PHYSICS S.S ONE 3RD TERM WEEK 9

Topic:                    MOMENT           

Sub topic:

Reference materials:

(1) ESSENTIAL PHYSICS, TONALD PUBLISHERS, EWELUKWA (2017)

(2) NEW SCHOOL PHYSICS, AFRICAN FIRST PUBLISHERS, ANYAKOHA M.W (2011)

(3) COMPREHENSIVE PHYSICS FOR SSS, EZEBUIRO G.N (2004)

(3) INTERNET

Instructional materials: Rubber ball, catapult, small pebbles

Entry behavior: The students have been taught expansion

Behavioural objective: At the end of the lesson the students should be able to:

I.                    Define moment

II.                  Explain principle of moment

III.                Define couple and centre of mass

IV.                Solve mathematical problems on moment

CONTENT

MOMENTS

The moment (or torque) of a force about a turning point is the force multiplied by the perpendicular distance to the force from the turning point.

Moments are measured in newton metres (Nm).

Moment = F d

F = the force in newtons (N)

d = perpendicular distance in metres (m)

Example; A 10N force acts at a perpendicular distance of 0.50m from the turning point. What is the moment of the force?

Moment = Fd
= 10 x 0.50
= 5.0 Nm

 The principle of moments.

” When an object is in equilibrium the sum of the anticlockwise moments about a turning point must be equal to the sum of the clockwise moments.”

sum of anticlockwise moments = sum clockwise moments

Example;


sum of anticlockwise moments = sum clockwise moments
F1 x d1 = F2 x d2

OR


sum of anticlockwise moments = sum clockwise moments
F1 x d1 =  (F2 x d2) +  (F3 x d3)

Couples

A couple is two equal forces which act in opposite directs on an object but not through the same point so they produce a turning effect.

The moment (or torque) of a couple is calculated by multiplying the size of one of the force (F) by the perpendicular distance between the two forces (s).

E.g. a steering wheel in a car;

OR

Moment of Couple = Fs

Centre of Mass

The centre of mass of an object is the point at which all the mass of the object seems to act. In a uniform gravitational field this is the same point as the centre of gravity of the object which is where all the weight of the object seems to act.

1. If FR is the net force of F1, F2, and F3, what is the magnitude of force F2 and x?

Known :

Net force (FR) = 40 N

Force 1 (F1) = 10 N

Force (F3) = 20 N

Wanted: The magnitude of force F2 and distance of x

Solution :

Find the magnitude of force F2 :

Force points to upward, signed negative and force points to downward, signed negative.

ΣF = 0

– FR + F1 + F2 – F3 = 0

– 40 + 10 + F2 – 20 = 0

– 30 + F2 – 20 = 0

– 50 + F2 = 0

F2 = 50 Newton.

Plus sign indicates that the direction of the force is upward.

Find x.

Choose A as the axis of rotation.

τ1 = F1 l1 = (10 N)(1 m) = 10 Nm

The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.

τ2 = F2 x = (50)(x) = 50x Nm

The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.

τ3 = F3 x = (20 N)(1.75 m) = -35 Nm

The torque 2 rotates beam clockwise so we assign negative sign to the torque 2.

The net of moment of force :

Στ = 0

10 + 50x – 35 = 0

50x – 25 = 0

50x = 25

x = 25/50

x = 0.5 m

2. Forces of F1, F2, F3, and F4 acts on the rod of ABCD as shown in figure. If rod’s mass ignored, what is the magnitude of the moment of force, about point A.

The axis of rotation = points A.

Known :

Force F1 = 10 N, the lever arm l1 = 0

Force F2 = 4 N, the lever arm l2 = 2 meters

Force F3 = 5 N, the lever arm l3 = 3 meters

Force F4 = 10 N, the lever arm l4 = 6 meters

Wanted : the moment of force about point A

Solution :

Moment of force 1 (τ1) = F1 l1 = (10)(0) = 0

Moment of force 2 (τ2) = F2 l2 = (4)(2) = -8 Nm

Moment of force 3 (τ3) = F3 l3 = (5)(3) = 15 Nm

Moment of force 4 (τ4) = F4 l4 = (10)(6) = -60 Nm

If torque rotates rod counterclockwise then we assign positive sign.

If torque rotates rod clockwise then we assign negative sign.

