Unit 2 mm9400 ver 1.1(2014)

Equilibrium Conditions 2 - 1

Statics & Dynamics (MM9400) Version 1.1

2. EQUILIBRIUM CONDITIONS

Objectives: At the end of this unit, students should be able to:

Understand the concept of "Force"

Define a vector quantity.

Distinguish between a vector and a scalar quantity.

Define "Resultant Force."

Understand the Addition of Forces

Resolve a force into 2 components analytically.

Determine analytically the resultant of several forces using rectangular

components.

Understand the Moment of a Force

Define the moment of a force about a point.

Calculate the moment of a given force about a point.

Sum up the moments due to several forces acting on body about a point.

2 - 2 Equilibrium Conditions

Version 1.1 Statics & Dynamics (MM9400)

Understand the concept of Free Body Diagrams

Define a Free Body Diagram (FBD).

Know the nature of different types of forces, namely, push, pull (e.g. tension),

weight and reaction.

Know the reactions present at various supports, e.g. rollers, pin-joints, fixed-ends.

Sketch FBDs of two-dimensional mechanical systems in equilibrium.

Understand the Equilibrium of coplanar, concurrent force systems

State the conditions of equilibrium for concurrent forces.

Determine if a body is in equilibrium.

Solve Statics problems involving concurrent force systems.

Understand the Equilibrium of coplanar, non-concurrent force systems

State the conditions of equilibrium for non-concurrent forces.

Determine if a body is in equilibrium.

Solve Statics problems involving non-concurrent force systems.



2.1 Vector Quantity

A force is a vector quantity that has magnitude(size) and direction. The direction is

indicated by an angle measured from a reference plane and the sense is shown by

the arrow head.

Fig. 2.1 shows a force vector lying in the plane of the paper with a magnitude of

5 N acting at an angle of counter-clockwise from the positive x-axis.

If several forces lie on the same plane, they are said to be co-planar. Fig. 2.2

shows three sets of co-planar forces, each on a different plane.

Fig. 2.2

Fig. 2.1

5 N



2.2 Resolution of a Force

Resolution of a force means “splitting or resolving” it into two or more component

forces. It is the reverse process of adding two given forces together.

The most common way to resolve a force is to split it into two mutually

perpendicular components, also known as rectangular components. See Fig. 2.3 a

(Parallelogram method) and Fig. 2.3 b (Triangle method)

Therefore F2 = F sin and F1 = F cos

where F1 and F2 are the rectangular components of F.

Example 2.1

Find the components for the given 100 N force along the x- and y- axes.

F

F1

F2

F2

F1

F

F

F1

F2 F2

F1

F

Fig. 2.3 a

Fig. 2.3 b

Fx = F cos = 100 cos 30° = 86.60 N

Fy = F sin = 100 sin 30° = 50 N

F

F

F

F

triangle shaded the From

cos ; sin

,

1 2

F = 100 N

Fx

Fy

x

y



Example 2.2

Find the magnitude of the rectangular components of the 100 N force along the x-

and y- axes and indicate their directions with respect to the horizontal.

TUTORIAL

2.1 Resolve the given forces into the specified directions.

2.2 Resolve the 300 N pulling force (tension) into rectangular components in

(a) the x and y directions.

(b) the vertical and horizontal directions

Answers :

2.1 (a) Fn = 433 N

Ft = 250 N

2.1 (b) Fx = 64.71 N

Fy = 241.5 N

2.2 (a) Fx = 281.9 N

Fy = 102.6 N

2.2 (b) Fv = 172.1 N

Fh = 245.7 N

F = 100 N

y

x

30° Horizontal

30°

weight

500 N

n

t

(a)

75°

250 N y

x

(b)

15°

20°

Tension

300 N

y

x

v

h



2.3 Resultant Force (Analytical Method)

We can use rectangular components, i.e., the x- and y- components, to determine

the resultant of any number of forces.

This method is illustrated using two concurrent

forces, F1 and F2 (Fig. 2.4a). Concurrent means the

lines of action of the forces meet at a point.

1. Resolve the given forces into x- and y-components (see Fig. 2.4b & Fig. 2.4c).

2. Sum up all x-components (i.e. Fx ), taking into account their directions

If direction is assigned as positive, then Fx = F2 cos - F1 cos .

3. Sum up all y-components ( Fy ), taking into account their directions.

If direction is assigned as positive, then Fy = F1 sin + F2 sin .

