forces act in the same direction.

Draw vector    ab = 2 cm to represent F1,

vector   bc= 1cm to represent F2 and

vector   cd = 3 cm to represent F 3 as shown in Fig. 1.40.

Measure vector ad which represents the resultant.

By measurement length ad = 6 cm

Resultant= Length ad x chosen scale

= 6 x 100 = 600 N. Ans.

(·: Chosen scale is 1 cm= 100 N)

(ii) When force 100 N = F2 , acts in the opposite direction

Draw length ab = 2cm to represent force Fr

From b, draw be = 1 cm in the opposite direction to represent F2 . From e draw cd = 3 cm to representF3 as shown in Fig. 1.40(a).

Measure length ad. This gives the resultant.

By measurement, length ad = 4 cm

:. Resultant = Length ad x chosen scale

= 4 x 100 = 400 N. Ans.

 

1.10.2. Resultant of Concurrent Coplanar Forces. As defined in Art. 1.9.2, concurrent coplanar forces are those forces which act in the same plane and they intersect or meet at a common point. We will consider the following two cases:

(i) When two forces act at a point

(ii) When more than two forces act at a point.

 

1. When two forces act at a point

(a) Analytical method

In Art. 1.1.4, we have mentioned that when two forces act at a point, their resultant is

found by the law of parallelogram of forces. The magnitude of resultant is obtained from equation (1.1) and the direction of resultant with one of the forces is obtained from equation (1.2).

Suppose two forces P and Q act at point 0 as shown in Fig. 1.41 and a is the angle between them. Let e is the angle made by the resultant R with the direction of force P ..

Forces P and Q form two sides of a parallelogram and according to the law, the diagonal through the point O gives the resultant R as shown.

 

The magnitude* of resultant is given by

The above method of determining the resultant is also known as the cosine law method.

The direction* of the resultant with the force P is given by

θ = tan -1 ( Q sin a / P + Q cos a)

 

(b) Graphical method

(i) Choose a convenient scale to represent the forces P and Q.

(ii) From point 0 , draw a vector Oa = P.

(iii) Now from point 0, draw another vector Ob = Q and at an angle of a as shown in Fig. 1.42.

(iv) Complete the parallelogram by drawing linesa || to Ob and be || to Oa.

(v) Measure the length Oc.

Then resultant R will be equal to length Oc x chosen scale.

(vi) Also measure the angle e, which will give the direction of resultant.

 

The resultant can also be determine graphically by drawing a triangle oac as explained below and shown in Fig. 1.43.

(i) Draw a line oa parallel to P and equal to P.

(ii) From a, draw a vector ac at an angle a with the horizontal and cut ac equal to Q.

(iii) Join oc. Then oc represents the magnitude and direction of resultant R.

Magnitude of resultant R = Length Oc x chosen scale. The direction of resultant is given by angle θ. Hence measure the angle θ.

 

2. When more than two forces act at a point

  (a) Analytical method. The resultant of three or more forces acting at a point is found analytically by a method which is known as rectangular components methods (Refer to Art. 1.8.1). According to this method all the forces acting at a point are resolved into horizontal and vertical components and then algebraic summation** of horizontal and vertical components is done separately. The summation of horizontal component is written as ΣH and that of vertical as ΣV. Then resultant R is given by

The angle made by the resultant with horizontal is given by

tan θ = ( ΣV / ΣH)

Let four forces F1 F2, F3 and F4 act at a point 0 as shown in Fig. 1.44.

The inclination of the forces is indicated with respect to horizontal direction. Let

θ1 =Inclination of force F1 with OX

θ2 = Inclination of force F2 with OX’

θ3 = Inclination of force F3 with OX’

θ4 =Inclination of force F4 with OX.

The force F 1 is resolved into horizontal and vertical components and these components are shown in Fig. 1.44(a). Similarly, Figs. 1.44(b), (c) and (d) show the horizontal and vertical components of forces F2, F3 and F4 respectively. The various horizontal components are:

F1 cos θ1 à  (+)

F2 cos θ 2 à (-)

F3 cos θ 3 à (-)

F4 cos θ4 à (+)

Summation or algebraic sum of horizontal components :

ΣH F1 cos θ1 -F2 cos θ2 -F3 cos θ3 + F4 cos θ4

Summation or algebraic sum of vertical components:

ΣH = F1 sin θ1 + F2 sin θ2 -F3 sin θ3 -F4 sin θ4

And the angle (θ) made by resultant with x-axis is given by tan θ = (ΣH / ΣH)

(b) Graphical method. The resultant of several forces acting at a point is found graphically with the help of the polygon law of forces, which may be stated as

“If a number of coplanar forces are acting at a point such that they can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.

