LIST OF EXPERIMENTS
SR.
NAME OF EXPERIMENT
NO.
1) SIMPLE BEAM REACTION
2) BELL CRANK LEVER
3) POLYGON LAW OF COPLANER FORCES
4) VERIFICATION OF LAMIS THEOREM
5) FRICTION PLANE
6) CO-EFFICENT OF FRICTION
7) COLLISION OF ELASTIC BODIES
Experiment No. 1
SIMPLE BEAM REACTION
AIM :To find experimentally the reactions at the supports of a simply supported beam
and compare the results with analytical values.
APPARATUS REQUIRED :Simply supported beam setup, Hangers, loads.
THEORY :Beam is horizontal and straight structural member which carries loads that are
vertical .
A simply supported beam is one whose ends are resting freely on the supports that
provide only vertical reactions. Simply supported beam becomes unstable if it is
subjected to oblique or inclined loads.
When simply supported beam is subjected to only vertical loads .its FBD forms a
system of parallel forces in equilibrium. Condition of equilibrium ∑y = 0 and ∑M =0can
be applied to determine the support reactions analytically.
PROCEDURE :1. Place the beam length of L on simple supports. Note that below both the simple
supports there is a spring arrangement. On loading, the spring gets compressed
and this compressive force is indicated on the dial.
2. Arrange the hangers arbitrarily on the beam and set the left and right dial pointers
to zero. This will nullify the effect due to self weight of the beam and the hangers.
3. Suspend the loads from the hangers. Note the load values W1, W2, W3 etc., their
distances X1, X2, X3 etc. from the left support.
4. Note the left and right support dial reading.
5. Repeat the above steps 1 to 4 by changing the weights on the hangers and also the
hanger position for two set of observations.
6. Compare the experimental values with analytical values obtained by applying
conditions of equilibrium. Also find out the percentage of error for both end
reactions.
FREE BODY DIAGRAM (FBD)
OBSERVATION TABLE :Sr.
No.
W1
(N)
W2
(N)
W3
(N)
X1
(m)
X2
(m)
X3
(m)
CALCULATION :Length of beam = L = _____________ Mts.
Applying conditions of equilibrium.
∑M1 = 0
+ve
-W1  X1-W2  X2-W3  X3+R2  L = 0
 R2 =
W1  X1  W2  X2  W3  X3
L
 Fy  0
+ve
R1 – W1 – W2 – W3 + R2 = 0
R1 = W1 + W2 + W3 – R2
% Error in R1 =
R1 cal. - R1 obs.
R1 cal
% Error in R2 =
R2 cal. - R2 obs.
R2 cal
RESULT :-
CONCLUSION :-
Observed
Analytical
% of Error
reactions
reactions
R1 (N) R2 (N) R1 (N) R2 (N) R1
R2
POLYGON FORCES TABLE APPARATUS
EXPERIMENT NO. :- 2
FORCE TABLE APPARATUS
Aim:To verify the equilibrium of concurrent coplanar force system.
APPARATUS REQUIRED :Force table with angle measurement, Slotted weights of 50 gms each, Weight
hangers, Metal ring of 2cm diameter.
Theory:The body is said to be at rest, if a resultant of several forces acting at a point on
the same plane. We will verify this by using force table. The body is said to be in static
equilibrium if a resultant of several forces acting at a point on the same plane is zero.
PROCEDURE :1. Tie four strings on the rim of a 2cm diameter metallic ring
2. Suspend the ring in a horizontal plane by passing four of these strings over the
four pulleys fixed at the circumference of the apparatus.
3. Attach the hangers at the end of these strings.
4. Insert slotted weights in these four suspended hangers such that the metallic ring
will occupy a disturbed position. The weights in the hangers 1 to 4 represent the
forces F1, F2, F3 and F4 respectively.
5. Measure the angles made by the strings with horizontal. Call theses angles as Ө1,
Ө2, Ө3 and Ө4 respectively.
6. Insert the fifth weight F5 (experimental) and adjust the angle Ө5 (experimental) in
such a way that pointer comes exactly to the center. When the pointer is at the
center of the ring, it is assumed that all the forces are concurrent and coplanar and
the body is in equilibrium.
7. Verify the equilibrium analytically and check the value of F5 and find the
percentage of error.
8. Repeat the above steps by changing the weights in the hangers and also the
position of hanger for two more sets of observations.
OBSERVATION TABLE :Sr. Forces ( N )
No.
F1 F2 F3
Angles ( Degrees )
F4
Analytical
Experimental
Ө1 Ө2 Ө3 Ө4 F5
Ө5
F5
analy analy exp.
Ө5
exp.