4. The magnitude of the Resultant FR (see Fig. 2.4d) is determined using

Pythagoras' Theorem:

2y

2xR ) F ( ) F ( F

5. The direction Resultant FR makes an acute angle

with the x-axis:

x

y

F

F θtan

F2

F2 sin

F2 cos

F1

F1 cos

F1 sin

F1

F2

Fig. 2.4 a

Fig. 2.4 b Fig. 2.4 c

Fig. 2.4 d

Fx

Fy

FR



The vector sum, or Resultant Force, FR, is a single force which has the same effect

as several forces combined. Each force can produce motion in the direction in

which it acts, but if several forces are acting at the same time, the direction of

motion will be in the direction of the resultant force.

Example 2.3

Two perpendicular forces of 30 N and 40 N are acting on a particle as shown below.

Find the resultant of the two forces.

Example 2.4

Find the resultant of the concurrent forces shown below using an analytical method.

Solution : Take and directions as positive

30°

50 N

60 N

Being vector quantities, the forces to be added

can be arranged to form a right-angled triangle.

The Resultant force is then represented by the

hypotenuse.

Its magnitude is determined using Pythagoras’

Theorem: FR = 22 3040 = 50 N

Its direction is defined by calculating the acute

angle that the resultant force makes with the

horizontal side as shown.

Thus, Resultant force has magnitude FR = 50 N

and direction :

37° FR

30

N

40 N

30 N

40 N

FR

60cos30

60sin30

50



Example 2.5

Find the resultant force for the given co-planar concurrent forces using an analytical

method.

Example 2.6

Find the resultant force for the given co-planar forces.

50 N

30°

60 N

100 N

45°

80 N 40°

30°

60 N

100 N

45°

50 N



TUTORIAL

2.3 A bracket is shown with 2 forces acting on it. Determine the resultant force

acting on the bracket. (See Example 2.4)

2.4 For each of the coplanar concurrent system of forces shown, find the resultant

force:

(a)

(b)

( Ans : 16.99 N 70.1 ° )

(Ans : 6.136 kN, 16.1°)

( Ans : 7.108 kN 85.77 ° )

45°

2.4 kN

4.2 kN

8 N

50°

60°

45°

15 N 12 N

75°

20°

3 kN

2 kN

4 kN 5 kN

30°



(c)

2.5 A vertical force of magnitude 400 N is required to drive a pole into the ground.

If two forces, each of magnitude 250 N, can be applied by means of cables as

shown below, determine analytically what the angle should be.

(Ans: 36.9 o )

30°

45°

30°

75°

20 kN

6 kN

16 kN

10 kN

(Ans : 19.95 kN 75.0° )

vertical

250 N 250 N



2.4 Moment of a Force

The moment of a force produces or tends to produce rotation about an axis normal to

the plane in which the force acts. A moment arises when we push a revolving door,

tighten a nut with a spanner or turn the steering wheel of a car. There may be no

motion, but we can still determine the moment if the force tends to cause rotation.

The moment of a force about a point is defined as the product of the force and

the perpendicular distance of the line of action of the force from that point.

(Since we are dealing with coplanar forces, a point actually represents an axis normal

to the plane in which the forces are acting.)

Moment of F about O, Mo = F.d

Moments are indicated as clockwise () or anticlockwise () sense, depending on

the direction in which the force tends to rotate the body about the given point or axis.

Example 2.7

Determine the moment of the 100 N force about points A, B, C and D.

M

M

M

M

D

C

B

A

M

M

M

M

D

C

B

A

F

d O

100 N

A B C

D

2 m 2 m 2 m

Fig. a

Fig. b

A B D 2 m 2 m 2 m

30°

C

100 N

100

cos3

0

0

100sin30’s line of action

passes through A, B, C &

D, hence no moment

about these points.

Alternative solution for Fig. b

Using the given force, perpendicular

distance from its line of action to the

point of consideration needs to be

calculated.

A B D 2 m 2 m

2 m

30°

C

100 N

line of action of the

100 N force

dA

dB

dC



2.5 Addition of Moments

If more than one force or its components act on a body, the resultant moment ( M ),

is the algebraic sum of all the moments acting about the same point. If clockwise

moment is taken as positive then anticlockwise will be negative, or vice versa.

Example 2.8

Determine the resultant moment about various points, neglecting the weight of the

structure. (Resolve the force where necessary).