 

Let the four forces F1, F2 , F 3 and F4 act at a point 0 as shown in Fig. 1.45. The resultant 1s obtained graphically by drawing polygon of forces as explained below and shown in Fig. 1.45(a).

(i) Choose a suitable scale to represent the given forces.

(ii) Take any point a. From a, draw vector ab parallel to force OF1. Cut ab =force F 1 to the scale.

(iii) From point b, draw be parallel to OF2. Cut be= force F2:

(iv) From point C, draw cd parallel to OF3. Cut cd =force F3.

(v) From point d, draw de parallel to OF4 Cut de= force F4 .

(vi) Join point a to e. This is the closing side of the polygon. Hence are represents theresultant in magnitude and direction.

 

Magnitude of resultant R = Length ae x scale.

The resultant is acting from a to e.

 

Problem 1.22. A force of 100 N is acting at a point making an angle of 30o with the horizontal. Determine the components of this force along X and Y directions.

 Sol. Given:

Force,                          F = 100 N

Angle made by F with horizontal, e = 30°

Let                               F x = Component along x-axis

FY = Component along y-axis

Then                                        F = F cos e = 100 cos 30°

= 100 x 0.866

= 86.6 N. Ans.

FY = F sin θ = 100 sin 30°

= 100 x 0.5 = 50 N. Ans.

 

Problem 1.23. A small block of weight 100 N is placed on an inclined plane which makes an angle 8 = 30° with the horizontal. What is the component of this weight; (ii) parallel to the inclined plane and (iii) perpendicular to the inclined plane ?

 Sol. Given:

Weight of block, W= 100 N

Inclination of plane, θ = 30°

The weight of block W = 100 N is acting vertically downwards through the C.G. of the block. Resolve this weight into two components i.e., one perpendicular to the inclined plane and other parallel to the inclined plane as shown in Fig. 1.50. The perpendicular (normal) component makes an angle of 30° with the direction of W.

Hence component of the weight perpendicular to the inclined plane

= W cos 30° = 100 x 0.866 = 8.66 N. · Ans.

Component of the weight (W) parallel to the inclined plane

= W sin 30° = 100 x 0.5 = 50 N. Ans.

 

Problem 1.24. The four coplanar forces are acting at a point as shown in Fig. 1.52. Determine the resultant in magnitude and direction analytically and graphically.

Sol. Given:

Forces,             F1 = 104 N

F2 = 156 N,

F3 = 252 N, and

F4 = 228N.

 

(a) Analytical method. Resolve each force along horizontal and vertical axes. The horizontal components along OX will be considered as +ve whereas along OX’ as – ve. Similarly, vertical components in upward direction will be +ve whereas in downward direction as -ve.

(i) Consider force F1 = 104 N. Horizontal and vertical components are shown in Fig. 1.52(a).

Horizontal component,

Fx1 = F1 cos 10° = 104 x 0.9848

= 102.42 N

Vertical component,

Fy1 = F1 sin 10o = 104 x 0.1736

= 18.06 N.

 

(ii) Consider force F2 = 156 N. Horizontal and vertical components are shown in Fig. 1.52(b):

Angle made by F2 with horizontal axis

OX’ = 90 – 24 = 66°

. . Horizontal components,

Fx2 = F2 cos 66° = 156 x 0.4067

= 63.44 N.

It is negative as it is acting along OX’.

Vertical component,

Fy2 = F2 sin 66o = 156′x 0.9135

= 142.50 N.

 

(iii) Consider force F3 = 252 N. Horizontal and vertical components are shown in Fig. 1.52(c).

Horizontal component,

Fx3 = F3 cos 3° = 252 x 0.9986

= 251.64 N.

Vertical component,

Fy3 = F3 sin 3o = 252 x 0.0523

= 13.18 N.

 

(iv) Consider force F4 = 228 N. Horizontal and vertical components are shown in Fig. 1.52(d).

Angle made by F4 with horizontal axis

OX’ = 90-9 =81°.

Horizontal component,

Fx4 = F4 cos 81o = 228 x 0.1564

= 35.66 N

Vertical component,

Fy 4 = F4 sin 81o = 228 x 0.9877

= 225.2N.

