The document discusses solving quadratic equations by factorizing. It provides examples of factorizing quadratic expressions and equations to find their roots. In one example, the quadratic equation x^2 - 2x + 2 = 4 is factored into (x - 2)(x + 2) = 0, showing it has only one real root of x = 2. Another example factors a quadratic expression f(x) = x^2 - x - 1 to find its two roots of 1 and -1. The document demonstrates how to factorize quadratic expressions and equations in order to solve for their real roots.
This is a 3 hour sample paper for cbse class 12 board exam. This covers all chapters of ncert 12th math book. For more such papers visit clay6.com/papers/.
This is a 3 hour sample paper for cbse class 12 board exam. This covers all chapters of ncert 12th math book. For more such papers visit clay6.com/papers/.
Chapter wise important questions in Mathematics for Karnataka 2 year PU Science students. This is taken from the PU board website and compiled together.
II PUC (MATHEMATICS) ANNUAL MODEL QUESTION PAPER FOR ALL SCIENCE STUDENTS WHO...Bagalkot
My dear Students,
Wishing you all happy SHIVRATRI. & ALL THE BEST IN YOUR ANNUAL EXAMS-2014
Here I have uploaded II- P.U.C MATHEMATICS MODEL QUESTION PAPER FOR the year 2014 Which i have designed according to New syllabus of CBSE. I hope this model paper will be helpful to all the students who are writing annual exams on 18-March-2014.
wish you all the best
Regards,
A. NAGARAJ
Director-Faculty
Shree Susheela Tutorials
BAGALKOT-587101
mob: 9845222682
Chapter wise important questions in Mathematics for Karnataka 2 year PU Science students. This is taken from the PU board website and compiled together.
II PUC (MATHEMATICS) ANNUAL MODEL QUESTION PAPER FOR ALL SCIENCE STUDENTS WHO...Bagalkot
My dear Students,
Wishing you all happy SHIVRATRI. & ALL THE BEST IN YOUR ANNUAL EXAMS-2014
Here I have uploaded II- P.U.C MATHEMATICS MODEL QUESTION PAPER FOR the year 2014 Which i have designed according to New syllabus of CBSE. I hope this model paper will be helpful to all the students who are writing annual exams on 18-March-2014.
wish you all the best
Regards,
A. NAGARAJ
Director-Faculty
Shree Susheela Tutorials
BAGALKOT-587101
mob: 9845222682
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
1) Use properties of logarithms to expand the following logarithm.docxdorishigh
1) Use properties of logarithms to expand the following logarithmic expression as much as possible.
Logb (√xy3 / z3)
A. 1/2 logb x - 6 logb y + 3 logb z
B. 1/2 logb x - 9 logb y - 3 logb z
C. 1/2 logb x + 3 logb y + 6 logb z
D. 1/2 logb x + 3 logb y - 3 logb z
2) Solve the following logarithmic equation. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, to two decimal places, for the solution.
2 log x = log 25
A. {12}
B. {5}
C. {-3}
D. {25}
3) Write the following equation in its equivalent logarithmic form.
2-4 = 1/16
A. Log4 1/16 = 64
B. Log2 1/24 = -4
C. Log2 1/16 = -4
D. Log4 1/16 = 54
4) Use properties of logarithms to condense the following logarithmic expression. Write the expression as a single logarithm whose coefficient is 1.
log2 96 – log2 3
A. 5
B. 7
C. 12
D. 4
5) Use the exponential growth model, A = A0ekt, to show that the time it takes a population to double (to grow from A0 to 2A0 ) is given by t = ln 2/k.
A. A0 = A0ekt; ln = ekt; ln 2 = ln ekt; ln 2 = kt; ln 2/k = t
B. 2A0 = A0e; 2= ekt; ln = ln ekt; ln 2 = kt; ln 2/k = t
C. 2A0 = A0ekt; 2= ekt; ln 2 = ln ekt; ln 2 = kt; ln 2/k = t
D. 2A0 = A0ekt; 2 = ekt; ln 1 = ln ekt; ln 2 = kt; ln 2/k = toe
6) Find the domain of following logarithmic function.
f(x) = log (2 - x)
A. (∞, 4)
B. (∞, -12)
C. (-∞, 2)
D. (-∞, -3)
7) An artifact originally had 16 grams of carbon-14 present. The decay model A = 16e -0.000121t describes the amount of carbon-14 present after t years. How many grams of carbon-14 will be present in 5715 years?
A. Approximately 7 grams
B. Approximately 8 grams
C. Approximately 23 grams
D. Approximately 4 grams
8) Use properties of logarithms to expand the following logarithmic expression as much as possible.
logb (x2 y) / z2
A. 2 logb x + logb y - 2 logb z
B. 4 logb x - logb y - 2 logb z
C. 2 logb x + 2 logb y + 2 logb z
D. logb x - logb y + 2 logb z
9) The exponential function f with base b is defined by f(x) = __________, b > 0 and b ≠ 1. Using interval notation, the domain of this function is __________ and the range is __________.
A. bx; (∞, -∞); (1, ∞)
B. bx; (-∞, -∞); (2, ∞)
C. bx; (-∞, ∞); (0, ∞)
D. bx; (-∞, -∞); (-1, ∞)
10) Approximate the following using a calculator; round your answer to three decimal places.
3√5
A. .765
B. 14297
C. 11.494
D. 11.665
11) Write the following equation in its equivalent exponential form.
4 = log2 16
A. 2 log4 = 16
B. 22 = 4
C. 44 = 256
D. 24 = 16
12) Solve the following exponential equation by expressing each side as a power of the same base and then equating exponents.