MA ( for 100N force) = _________________

MA ( for 25N force) = _________________

MA (Resultant) = _________________

MD ( for 100N force) = _________________

MD ( for 25N force) = _________________

MD (Resultant) = _________________

MD ( for 100N force) = _________________

MD ( for 25N force) = _________________

MD (Resultant) = _________________

MA = _____________________________

MC = _____________________________

ME = _____________________________

MF = _____________________________

100 N

A C B 4 m 2 m

30°

25 N Fig. a

100 N

A C B 4 m 2 m

30°

25 N

1.5 m

Fig. c

D

100 N

A C B 4 m 2 m

30°

25 N D

1.5 m

Fig. b

100 N

A

B

C

D

2 m

2 m 2 m

Fig. d

30°

F

3 m

E

3 m



Example 2.9

Determine the resultant moment about point A of the structure due to the 2 forces.

Given : AB = BC = 3 m; CD = 1.5 m.

Example 2.10

Calculate the resultant moment about point B for the rectangular plate which weighs

1 kg at G. AB = 100 mm; BC = 160 mm.

TUTORIAL

2.6 Determine the sum of moments of the given forces about the fixed end A of the

cantilever. ( Ans : 557 Nm )

B

500 N

D C

100 N

80°

A

A B

40°

D C

4 kN

2 kN 8 kN

AB = CD = 0.2 m

BC = 0.3 m

30°

A C

B

110° 20 N 30 N

G G



2.7 Calculate the moment of the given force about point A in the following cases,

showing the resolved components: ( Ans : 58.99 Nm , 953.5 Nm )

2.6 Couples

Couples are commonly encountered in engineering. A couple consists of two equal

and opposite forces having separate lines of action. We often represent a couple with

a curved arrow, i.e. () or ()

A couple has the following characteristics:

1. The resultant force of a couple is zero.

2. The moment of a couple is the product of one of the forces and the perpendicular

distance between their lines of action.

3. The moment of a couple is the same for all points in the plane of the couple.

Example 2.11

Calculate the moment of the

couple applied to the steering

wheel shown. Diameter of the

steering wheel is 45 cm.

For the die-holder shown,

calculate the moment of the

couple about points A, B, C

& D.

MA = ______________

MB = ______________

MC = ______________

MD = ______________

_______________________

800 N

38° A

B

AB = 1.2 m

BC = 0.4 m

Fig. b

C B

Fig. a

100 N

10°

25° A

C

AB = 0.6 m

BC = 0.3 m

3 N

3 N

b

A C B

D

a/2 a/2 F

F



Tutorial

2.8 For the piping system shown below, calculate the moment

(a) due to the couple. ( Ans : 9 Nm )

(b) of the 20 N force about points A and B.

( Ans : 17.18 Nm , 22.18 Nm )

2.9 Determine the sum of the moments about fulcrum B of the bell-crank due to

the applied forces and the couple.

( Ans : 18.60 Nm )

10 N

20 N

A

60°

AB = 0.5 m

BC = 1.8 m

CD = 0.9 m

10 N

B

C

D

AB = 0.2 m

BC = 0.18 m 130 N

75°

100 N

kg

A B

C

60°

16 Nm



2.10 For the framework shown below, calculate the sum of moments of the 3 given

forces about point A. Each triangular frame is an equilateral triangle of side 1 m.

( Ans : 13.77 kNm )

2.11 Determine the sum of moments about point A due to all forces and couple

acting on the 10 kg plate, which has its centre of gravity at G. The plate

measures 0.8 m by 1.2 m. ( Ans : 217.3 Nm )

B

10 kN

2 kN

5 kN

A

3 kNm

40°

25°

100 N

70 N

90 N

A

30 Nm

Couple

G



2.7 Free Body Diagram

A mechanical system can be a single body or a combination of bodies. The first

important step to analyzing a system is to isolate it or its component parts. Applied

loads (forces & couple/moment) and reactions (support forces/moments) are then put

in.

Then you have what is called a Free Body Diagram(s). See illustrations below.

Free body diagram of a single body is usually used to analyze reactions at

supports, for example Ax, Ay and M above.

Free body diagrams of the component bodies will be required when reactions at

the mating surfaces are required. An example of this can be seen above, when Bx

and By at point B are required.

Support reactions are represented by arrows with arbitrarily assumed

direction/sense. See table on the next page for guidance. The correct sense of the

reactions can be confirmed through calculations.