Now algebraic sum of horizontal components is given by,

ΣH = Fx1 -Fx2 -Fx3 -Fx4

= 102.4- 63.44-251.64-35.66

= – 248.32 N.

- ve sign means that m is acting along OX’ as shown in Fig. 1.52(e).

Similarly, the algebraic sum of vertical components is given

ΣV= 18.06:: 142.50 + 13.18-225.2

= -77.82 N.

- ve sign means that LV is acting along OY’ as shown in Fig. 1.52(e).

The magnitude of resultant (i.e., R) is obtained by using equation (1.20).

= 260.2 N.

 

The direction of resultant is given by equation (1.21).

tan θ = ΣH / ΣH = 77.82 / 248.32 = 0.3134

θ = tan -1 0.3134 = 17.4

 

(b) Graphical method. Fig. 1.53(a), shows the point at which four forces 104 N, 156 N, 252 N and 228 N are acting. The resultant force is obtained graphically by drawing polygon of forces as explained below and shown in Fig. 1.53(b):

(i) Choose a suitable scale to .represent the given forces. Let the scale is 25 N = 1·cm. Hence the force 104 N will be represented by 104 / 25 = 4.16 cm force 156 N will be represented by 156 / 25 = 6.24 cm force 252 N will be represented by = 252 / 25 = 10:08 cm and the force 228 N will be represented by  228 / 25 =  9.12 cm.

 

(ii) Take any point a. From point a, draw vector ab parallel to line of action of force 104 N. Cut ab = 4.16 cm. Then ab represents the force 104 N in magnitude and direction.

(iii) From point b, draw vector be parallel to force 156 N and cut be= 6.24 cm. Then vector cd represents the force 156 N in magnitude and direction.

(iv) From point c, draw a vector cd parallel 252 N force and cut cd = 10.08 cm. Then vector cd represents the force 252 N in magnitude and direction.

(v) Now from point d, draw the vector de parallel to 228 N force and cut de= 9.12 Cm. Then vector de represents the force 228 N in magnitude and direction.

(vi) Join point a to e. The line ae is the closing side of the polygon. Hence the side ae represents the resultant in magnitude and direction. Measure the length of ae.

 

By measurement, length ae = 10.4 Cm

. . Resultant, R =Length ae x Scale= 10.4 x 25 (·: 1 Cm= 25 N)

=260N. Ans.

 

Problem 1.25. Determine the magnitude, direction and position of a single force P, which keeps in equilibrium the system of forces acting at the corners of a rectangular block as shown in Fig. 1.55. The position of the force P may be stated by reference to axes with origin 0 and coinciding with the edges of the block.

Sol. Given:

Length             OC = 4 m,

Length             BC = 3 m

Force at           O = 20 N

Force at           C = 35 N

Force at           B = 25 N

Force at           A= 50 N

Let 0 be the origin and OX and OY be the reference axes as shown in Fig. 1.56.

Forces 50 N and 20 N form a concurrent system and their line of action intersect at 0.

Similarly the forces 35 N and 25 N form a concurrent system and their line of action intersect at B.

 

These two forces R1 and R2 intersect at D. The angle between these forces is θ1 + θ2 i.e.,

angle R1DR2 = θ1 + θ2 = 21.8° + 35.53° = 57.33°.

 

The angle made by the resultant P with R1 is given by

tan  a = R2 sin 57.33 / R1 + R2 cos 57.33 = 43.01 sin 57.33 / 53.85 + 43.01 x cos 57.33

= 43.01 x0.8418 / 53.85 + 23.21 = 36.2058 / 77.06 = 0.4698

a  = tan -1 0.4698 = 25 . 16

 

Hence the resultant P makes (a– 91) angle with vertical in anti-clockwise direction i.e., P makes <25.16- 21.8 = 3.36°). Ans.

 

Position of the Force P

The position of the force P is obtained by equating the clock wise moments and anticlockwise moments about 0 (Refer Fig. l.S6).

Let                   OE = Perpendicular distance between 0 and line of action of the force P.

Taking moments of all forces about 0,

20 x 0 + 50 x 0 + 35 x 4 + 25 x 3 = p x OE

0 + 0 + 140 + 75 = 85.147 x OE or 215 / 85.147 = 2.525 m

From right angled triangle OED, sin a= OE / OD

OD = OE / sin a = OE / sin 25.16 = 2.525 / 0.4241 = 5.939

Let x and y are the co-ordinates of the force P with reference to the axes with origin 0.

