Week 7 – SS3 First Term Further Mathematics Notes

WEEK SEVEN
STATICS – CONTINUATION
Definition of Moment of a Force
Principles of Moments
Application of the Principle in Solving Problems

Definition
Moments of a Force: The moment of a force about a reference point is defined as the product of the force and the force arm. Moment of a force is a vector quantity and its units is Nm.

The direction of movement or sense of movement about the point is duly considered.
The object can move in clockwise moment and ant-clockwise moments about the given point.

Suppose we have an object acted upon by two forces, F1 and F₂ in the opposite direction, then
M₁ = F₁ x di M₂= F₂ x d₂
Where M₁ = Magnitude of F₁ M₂ = Magnitude of F₂
d1 = Force arm of d₁ d₂ = Force arm of d₂

PRINCIPLE OF MOMENTS
1. When a system of coplanar forces is in equilibrium, then the sum of the clockwise moment is equal to the sum of the anti-clockwise moments about the same point in the plane.
2. When the system is not in equilibrium, then the resultant of two coplanar forces F₁ and F₂ denote by R is represented by the relationship below:
M₁ + M₂ + M_R
Where M₁ = Moment of F1 M₂ = Moments of F₂ M_R = Moments of T.

Centre of Gravity: The centre of gravity of a uniform plank or rod is the midpoint of the plant or rod

Examples
1. A uniform rod PQ is 15m long and has mass 20kg. The rod rests on two supports at P and Q. An object of mass 5kg is suspended at a point R on the rod 5m from the end P. Calculate the reaction of the supports P and Q (take g = 10ms^-2)
Solution
    Kpkq

    7.5m
     5m
P Q


5kg 20kg
Let the reaction at P be Kp and at Q beK_q
NB: The weight of the rod acts downward through the midpoint of the rod.

Moments about the point P, M_R = M₁ + M₂
But M₁ = F₁ x d₁ and F = Mg.
Kq x 15 = (5 x 10 ) x 5 + ( 20 x 10 ) x 7.5
15kq = 50 x 5 + 200 x 7.5
K_q = 250 + 1500
15.
Kq = 116. 7N.
Moment about the point Q,
Kp x 15 = ( 5 x 10 ) x 10 + (20 x 10 ) x 7.5.
15Kp = 500 + 1500
Kp = 2000
15.
Kp = 133.3N

A uniform beam AB of length 6m and mass 20kg rests on support P and Q placed 1m from each end of the beam. Masses of 10kg and 8kg are placed at A and B respectively. Calculate the reactions at P and Q (g = 9.8ms^-2)

Solution
8kg
10kg

20kg

Let reaction at point P be Rp and at point Q be Rq.
:. Moment about the point Q
Rp x 4 = ( 10 x 9.8) x 5 + (20 x 9.8 ) x 2 – ( 8 x 9.8 x 1 )
4Rp = 490 + 392 – 78.4
Rp = 803.6
4. Rp = 200.9N

Moment about the point P:
Rq x 4 = ( 8 x 9.8 ) x 5 + (20 x 9.8 ) x 2 – ( 10 x 9.8) x 1
4Rq = 392 + 392 – 98
Rq = 689
4
Rq = 171.5N

EVALUATION
A uniform rod PQ, is 20 m long and weighs 80N, has weights 20 N and 50N suspended at P and Q respectively. Find the distance from P where the rod must be supported so that it will rest horizontally.

GENERAL EVALUATION
A uniform rod PQ of length 10 m and mass 2kg rest on two supports at x and y. If PX = 2m and QY = 1m, find the reaction of X. (Take g = 10ms^-2)
WEEKEND ASSIGNMENT
A uniform beam PQ of length 100cm and of weight 35N lies on a support 40cm from the end P. Weights 54N and W are attached to the ends P and Q respectively to keep the beam in equilibrium. Find the value of W, to the nearest whole number.