By

Bx

D1

D2

D3 C1

C2

C3

C

B D

Bx

By

Ay

Ax

M

A

B C D

Mechanical System

Free Body Diagram of single body

W

C3 C2 C1

WD

B

Ay

Ax M

WC

Free Body Diagrams of component bodies



Table Showing Reactions for Various Support Types

Type of Support Direction of reaction No. of unknowns

1. Rollers

1 unknown reaction.

Direction is normal to

the supporting surface.

2. Smooth surfaces

3. Cables/chain

1 unknown tension F pointing away from the

body.

4. Rigid links or bars

1 unknown reaction.

Direction along the link,

either in compression or

tension.

50°

rigid link

R

R

R

R

50° R

50o

50o R

R

F

F F

F



Type of Support Direction of reaction No. of unknowns

5. Pinned or Hinged

2 unknown rectangular

components of reaction.

Directions are not

restricted to just vertical

& horizontal

6. Rough Surfaces



Directions are normal

and tangential to the

supporting surface.

7. Fixed or Built-in Support



1 unknown fixing

(resisting) moment M

Ry

Rx

Rx

Ry

M

Rx

Ry

Rx

Ry

Rx

Ry

Ry Rx



TUTORIAL (Sketching Free Body Diagrams)

2.12 Complete the free body diagram of the body to be isolated:

Body to isolate Mechanical system FBD of body

a)

Bell

crank

b)

Structure

ABCD

c)

Beam

d)

Crate

e)

Beam

&

Pulley

B 110°

30° 20 N 30 N

A C

A B

40°

D C

4 kN

2 kN 8 kN

AB = CD = 0.2 m; BC = 0.3 m

800 N 38° A

B

AB = 0.8 m; BC = 0.4 m; CD = 0.4 m

C

D 70°

50°

Rough ground

Sm

oo

th w

all

AB = 5 m BC = 2 m

Weight mg at G

G

A

B

C

15 °

3 kN

30 ° cable

A O

T

pulley

beam

A B D C

A

B

C

D

B

A C



2.8 Equilibrium Conditions

For a body to be in a state of balance or equilibrium, all forces and moments acting

on it are in balance. In other words, resultant forces as well as moments equal zero.

F = 0 and M = 0

For a 3-dimensional or spatial force system, the above conditions must be applied

with respect to x, y and z directions which are mutually perpendicular.

For a 2-dimensional or coplanar force system, the following conditions are

necessary and sufficient for complete equilibrium.

where subscript “o” refers to any point on or off the body.

Note : Only the first 2 conditions need to be used when considering the equilibrium

of a point where all forces meet i.e. concurrent force system.

Example 2.12 ( Equilibrium of coplanar concurrent forces )

Figure shows concurrent forces acting on a truss joint. With the help of a FBD of the

joint, determine the magnitude of the forces A and B for equilibrium.

( Hint: Resolve all forces along the directions of A & B )

10 kN

30° 5 kN

2 kN

B

A

Fx = 0

Fy = 0

Mo = 0

B

A

10

5cos30

5sin30

2



Example 2.13

Three cables supporting a weight D form a

concurrent force system at B. Sketch the relevant

FBDs and determine the force in each cable.

Example 2.14

For the magnified joint B of the plane truss shown, find the forces in the members AB

& BE. The forces in the 2 members are assumed as indicated (i.e. under tension)

5 kg

30°

A

45°

C

B

D

B

5 kN

FAB FBE

5.4375 kN

33.69°

36.87°

D

C

F

A

B

E

5 kN 8 kN

AF = 3 m; FE = 6 m; FB = 4 m

D

FBD

mg

FBCcos30

FBCsin30 FABsin45

FABcos45

FBD

B

B



TUTORIAL (Equilibrium of Coplanar, Concurrent Forces)

2.13 Force A = 3.5 kN and B = 2 kN are applied to the gusset plate shown below.

Knowing that the plate is in equilibrium, determine the magnitude of forces C

and D.

( 1.73 kN , 2.5 kN )

2.14 The members of a plane truss are connected to a gusset plate shown below.

Find the magnitude of the force T and P to maintain equilibrium.