Then                            x = OF and y= DF

In right angled triangle OFD,

OF= OD X sin el = 5.939 X sin 21.8°

= 2.20m

Also                 FD = OD x cos cl = 5.939 x cos 21.8° = 5.514 m

x = OF = 2.20 m. Ans.

and                      y = FD = 5.514 m. Ans.

Graphical method [Refer to Fig. 1.57(a)]

(i) To a suitable scale, take OG =50 N and GH = 20 N. Join OH. Then OH represents the resultant R1 in magnitude and direction. Produce the line HO backward.

(ii) From point E, Take BJ = 35 N and JK = 25 N. Join BK, which represents the resultantR2 in magnitude and direction. Produce KB in the backward direction to interest the line of action of R1 at point D.

(iii) To find the resultant of R1 and R2 (i.e., force P) refer to Fig. 1.57(b).

(iv) Take any point ‘a’. From this point draw line ab parallel toR1 and equal toR1. From point ‘b’, draw line be parallel to R2 and equal to R2 Join the point c to a.

(u) Then ca represents in magnitude and direction the force P. Hence measure ca. Then P = ca = 85.15 N. Ans.

(vi) From point D, draw the line DL parallel to ca. Hence DL represents the direction of the force P. ·

(vii) To find the position of the force P which is acting at point D, draw DF parallel to axes OY. Then OF represents the x-coordinate and FD represents they-coordinate of the force P.

Measure OF and FD. Then by measurement,

OF = x = 2.20 m. Ans.

and                                          FD = y = 5.514. Ans.

1.10.3. Resultant of Parallel Coplaner Forces. The forces, which are having their line of action parallel to each other, are known parallel forces. The two parallel forces will not intersect at a point.

The following are the important types of parallel forces :

1. Like parallel forces,

2. Unlike parallel forces.

1. Like parallel forces. The parallel forces which are acting in the same direction, are known as like parallel forc.es. In Fig. 1.58, two parallel forces F1 and F2 are shown. They are acting in the same direction. Hence they are called as like parallel forces. These forces may be equal or unequal in magnitude.

2. Unlike parallel forces. The parallel forces which are acting in the opposite direction, are known as unlike parallel forces. In Fig. 1.59, two parallel forces F1 F2 are acting in opposite direction. Hence they are called as unlike parallel forces. These forces may be equal or unequal in magnitude.

The unlike parallel forces may be divided into : (i) unlike equal parallel forces, and (ii) unlike unequal parallel forces.

Unlike equal parallel forces are those which are acting in opposite direction and are equal in magnitude.

Unlike unequal parallel forces are those which are acting in opposite direction and are unequal in magnitude.

The resultant of following two parallel forces will be considered:

1. Two parallel forces are like.

2. Two parallel forces are unlike and are unequal in magnitude.

3. Two parallel forces are unlike but equal in magnitude.

 

1. Resultant of two like parallel forces. Fig. 1.60 shows a body on which two like parallel forcesF1 andF2 are acting. It is required to determine the resultant (R) and also the point at which the resultant R is acting. For the two parallel forces which are acting in the same direction, obviously the resultant R is given by,

R =F1 +F2

In order to find the point at which the resultant is acting, F Varignon’s principle (or   method of moments) is used. According 1 60 IQ to this, the algebraic sum of moments of F 1 and F2 about any point should be equal to the moment of the resultant (R) about that point. Now arbitrarily choose any point 0 along line AB and take moments of all forces about this point.

 

Moment of F 1 about 0 = F 1 x AO (clockwise)(-)

Moment of F2 about 0 = F2 x BO (anti-clockwise)(+ ve)

. . Algebraic sum of moments of F1 and F2 about O= – F 1 x AO + F 2 x BO

 

 

Moment of resultant about 0 = Rx OC (anti-clockwise){+)

But according to principle of moments the algebraic sum of moments of F 1 and F2 about

0 should be equal to the moment of resultant about the same point 0.

 

 

 

F 1 x AO + F2 x BO = +R x CO= (F1 + F2) x CO

 

or .                                F1(AO +CO)= F2(BO- CO)

F l x AC = F 2 x BC

F1 / F2 = BC / AC

 

The above relation shows that the resultant R acts at the pointe, parallel to the lines of action of the given forcesF1 andF2 in such a way that the resultant divides the distance AB in the ratio inversely proportional to the magnitudes of F 1 and F2 Also the point C lies in line AE i.e., point C is not outside AB.