%
error
F5
CALCULATION: f x=0
F1cosӨ1+ F2cosӨ2+ F3cosӨ3+ F4cosӨ4 + F5cosӨ5 = 0
 f y=0
F1sinӨ1+ F2sinӨ2+ F3sinӨ3+ F4sinӨ4+ F5sinӨ5 = 0
F5 =
(F5sin  5) 2  (F5cos 5) 2
 = tan -1 (
% Error
( F5sin  5)
( F5cos 5)
=
RESULT:-
CONCLUSION:-
F5 analy - F5 exp.
F5 analy
x 100
BELL CRANK LEVER APPARATUS
Experiment No. 3
BELL CRANK LEVER
AIM :A)
B)
To verify the law of moments using Bell Crank Lever.
To determine percentage error in the spring balance.
APPARATUES :Bell Crank Lever apparatus, Weights, Meter scale.
Theory :The law of moment sates that, if number of co-planar forces acting on a body
keeps the body in equilibrium, then the algebraic sum of all their moments about any
point in the plane is zero.
PROCEDURE :1.
2.
3.
4.
5
6.
Place the load on the lever at a suitable distance (W)
Measure the distance of the applied load from the center of the hinge (X)
Note the reading ( magnitude of the force developed in the spring ) in the spring
balance (TE)
Measure the perpendicular distance of the spring balance from the center of the
hinge (Y).This distance is constant for all the reading
Repeat the same procedure by changing weights and distances.
Clockwise moment produced by the weight and anticlockwise moment produced
by the tension in the spring if they are not equal then find out the percentage error
between the clockwise and anticlockwise moments.
OBSERVATION TABLE :a)
Verification of Law of Moment :Initial reading in the spring balance is IE = ------------Sr. No.
W (N)
X (m.)
TE = FR - IE
(N)
Y (m.)
M+
(A.C.W)
M(C.W)
Me
% Error
1
2
3
b)
Sr. No.
1
2
3
Error in Spring Balance
Force in spring balance (T = P x g ) N
Experimental Value
Analytical Value
TE
TA
% Error =
TA -TE
TA
*100
CALCULATION :a)
Verification of Law of Moments :Weight, W = M  g = ____________ N
X = ____________ m
Y = ____________ m
Tension in the spring, TE
= P g
= ____________ N
Clockwise moment M-
= WX
= __________ N-m
Anticlockwise moment M+
= TA  Y
= _______ N-m
Total moment, | M | =
M + - M-
Percentage Error, Me =
b)
= __________ N-m
|M|
 100
M
Error in spring balance :-
FBD of Bell Crank Lever
TA =
WX
Y
(N)
TE = Experimental / Observed Value
% Error =
TA -TE
TA
 100
RESULT :-
CONCLUSION :-
LAMI’S THEOREM VERTICAL BOARD
Experiment No. 4
VERIFICATION OF LAMI’S THEOREM
AIM :A)
B)
To understand and verify Lami’s theorem using force table/ Vertical Board
apparatus
To find the unknown force in a concurrent balance force system using Lami’s
Theorem.
APPARATUES :Vertical Board, force table apparatus, angle measuring instrument, slotted
weights.
Theory :When three concurrent coplanar forces are in equilibrium, then each force is
proportional to the sine of the angle between the other two forces. Also the force triangle
is closed one.
If two forces and three angles are known, the third unknown force can also be calculated.
PROCEDURE :1) Three strings are tied to a small metallic ring of 2cm diameter.
2) The ring is suspended in a vertical plane by passing the strings over three pulleys
witch can be fixed at three sides of apparatus as shown in the figure.
3) Weight hanger is attached at the end of these strings.
4) The weights are added on the three hangers in such a way that the metallic ring
comes to an equilibrium position.
5) The angles made by the strings are measured either by an instrument or by tracing
it on plain paper.
6) The experiment is verified if the ratios of forces to the respective sine angles are
found to be approximately equal.
OBSERVATION TABLE :Sr. No.
Forces (gms).
Experimental
F1
F2
F3
Angles (Degrees)
θ1
θ2
θ3
Forces (gms)
Analytical
F2
F3
% Error
F2
F3
CALCULATION :Reading 1
F
F1
F2

 3
sin 1 sin  2 sin  3
F1
F Ana F3 Ana
 2

sin 1 sin  2
sin  3
% Error F2=
F2 Ana  F2 Exp
X 1OO
F2 Ana
% Error F3=
F3 Ana  F3 Exp
X 1OO
F3 Ana
RESULT :-
CONCLUSION :-
FRICTION PLANE APPARATUS
EXPRIMENT NO. 5
FRICTION PLANE
AIM :To determine the coefficient of static friction between two surfaces by
1) Angle of Repose method
2) Friction plane method
APPARATUS REQUIRED :Friction plane apparatus with glass top and graduated quadrant, Weights, Box
with wooden base, String effort pan, fractional weight.