(See Example 2.12)

( 13.3 kN , 10.2 kN )

A = 3.5 kN D

C

B = 2 kN

30 °

O

T

O

5 kN

30° O

8 kN

T

P

45 °



2.15 A street lamp of mass 15 kg is supported at mid-point between two poles by a

single cable ABC. If the length of the cable is 20 m and the deflection BD at

mid-point is 0.2 m, sketch the free body diagram of point B and determine the

force in the cable.

( 3.68 kN )

2.16 A 100 N force pulls at point B of a cable ABC as shown. Sketch the free body

diagram of point B and determine the tensions in AB and BC. (See example

2.14)

(75.25 N, 96.45 N )

D C

B

A

A 40° C

B 60°

100 N

15°



2.17 Shown below is a simple framework consisting of light metal rods pinned

together at their ends. At one joint A hangs a chandelier of mass 25 kg.

Calculate the forces acting in AB & AC pinned at the joint A. Assume that both

rods are in tension. (See Example 2.14)

(381.5 N ; – 292.3 N )

40°

A

B

C



Worked example ( Equilibrium of non-concurrent forces)

For the beam shown,

a) sketch a free body diagram of the

beam ABC, given that its weight is

negligible.

b) determine the reactions on the beam at

roller A and pin-joint B.

Solution:

a) Indicate the dimensions of the body

for computing the components and

moments of forces.

b) When applying equilibrium conditions, note that MC = 0 will result in 2

unknowns ( RA & By ) in the equation.

* Summing moments about the point where an unknown acts will

enable the other unknown to be found.

40sin50o(0.3) – RA(0.2) = 0

(Roller reaction) RA = 45.96 N

YB – RA – 40sin50 = 0

YB – 45.96 – 30.64 = 0

YB = 76.60 N

XB – 40cos50 = 0

XB = 25.71 N

RB = 2B

2B XY = 2 2 76.60 25.71 = 80.8 N

B = tan-1(B

B

X

Y) = tan-1(

71.25

60.76) = 71.4°

A

B C

40 N

50 °

50°

0.2 m

A C B

0.3 m

40 N

XB

YB RA

MB = 0

Fy = 0

Fx = 0



Example 2.15

The 500 mm x 400 mm plate is subjected to forces as shown. Prove that the plate is in

equilibrium. Neglect the weight of the plate.

Example 2.16

Determine the reactions at A and B for the light beam shown below.

(The reactions at the supports keep the beam at rest)

500 mm

400 mm

50 N

40 N

30 N 80 N

60 N

100 N

A

45° C D

2 m 3 m 2 m

100 N 600 N

B A C D B



60°

A B

C

25°

D

100 N

AB = 1.5 m

BC = 1.5 m

Example 2.17

For the piping system shown below,

a) sketch a free body diagram of the piping.

b) calculate reactions at the fixed end A.

Example 2.18

An awning structure is supported as shown. It weighs 500 N and the centre of gravity

is at point B. A load of 100 N hangs from C. Determine

a) the tension in the tie rod BD.

b) the reaction at the pin support A.

10 N

20 N

A

60°

AB = 0.5 m

BC = 1.8 m

CD = 0.9 m

10 N

B

C

D

FBD

FBD

A

B

C

D



Example 2.19

A light beam and pulley system shown below is at rest. Determine

a) the reaction acting at the centre B of the pulley.

b) the reactions at A, B and C on the beam.

Solution: Pulley has to be isolated from the pin support at B. FBDs of the separate

bodies are thus required. Note the opposing sense of the component

reactions at B on both the bodies.

a)

b)

A

B

D

30°

1.5 m 1.5 m

600 N

C B

A C B



TUTORIAL (Equilibrium of Coplanar, Non-concurrent, Forces)

2.18 Calculate the support reactions for the beams/structure shown below. Neglect

the mass of the beams.

i)

ii)

iii)

AB = CD = 0.2 m

BC = 0.3 m

A B

40 °

D C

4 kN

2 kN 8 kN

Ans ( 5.5 kN 73.8 o ; 557 Nm )

Ans ( 645.7 N 88.12° ; 54.91 Nm )

5 kN

19 kN

B

2 m 2 m 2 m

Ans ( 11 kN , 3 kN )

B

65° A

D C

100 N

500 N

50 N

AB = 4.5 m

BC = 1.5 m

CD = 1.5 m



2.19 Calculate the tension in the chain at B and the reaction at the pin support A.

( 1268.4 N, 1083.2 N , 6.779° )

2.20 An advertising sign of mass 10 kg is supported on a light beam AB as shown.

Determine, with the aid of a free body diagram of AB:

a) cable tension at B.

b) magnitude and direction of the reaction at A.