 

The location of the pointe, at which the resultant R is acting, can also be determined by taking moments about points A of Fig. 1.60. As the forceF1 is passing through A, the moment of F 1 about A will be zero.

 

 

The moment of F 2 about A= F2 x AB (anti-clockwise)(+)

Algebraic sum of moments of F 1 and F2 about 0

= 0 + F2 x AB = F2 x AB (anti-clockwise) (+)

 

The moment of resultant R about A

= R x AC (anti-clockwise)(+)

But according to the principle of moments, the algebraic sum of moments of F1 and F 2

about A should be equal to the moment of resultant about the same point A. Hence equating

equations (i) and (ii),

F2 x AB =R x AC

But R = (F 1 + F2 ) hence the distance AC should be less than AB. Or in other words, the

point C will lie inside AB.

 

2. Resultant of two unlike parallel forces (unequal in magnitude). Fig 1.61 shows

a body on which two unlike parallel forces F 1 and F2 are acting which are unequal in magnitude. Let us assume that forceF1 is more than F2• It is required to determined the resultant and also the point at which the resultant R is acting. For the two parallel forces, which are acting in opposite direction, obviously the resultant is given by,

R=F1 -F2

Let the resultant R is acting at C as shown in Fig. 1.61.

In order to find the point C, at which the resultant is acting, principle of moments is used.

Choose arbitrarily any point 0 in line AB. Take the moments of all forces (i.e., Fp F2 and R) about this point.

Moment of F 1 about 0 = F 1 x AO (clockwise)

Moment of F2 about 0 = F2 x BO (clockwise)

Algebraic sum of moments of F 1 and F2 about 0

 

=F1 x AO +F2 x BO

 

Moment of resultant force R about 0

= R x CO (clockwise)

= (F1 – F2) x CO

= F1 x CO –F2 x CO

 

But according to the principle of moments, The algebraic sum of moments of all forces about any point should be equal to the moment of resultant about that point. Hence equating equations (i) and (ii), we get

 

F 1 x AO + F2 x BO = F 1 x CO – F 2 x co

F2(BO +CO) = F 1(CO -AO)

F2 x BC=F1 x AC

BC / AC = F1 / F2 or F1 / F2 = BC / AC

 

But F1 >F2, hence BC will be more than AC. Hence point O lies outside of AB and on the same side as the larger force F1 Thus in case of two unlike parallel forces the resultant lies de the line joining the points of action of the two forces and on the same side as the larger

 

The location of the Point C, at which the resultant R is acting, can also be determined by taking moment about point A, of F1g. 1.61. As the force F1 is passing through A, the moment of F1 about A will be zero.

The moment of F2 about A= F2 x AB (clockwise)(-)

Algebraic sum of moments of F 1 and F2 about A

= O + F2 x AB = F2 x AB (clockwise)(- ) … (i)

The moment of resultant R about A should be equal to the algebraic sum of moments of F1 and F2 (i.e.,= F2 x AB) according to the principle of moments. Also the moment of resultant R about A should be clockwise. As R is acting upwards [ ·: F1 > F2 and R = (F1 - F2) so R is acting in the direction ofF1], the moment of resultant R about A would be clockwise only if the points C is towards the left of point A. Hence the point C will be outside the line AE and on the side of F 1 (i.e. , larger force).

 

Now the moment of resultant R about A

= R x AC (clockwise)(-) … (ii)

Equating equations (i) and (ii),

F2 x AB =R x AC

= (F1 -F2) x AC (·: R=F1 – F2)

As F1, F2 and AB are known, hence AC can be calculated. Orin other words, the location of point C is known.

 

3. Resultant of two unlike parallel forces which are equal in magnitude. When two equal and opposite parallel forces act on a body, at some distance apart, the two forces from a couple which has a tendency to rotate the body. The perpendicular distance between the parallel forces is known as arm of the couple.

Fig. 1.62 shows a body on which two parallel forces, which are acting in opposite direction but equal in magnitude are acting. These two forces will form a couple which will have a tendency to rotate the body in clockwise direction. The moment of the couple is the product of either one of the forces and perpendicular distance between the forces.

 

Let       F =Force at A or at B

a= Perpendicular distance (or arm of the couple)

The moment (M) of the couple is given by, M = F x a.

The units of moment will be Nm.

