Department of Civil Engineering
S.V. National Institute of Technology, Surat.
7-11 December 2020
Engineering Mechanics (AM 108)
Odd Semester (2020-2021)
Practicals
B.Tech. I (I Semester)
Name.:
Roll No:
Date of Submission:
/ /2020
ASSIGNMENT 1: PLANE FORCE POLYGON
AIM:
To verify the law of polygon of coplanar, concurrent forces in equilibrium using
universal force table.
Figure 1: Universal Force Table for verifying the law of polygon of forces
1
THEORY:
(1) Law of polygon of forces: “More than three coplanar forces acting at a point
which are in equilibrium can be represented in magnitude and direction by
sides of a closed polygon taken in order”;
(2) Concurrent forces: Forces whose line of action passes through a point are called
concurrent forces;
(3) Coplanar forces: Forces acting in one plane are called coplanar forces;
(4) Equilibrium: When the resultant of forces acting on a particle is zero, the
particle will not be displaced and hence the particle can be called to be in
equilibrium under the action of this system of forces;
Let a system of five coplanar forces acting at a point O is in equilibrium
as shown in Figure 1. Let the force polygon of this system of forces has
closing error fa. The resultant of forces F2 and F3 may be represented by side
___
___
___
bd . Similarly, resultant of bd and F4 is represented by be . If the system of
___
forces is in equilibrium, then the force triangle formed by forces be , F5 and F1
must be close (Law of triangle). As a result, the polygon of given system of
forces must also be close. There should not be a closing error, end points a
and f of force polygon must be coincide; and
(4) Resolution of forces: Under equilibrium condition, resultant of given force
system, R = 0.
Burt, R = RX2 + RY2 ,
where, RX and RY are rectangular components of R.
Hence, RX = 0 and RY = 0;
But, RX = ΣFX = 0, and RY = ΣFY = 0.
Σ FX = F1X + F2X+ F3X+ F4X +F5X = Sum of x components of all forces.
Σ FY = F1Y + F2Y+ F3Y+ F4Y +F5Y = Sum of y components of all forces.
Signature of student: _____________
2
PROCEDURE:
(1) Level the force table with the help of spirit level by adjusting foot screws;
(2) Adjust the weights and subtended angles such that the center of ring
coincides with central pivot;
(3) Note down the magnitude of forces and the subtended angles between them;
(4) Repeat the experiment with different weights and subtended angles;
(5) Draw space diagram for these sets of observations to a suitable scale;
(6) Use Bow’s notation to identify forces;
(7) Draw polygon of forces by drawing its sides parallel to corresponding forces
in order to represent them both in magnitude and direction;
(8) Join the tail of first forces vector to tip of last force vector by dotted closing
line;
(9) Measure the length of closing line which gives the magnitude of error;
(10) (a) Determine the unknown weight of block, P; and
(b) Determine the unknown weight of block, P and Q.
EXERCISES:
(1) Using Universal force table, verify the equilibrium condition;
(2) Determine the unknown weight of block, P and Q jointly; and
(3) Verify the equilibrium condition by the method of resolution of forces.
Signature of student: _____________
3
OBSERVATION TABLE:
Magnitude of Forces (N) (Include
weight of hooks)
S.No.
Subtended angles (Degree)
(Between forces)
F1
F2
F3
F4
F5
F1-F2
F2-F3
F3-F4
F4-F5
F5-F1
1
X.0
X.5
X.7
X+1
X.0
50º
80º
100º
60º
70º
2
X.5
X.5
X.0
X.6
X.1
83º
61º
75º
71º
70º
3
X.5
X.0
X.1
X.7
X.0
60º
90º
50º
100º
60º
4
X.5
X+1.2
X.5
X+1
X.0
30º
130º
40º
70º
90º
5
X.0
X.0
X.0
X.0
X.0
72º
72º
72º
72º
72º
F1
F2
F3
F4
P
F1-F2
F2-F3
F3-F4
F4-F5
F5-F1
6
X.0
X.3
X.0
X.0
P
65º
85º
60º
70º
80º
7
X.1
X.2
X.0
X.6
P
80º
40º
100º
70º
70º
F1
F2
F3
P
Q
F1-F2
F2-F3
F3-F4
F4-F5
F5-F1
8
X.5
X.5
X.5
P
Q
90º
65º
45º
100º
60º
9
X.0
X.7
X.7
P
Q
90º
80º
50º
100º
40º
Note: P and Q are unknown weights; X is the last digit of Student’s Roll No., e.g.,
U20CE001, X.0 = 01.0, X.5 = 1.5, X + 1.2 = 2.2. Roll number ending with Zero may
consider the second last digit, U20CE010, X.0 = 1.0, X.5 = 1.5, X + 1.2 = 2.2.
Signature of student: _____________
4
GRAPHICAL METHOD (SAMPLE DIAGRAM)
Signature of student: _____________
5
GRAPHICAL METHOD
Signature of student: _____________
6
ANALYTICAL METHOD (Calculations)
Signature of student: _____________
7
RESULTS AND CONCLUSIONS:
P=
N (by force polygon)
P=
N (by method of resolution of
forces)
P=
N (Actual)
Signature of student: _____________
8
Q=
Q=
forces)
Q=
N (by force polygon)
N (by method of resolution of
N (Actual)