THEORY :Friction force is developed whenever there is motion or tendency of motion of
one body with respect to the other body involving rubbing surfaces of contact. Friction is
therefore a resistance force to sliding between two bodies produced at a common contact
surfaces.
Friction occurs because of roughness of the surface even though surface may
appear to be smooth to the naked eyes. On every surface there are microscopic hill and
valleys and due to this, the surface gets interlocked making it difficult for one surface to
slide over the other. During static state the friction force developed at the contact surface
depend on the magnitude of the disturbing force. When the body is on the verge of
motion, the contact surface offers maximum friction force called as ‘Limiting Frictional
Force’.
In 1781 the French physicist Charles de Coulomb found that the limiting friction
force did not depend on the area of contact but depends on the material involved and the
pressure ( normal reaction) between them.
Thus friction force
F αN
Or
F =  s N where
 s is the coefficient of static friction, a term introduced by coulomb. The value of  s
depends on the surfaces of contact of two materials and for most of the materials the
value of coefficient of friction is less than 1.
Coefficient of static friction  s between two surfaces can be found
experimentally by two methods. viz. Angle of Repose method and Friction plane
method.
Angle of Repose method :
The maximum angle of an inclined plane at which a body kept on it slides down
the plane without the application of an external force is known as Angle of Repose. It is
denoted by letter  . Angle of repose depends only on the coefficient of static friction
and is independent of the weight of the body. Let’s derive the relation between the two.
Let a block of a weight W just starts to slide down the inclined plane at a
maximum angle  of an inclined plane. Refer F.B.D.
Angle of Repose Method
Applying Condition of Equilibrium.
 Fx  0
 s N - Wsin   0 ……………….(1)
 Fy  0
N - Wcos  0 ………………....(2)
From equation (1) and (2) we get
 s (Wcos ) - Wsin   0
sin 
 tan 
 s 
cos
PROCEDURE :1) Initially keep the inclined plane with glass top at horizontal level. Take box of
known weight. Note its bottom surface (whether the surface is soft wood or sand
paper or of a card board etc) and place the box on the glass surface. Put some
weight in the box.
2) Slowly raise the inclination of the plane till the time when the box starts to slip on
the plane.
3) Note the inclination of the plane on the graduated quadrant scale. This is the angle
of repose.
4) Find the coefficient of friction ‘  s ’ using  s  tan  .
5) Repeat the above steps 1 to 3 by changing the weights in the box for two more
observations.
Friction Plane Method.
In this method we find the minimum force required to slide the body of weight W
up the inclined plane. Knowing the value of P we relate it to ‘  s ’as below. Refer F.B.D.
Applying Condition of Equilibrium.
 Fx  0
P - Wsin    s N  0 ……………….(1)
 Fy  0
N - Wcos  0 ………………....(2)
From equation (1) and (2) we get
P - Wsin    s (Wcos )  0
P - Wsin 
s 
Wcos
PROCEDURE :1. Set the inclined plane with glass top at some angle with the horizontal. Note
the inclination  of the plane on the quadrant scale. Take a box of known
weight. . Note its bottom surface(whether the surface is soft wood or sand
paper or of a card board etc.) and place the box on the glass surface. Put some
weight in the box. Find total weight W ( weight of box + weights in box)
2. Tie a string to the box and pass the string over a smooth pulley. Attach an
effort pan at the end of string.
3. Slowly add weights in the effort pan. A stage would come when the effort pan
just starts moving down pulling the box up the plane. Using fractional weights
up to 5 gm. find the least possible weight in the pan that causes the box to just
slide up the plane. Note the weight in the effort pan. This is force ‘P’.
4. Repeat the above steps 1 to 3 by changing the weights in the box for two more
sets of observations.
OBSERVATION TABLE :Surface of contact:________________________ and ________________________
Angle of repose
Sr. No.
Total Weight of Box
W(N)
Inclination of Plane
 ( Degrees)
Coefficient of friction
 s  tan 
Mean  s
Surface of contact:________________________ and ________________________
Friction Plane Method
Sr. No.  (Degrees) P (N)
Mean  s
CALCULATION :-
CONCLUSION :-
RESULT :-
W (n)
Coefficient of friction
P - Wsin 
s 
Wcos
BELT FRICTION APPARATUS
Experiment No. 6
BELT / ROPE FRICTION
AIM :To find the coefficient of friction between belt and pulley in a belt friction set up.
