WEEK FIVE
STATICS – CONTINUATION
Definition of equilibrium
Condition of equilibrium of rigid body
Application of the condition to solve problems
Lami’s theorem and application
Definitions
Equilibrium is when a body remains at rest under the action of given forces.
Translational Equilibrium: The state of equilibrium of bodies which remain at rest under the action of forces have tendency to cause translation.
Condition of Equilibrium
When a block is placed on a table as shown below and force F1 and F2 are applied to the block.
The block remains in translational equilibrium if the magnitude of F1 and F2 are equal.
Since F1 and F2 are acting in opposite direction, but have equal magnitudes.
Then, F1 = -F2
F1 + F2 = 0
Also, the upward force N balances the downward force mg on the block.
:. N = -mg, N + mg = 0
Hence, the sum of the vertical components and horizontal components of forces acting on a body in translational equilibrium is equal to zero.
Example
1. A particle of mass 5kg is supported by two light inelastic strings inclined at angles 300 and 450 respectively to the horizontal. If the system is in equilibrium, calculate the tension in each string.
Solution:
Let the two inelastic strings to be DP and OQ
Resolve each string vertically and horizontally:
Horizontal : Efx = T1Cos 300 + T2cos 450 + 0
Vertical : Efy = T1 Sin 300 + T2 sin 450 – 50
:. Efx = 0.7061T2 – 0.866T1 = 0 equation I
Efy = 0.7071T2 + 0.5T1 = 50 equation II
Solving equation I and II simultaneously:
- 1.366T1 = -50
T1 = -50/-1.366 T1 = 36.6N
Substitute T1 in equation I or II using equation 1, 0.7071T1 – 0.866T1 = 0.
0.7071T2 = 0.866 x 36.6
T2 = 31.6956
0.7071
T2 = 44.82N
:. The tensions in the strings are T1 = 36.6N, T2 = 44.82N
2. A particle of mass 5kg is suspended by a light in extensive string which makes an angle of 300 with the downward vertical and horizontal forced F. If the system is in equilibrium, calculate
- The tension in the string (ii) The magnitude of the force F (take g = 10ms-2)
Solution
I using Sin θ = AB
OA
Sin 600 = 5 x 10
T
T = 50/sin 600, T = 57.74N. The tension in the string is 57.74 N
II. Magnitude of F1 Tan 300 = OB
AB
50 x Tan 300 = F :. F = 28.87N
EVALUATION
1.A street lamp of mass 10kg is suspended at a position ) by two wires OP and OQ across a rood such that each wire is inclined at an angle of 800 to the upward vertical, If the system is in equilibrium, calculate the tension in one of the two wires. (Take g = 10ms-2)
TRIANGLE OF FORCES
If three coplanar forces act on a body in such a way that the system is in equilibrium, then the forces can be represented in magnitude and direction by the sides of a triangle taken in order
The triangle representing the three coplanar forces is called a triangle of forces.
Example:
1.A body of mass 6.5kg is supported by two strings, One of the stings is inclined at an angle of 300 and the other 400 to the horizontal . Find the tension in each strings, if the system is in equilibrium (take g = 10ms-2)
Solution
Using Sine rule;
T1 = 65
Sin 50o Sin 70o
T1 = 65 x sin 50o
Sin 50o Sin 70
T1 = 65 x 0.766
0.9397
T1 = 52.98N
Similarly:
T2 = W
Sin 600 Sin 700
T2, = 65 x sin 600 = 65 x 0.8660
Sin 700 0.9397
T2 = 59.9N
EVALUATION
A body of mass s10kg is suspended by means of two light inextensible strings. AP and BP which are inclined at angles 600 and 300 respectively to the downward vertical. If T1 and T2 are the magnitude of the tension AP and BP respectively, calculate the values of T1 and T2.
LAMI’S THEOREM
This theorem states that if three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the liens of action of the other two forces.
Consider the forces F1, F2 and F3 below
By Lami’s theorem: F1 α Sin BF2 α sin γF3 α sin θ
Since the forces are proportional to the sine of the angle: then F1 = F2 = F3
Sin Bsinγ sin θ
Example:
A body of weight 91N is suspended by two inelastic string 5m and 12 m long attached to two points on the same horizontal level, whose distance apart is 13m. Using Lami’s theorem or otherwise, find the tension along the strings.
Solution
s
Le the tension in the strings be T1nd T2 and x, B be the angles made with the horizontal.
AB2 = OA2 + OB2 = 52 +122
= 25 + 144 =169
but AB = 13
AB2 = 132 = 169
:. OA2 + OB2 = AB2
hence AOB is a right angle triangle
:. Sin B = 5/13, sin x = 12/13
B = sin -1 0.3845 Ө = sin-1 12/13
B = 22.60 = 0.9231 = 67.40
Using Lami’s theorem:
T1 = W
Sin 157.4 sin 900
T1 = 91 x sin 157.40 = 91 x 0.3843
Sin 900 1
T1 = 34.97N
T2 = W
Sin 112.6 sin 900 T2 = 91 x sin 112.6
Sin 900
T2 = 91 x 0.9232
1
T2 = 84.01N.
EVALUATION
A particle of mass 10kg is connected by two strings of length 3m and 4m to two points on the same horizontal level and 5m apart, find the tension in the strings.
GENERAL EVALUATION
- A particle of mass 98kg is suspended by two light inelastic strings of length 9m and 12m from two fixed point P and which are 15m apart. Calculate (i) the angles made by the strings with the upward vertical (ii) the tension in the strings.
- The ends P and Q of an inextensible string 17m long are attached to two fixed points 13m apart on the same horizontal level. A body of mass 20kg is suspended from a point C on the string 5m from P. Calculate (a) the angle which each part of the string makes with the horizontal (b) the tension in each part of the string.
READING ASIGNMENT
Read Equilibrium “pages” 170-177 of Further Mathematics Project III.
WEEKEND ASSIGNMENT
Two forces (8N, 0300) and (10N, 1200) act on a body; find the magnitude of the force that would be applied to keep the system in equilibrium.