The resultant of the moment of force :

τ = 0 – 8 Nm + 15 Nm – 60 Nm

τ = -68 Nm + 15 Nm

τ = -53 Nm

Minus sign indicates that the moment of force rotates rod clockwise.

3. Three forces act on a rod, FA = FC = 10 N and FB = 20 N, as shown in figure below. If distance of AB = BC = 20 cm, what is the moment of force about point C.

Known :

The axis rotation at point C.

Distance between FA and the axis of rotation (rAC) = 40 cm = 0,4 meters

Distance between FB and the axis of rotation (rBC) = 20 cm = 0.2 meters

 Distance between FC and the axis of rotation (rCC) = 0 cm

FA = 10 Newton

FB = 20 Newton

FC = 10 Newton

Wanted : The resultant of the moment of force about point C.

Solution :

Moment of force A :

ΣτA = (FA)(rAC sin 90o) = (10 N)(0,4 m)(1) = -4 N.m

Minus sign indicates that the moment of force rotates rod clockwise.

Moment of force B :

ΣτB = (FB)(rBC sin 90o) = (20 N)(0,2 m)(1) = 4 N.m

Plus sign indicates that the moment of force rotates rod counterclockwise.

Moment of force C :

ΣτC = (FC)(rCC sin 90o) = (10 N)(0)(1) = 0

The resultant of the moment of force :

Στ = Στ1 + Στ2 + Στ3

Στ = -4 + 4 + 0

Στ = 0 N.m

PRESENTATION

Step I: The teacher explains the principle of moment

Step II: The teacher explain couple and centre of mass

Step III: The teacher leads the students in solving problems on moment

EVALUATION

The teacher evaluates the lessons by asking the following questions:

I.                    Define moment

II.                  Explain principle of moment

III.                Define couple and centre of mass 

Comments

Post a Comment

Popular posts from this blog

POST HARVESTING PROCESSES

CHAPTER TWELVES POST HARVESTING PROCESSES 12.1             CONCEPTS AND PHILOSPHY OF POST HARVEST TECHNOLOGY Post-harvest fisheries entails all steps, process and or activities immediately after catch handling and transportation processing and distribution of fish and fishery product where if control is not applied post-harvest losses will occur.         Post –harvest fish losses are a major concern and occur in most fish distribution chains throughout the world. Not only do losses constitute lost income to fishers, processors and traders, they also contribute food insecurity i.e a loss of fish means less fish available for the consumer. POST HARVEST PROCESSES:         Fish are highly perishable. A central concern of fish processing is to prevent fish from deteriorating . Causatic factors of fish spoilage 1. High moisture content.   ...
Read more

NICOM SAT 1

TOPIC: - NICOM-SAT 1    INSTRUCTIONAL MATERIALS: -         Satellite dish REFERENCES BOOK: -     1.      FARINDE O. E e tal, ESSENTIAL PHYSICS FOR SSS, Tonad Publishing Limited. 2.      M. W. ANYAKOHA (2011), NEW SCHOOL PHYSICS FOR SENIOR SECONDARY SCHOOLS, Africana first publishers.    3.      INTERNET PREVIOUS KNOWLEDGE: -     Students have been taught Niger-sat 1 INSTRUCTIONAL OBJECTIVES: - At the end of the lesson students should be able to: - 1.        Explain the concept of NICOM-SAT 1 2.        State the features of NICOM-SAT 1 3.        Uses of NICOM-SAT 1 CONTENT: - NICOM- SAT 1 NigComSat-1 is a Nigerian Communication Satellite. It was launched on 14 may, 2007 in China by China Great Wall Industry...
Read more

FISH SEED PRODUCTION

CHAPTER THIRTEEN FISH SEED PRODUCTION BROOD STOCK SELECTION             Brood stock means a group of sexually mature individuals of a culture species that is kept separate for  breeding purpose. 13.1     DIFFERENCES  BETWEEN MALE  AND FEMALE SEXUALLY MATURED FISHES             Male and female breeders may be easily differentiated by the shape of the body and the relative position of genital papilla. In females, the body is plump and the genital opening is situated above the genital papilla. In males, the body is slender and the genital opening is found behind the genital papilla. External signs of the brood stock A male brood stock will release milt under a slight abdominal pressure: Its belly is not blown up but rather slim; it sometimes has callosities on the head. A female brood stock has a well-rounded and soft  or semi...
Read more