APPARATUS REQUIRED :Belt Friction Apparatus, Leather Belt, Rope, Standard weights.
THEORY :Flexible members such as belt and ropes are used for transmission of power from
one shaft to another and these drives are called non positive drives because of the
possible to slip. The various types of belts are Flat belts, V belts and circular belts or
ropes.
The ratio of tension in the belts is given by T1
= e µӨ,
T2
Where T1 = is the tension on the tight side,
T2 = is the tension on the slack side,
µ = is the coefficient of friction,
Ө = is the angle of contact or the angle of lap subtended by the portion of the belt
in contact with the pulley.
For a rope or a v-belt the ratio of tensions is given by
T1
 e  cos ec
T2
 = is the semi groove angle of the V-belt.
PROCEDURE :The belt is wound around the pulley and weights are kept on both sides. Weights
are gradually added on one side until the belt start moving down (impending motion).
The lap angle is varied for different readings by changing the position of movable pulley.
Then the experiment is repeated for rope also.
OBSERVATION TABLE :Sr. No.
Type of Belt
FLAT BELT
ROPE
T1
T2
θ
µ
CALCULATION :T1
 e  cos ec
T2

  cos ec  log e T1T 2

log e T1
T2



RESULT :-
CONCLUSION :-

EXPRIMENT NO. 7
COLLISION OF ELASTIC BODIES.
AIM :To understand and perform a direct central impact between two steel balls and
hence determine the coefficient of restitution of the steel ball.
APPARATUS REQUIRED :The Impact Apparatus, Steel ball, Meter Scale, Firm table with smooth top, Chalk
pieces.
THEORY :A collision between two bodies which occur in a very small interval of time and
during which two bodies exert on each other relatively large forces is called an Impact.
The line joining the common normal of colliding bodies is called the line of impact.
When the mass centers of colliding bodies lie on the line of impact, the impact is known
as central impact. When the velocities of both the colliding bodies are along the line of
impact, the impact is referred to as a Direct Central Impact.
Coefficient of restitution ‘e’ is the ratio of relative velocities after impact to the relative
velocities before impact. If VA and VB are relative velocities of colliding bodies A and B
V '  V A'
before impact and VA’ and VB’ are velocities after impact then, e  B
V AVB
The value of coefficient of restitution ‘e’ is always between 0 and 1 and depends
to a large extent on the two materials involved, the velocities of impact, shape and size of
the colliding bodies also effect ‘e’.
PROCEDURE :1. Fix the impact apparatus on the edge of a firm table.
2. Place the steel ball B on the holder. Adjust the height of the holder such that the
collision of the steel ball A with the steel ball B is direct central.
3. Note the height y of the holder from the ground using meter scale.
4. Using a plumb bob mark ‘O’ original on the ground just below the holder.
5. Place the steel ball on the side at a certain vertical height ‘h’ from the holder.
Note the height ‘h’ with the meter scale.
6. Let the initial location of steel balls be position 1, position 2on the holder and
ground be the position 3.
7. Release the steel ball from position 1 and let it slide down and strike the
secondary ball at position 2. Both the steel balls after impact undergo projectile
motion, falling through a height y land at a different spots on the ground i.e.
position 3 Mark these spots by a chalk and measure the horizontal range ‘x’ from
the ‘o’ origin to the spot. Thus the steel balls travel same vertical distance but
different horizontal distances.
8. Repeat the above steps by changing the height ‘h’ of the steel ball for two more
set of observations.
OBSERVATION TABLE :Sr.
No.
Height
‘h’ (m)
Ball A
Ball B
XA (m)
XB (m)
YA =
YB
(m)
Velocity before
Impact
Ball A
Ball B
VA (m/s) VB (m/s)
Velocity After
Impact
Ball A Ball B
VA’
VB’
(m/s)
(m/s)
Coefficient
of
Restitution
‘e’
CALCULATION :A) Using work energy principle applied to ball A between position 1 and 2 we have,
U12  T2  T1
1
1
mgh  mv A2  mvB2 Ball B is at rest hence VB= 0
2
2
1
 mgh  mv A2  0
2
v A  2 gh
B) Applying the equation of projectile motion to ball A and ball B from the position 3 to
the ground.
i.e. y  x tan  
gx 2
2u 2 cos 2 
h  0
gx 2
or u 
2u 2
g( X )2
2h
(Since for both the ball A and Ball B, the angle of projection   0 )
Using the above equation we get velocities of ball A and ball B after impact is given by
g( X B )2
and V =
2h
V '  V A'
Now coefficient of restitution e  B
V AVB
g( X A )2
V =
2h
'
A
RESULT :-
CONCLUSION :-
'
B