( 125 N ; 114 N , 18.3o )

800 N 38° A

B

AB = 0.8 m; BC = 0.4 m; CD = 0.4 m

C

D 70°

30° B

A

10 kg

1.4 m 0.8 m



2.21 A 4 kN crate with its mass centre at G rests against a smooth wall as shown.

Determine with the aid of a free body diagram,

a) the reaction at point A of the crate.

b) magnitude and direction of the ground

reaction at B.

2.22 The flip-up table-top ABC weighs 20 N and

its CG is at the middle of AC. It is supported

by a hinge at A and a pin-ended bar at B

which is anchored at the wall.

Calculate the reactions at A and B of the

table-top.

( 219 N, 36.4o , 274.1 N , 50o )

AB = 0.2 m

BC = 0.4 m

60 N

50 ° A B C

50°

AB = 5 m

BC = 2 m

A

B

C

G

Rough ground

Sm

oo

th w

all

( 0.8782 kN ; 4.095 kN ; 77.62° )



2.23 A car park gate control has a beam CB of negligible mass and a counter weight

C of 200 kg. Calculate the tension T in the rope at B and the reaction at the

pivot A.

(435.2 N ; 2349 N, 84.7o )

45°

60°

T

A

B

C

A

3.3 m

1 m



2.24 The resistance of the bracket to bending is tested under a 2 kN load as shown.

Determine the reactions acting on the bracket at:

a) A. ( 1.25 kN ) b) the hinge O. (2.358 kN 58o )

2.25 For the bell crank lever shown, determine:

a) force acting on the roller at R if the pull at P is 80 N. (128.8 N )

b) reaction at the hinge Q. (202.1 N, 11.4o )

250 mm

40

0 m

m

2 kN

O

A

PQ = 0.3 m

QR = 0.18 m

75°

Q

P

Roller

80 N

45o

R



2.26 A load of 3 kN acts on a light beam 2.2 m long which is pivoted at O as shown

below. The part of the cable between the pulley and the beam is vertical.

a) Sketch separate free body diagrams of the beam and the pulley.

b) Find the tension F in the cable required to hold the beam

in equilibrium. (9.76 kN)

c) Determine the reaction at O on the beam. (6.91 kN 83.5o )

d) Determine the reaction at A on the pulley. (16.9 kN , 60o )

15 °

3 kN 30°

0.7 m 2.2 m

cable

O

F

A



2.27 A load of 50 kg is held in equilibrium by a light angle bracket ABC, pulley and

cable as shown. Assuming that all the forces are coplanar, a) determine the tension in the cable; b) sketch the free body diagram of the pulley alone; c) sketch the free body diagram of the angle bracket ABC alone; d) determine the magnitude and direction of the force exerted on the bracket by the pin at C; e) determine the reaction at A on the bracket.

( 490.5 N , 849.6 N 60º , 708.0 N )

Cable

Roller

Angle

Pulley 75 mm

500 m

m

300 mm

B A

60O

50 kg

C



*2.28 Figure shows a portable mobile crane lifting a cargo weighing half a tonne. The

boom ABC weighs 50 kg and the centre of gravity is at point G. Find:

(a) force in the hydraulic cylinder BD which acts at point B of the boom.

(b) reaction at the hinge A.

Given: AC = 2.0 m

AB = 1.2 m

AG = 0.9 m

( 14.80 kN ; 10.48 kN , 45.08o)

C

A

30°

60°

B G

D



*2.29 A mass of 10 kg is supported by two cables DB and DE as shown in the figure.

The beam ABC has a mass of 3 kg and is held in equilibrium by a pin at A and

a frictionless roller at C.

a) Sketch a free body diagram of the point D and show that the tensions in

the cables DB and DE are 67.1 N and 52.2 N respectively.

b) Sketch a free body diagram of the uniform beam ABC and determine: [5 marks]

i) the reaction at C; [4 marks]

ii) the reaction at A.

( (i) Rc = 204.8 N, 45º (ii) 187.4 N, 17.8º )

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frictionless

roller E

A

D

45°

10 kg

60°

40°

60°

G

A

B

C

0.6 m

0.4 m

Unit 2 mm9400 ver 1.1(2014)

Engineering

Transcript of Unit 2 mm9400 ver 1.1(2014)