 

Problem 1.26. Three like parallel forces 100 N, 200 N and 300 N are acting at points A, Band C respectively on a straight line ABC as shown in Fig. 1.63. The distances are AB = 30c and BC = 40 cm. Find the resultant and also the distance of the resultant from point A on line

ABC.

Sol. Given:

Force at           A= 100 N

Force at           B = 200 N

Force at           C = 300 N

Distance AB = 30 cm, BC = 40 cm. As all the forces are parallel and acting in the same direction, their resultant R is given by

R = 100 + 200 + 300 = 600 N I

Let the resultant is acting at a distance of x cm from the point A as shown in Fig. 1.63.

Now take the moments of all forces about point The force 100 N is passing A, hence its moment about A will be zero.

Moment of 100 N force about A= 0

Moment of 200 N force about A = 200 x 30 = 6000 N cm (anti-clockwise)

Moment of 300 N force about A = 300 x AC

= 300 x 70 = 21000 N cm (anti-clockwise)

Algebraic sum of moments of all forces about A

= 0 + 6000 + 21000 = 27000 N cm (anti-clockwise)

Moment of resultant R about A=R x x

= 600 x x N cm

But algebraic sum of moments of all forces about A Moment of resultant about A

2.7000 = 600 x x or x = 27000 / 600 = 45 cm

 Problem 1.27. The three like parallel forces of magnitude 50 N, F and 100 N are shown in Fig. 1.64. If the resultant R = 250 N and is acting at a distance of 4 m from A, then find

(i) Magnitude of force F.

(ii) Distance of F from A.

Sol. Given :

Forces at A= 50 N, at B = F and D = 100 N

R = 250 N, Distance AC = 4 m, CD= 3m.

(i) Magnitude of force F

The resultant R of three like forces is given by,

R = 50+F+ 100

250 = 50+F+100 (·: R=250)

F = 250- 50- 100 = 100 N. Ans.

(ii) Distance of F from A

Take the moments of all forces about point A.

Moment of force 50 N about A= 0

Moment of force F about        A = F x x

Moment of force 100 N about A= 100 x AD= 100 x 7 = 700 N m (anti-clockwise)

Algebraic sum of moments of all forces about A

= 0 + F x x + 700 N m

=F x x + 700 N m

Moment of resultant R about A = R x 4 = 250 x 4 = 1000 N m

But algebraic sum of moments of all forces about A must be equal to the moment of Resultant R about A.

 

F x x + 700 = 1000 or F x x = 1000 – 700 = 300

x = 300 / F = 300 / 100

= 3 m.

 

Problem 1.28. Four parallel forces of magnitudes 100 N, 150 N, 25 N and 200 N are shown in Fig. 1.65. Determine the magnitude of the resultant and also the distance of the resultant from point A.

Sol. Given:

Forces are 100 N, 150 N, 25 N and 200 N.

Distances                     AB = 0.9 m, BC = 1.2 m, CD = 0.75 m.

As all the forces are acting vertically, hence their resultant R is given by

R = 100 -150-25 + 200

(Taking upward force + ve and downward as – ve)

= 300 – 175 = 125 N

 

+ve sign shows that R is acting vertically upwards. To find the distance of R from point

A, take the moments of all forces about point A.

Let                   x = Distance of R from A in metre.

As the force 100 N is passing through A, its moment about A will be zero.

Moment of 150 N force about A = 150 x AB

= 150 x 0.9 (clockwise)(-)=- 135 Nm

Moment of25 N force about A = 25 x AC = 25 x (0.9 + 1.2)

= 25 x 2.1 (clockwise)(- ) =- 52.5 Nm.

Moment of 200 N force about A = 200 x AD

= 200 X (0.9 + 1.2 + 0.75)

= 200 x 2.85 (anti-clockwise) (+) = 570 Nm

 

Algebraic sum of moments of all forces about A

= – 135 – 52.5 + 570 = 382.5 Nm … (i)

+ ve sign shows that this moment is anti-clockwise. Hence the moment of resultant R About A must be 382.5 Nm, i.e., moment of R should be anti-clockwise about A. The moment of

R about A will be anti-clockwise if R is acting upwards and towards the right of A.

Now moment of R about A  = R x x. But R = 125

= 125 x x

= + 125 x x

Equating (i) and (ii), 382.5 = 125 x x or x = 382.5 / 125 = 3.06 m

 

Resultant (R = 125 N) will be 125 N upwards and is acting at a distance of3.06 m tothe right of point A.