31-x = 1/27
A. {2}
B. {-7}
C. {4}
D. {3}
13) Use properties of logarithms to expand the following logarithmic expression as much as possible.
logb (x2y)
A. 2 logy x + logx y
B. 2 logb x + logb y
C. logx - logb y
D. logb x – ...
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Dr. Vinod Kumar Kanvaria
Exploiting Artificial Intelligence for Empowering Researchers and Faculty,
International FDP on Fundamentals of Research in Social Sciences
at Integral University, Lucknow, 06.06.2024
By Dr. Vinod Kumar Kanvaria
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
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Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
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Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Advantages and Disadvantages of CMS from an SEO Perspective
Additional mathematics
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I!
I:,
i
I
If b' < 4ac and a > 0, f(x) is always > 0.
If b' < 4ac and a < 0, f(x) is always < 0.
(a) D<O,a>O
:)
----~------------~x
Fig.4.21
1(b) D<O,a<O
REVISION EXERCISE 4 (Answers on page 617.) ·
A
'.
1 Without solving the following equations, state the nature of their roots:
(a) 3r-x= I (b) (x+ l)(x-2)=:S
(c) (I -x) = __±._2 (d) ! + 3 = - 1-
1X+ X X-
(e) (2x + 5)(2x + 3) = 2(6x + 7)
2 Find the range of values of x for which 3x' < lOx- 3.
3 Show that the equation (t- 3)x' + (2t- l)x +(t + 2) = 0 has rational roots for all values
oft.
4 Show that the equation (p + l)x' + (2p + 3)x + (p + 2) = 0 has real roots for all values
of p. (C)
5 The quadratic equation x' +px + q = 0 has roots -2 and 6. Find (i) the value ofp and
of q, (ii) the range of values of r for which the equation x' + px + q = r has no real
roots. (C)
6 Express 8 + 2x- x' in the form a- (x + b)'.Hence or otherwise find the range of
8 + 2x- x' for -I ,;; x,;; 5.
7 (a) Find the range of values of x for which 6x'- llx ~ 7.
(b) Find the coordinates of the turning point of the curve y = (2x :- 3)2
+ 6 and sketch
the curve. (C)
8 Find the range of the function 2x'- 7x + 3 for the domain 0,;; x,;; 4.
9 State the range of values of k for which 2k- I and k + 2 are (i) both positive, (ii) both
negative. Hence, or otherwise, find the range of values of k for which 2k2 + 3k < 2.
(C)
96
107.
108.
109.
110. Example 1
Expand (a+ b)8
•
From the triangle the coefficients are I 8 28 56 70 56 28 8 I.
Then (a+ b)'= !a8 + 8a1b + 28a6
b2
+ 56a5b3 + 70a''b' + 56a3b5
+ 28a2
b6
+ 8ab1
+ !b8
Note that the powers of a decrease from 8 to 0 while the powers of b increase from 0
to 8. The sum of these powers is always 8.
(a+ b) is the model binomial but we can replace a orb by other expressions.
Example2
Expand (2x - 1)4
•
The initial coefficients are I 4 6 4 I. Here a = 2x, b = -I.
Then (2x- 1)4 = 1(2x)4 + 4(2x)3(-l) + 6(2x)2
(-1)2
+ 4(2x)(-1)3
+ 1(-1)4
= !6x'- 32x' + 24x'- 8x + I
The coefficients are now quite different. The powers of x are in descending order.
Example 3
Find in ascending powers ofx the expansion of(2- ~ )6
•
The initial coefficients are I 6 15 20 15 6 I. The expansion is
2' + 6(2') (- D+ 15(2') (- ~)' + 20(2')(- ~)' + 15(2')(- ~y + 6(2)(- ~)' + (- D'
= 64- 6(24
)x + 15(22
)x2
- 20x3
+ 15(~) - 6(;) + ~
- 64 - 96x + 60x2 - 20x' + lSx' - 3x' + x'
- 4 8 64
Exercise 5.1 (Answers on page 618.)
1 Find, in descending powers of x, the expansions of:
(a) (x- 2)4 (b) (2x- 3)3 (c) (2x + 1)5
(d)(x-!)' (e)(x+k)' (t) 0-2)'
2 Expand, in ascending powers of x:
(a) (I - 2x)5
(b) (2- 3x)' (c) (2-D' (d) (I - x')'
3 Find, in ascending powers of x, the first four terms in the expansion of:
(a) (2- x)5 (b) (I - 2x)7
(c) (I - ~)' (d) (4- ~)'
100
111.
112. (b) (l - 2x)3(2 + x)4 ; (l - 6x + 12x2
- Bx')(l6 + 32x + 24x2
+ Bx' + x')
The first four terms will go up to the power of x'. So we multiply the terms in the
first bracket by 16, 32x, 24x' and Bx' and leave out any terms higher than x'.
Multiplying by 16 16- 96x + 192x2- 12Bx'
Multiplying by 32x 32x- 192x' + 3B4x'
Multiplying by 24x'
Multiplying by &x'
24x2- l44x'
Bx'
Adding 16- 64x + 24x' + 120x'
Example 6
(a) Find the terms in x' and X' in the expansion of(3-} )6 in ascending powers ofx.
(b) Hence find the coefficient ofx:' in the expansion of(I- ~)(3- ')f
(a) (3- ~)6 ; 36 + 6(35)(- ~) + 15(34
)(- ~)' + 20(33
)(- ~)3
+ 15(32)(- ~)4
...
So the x' term is -20x' and the X' term is + sr.
(b) Then (1- })(3- ~)6 ; (1- 1)(...- 20x' + Sf...)
The term in X' is found by multiplying the relevant terms as shown, and is
lOX'+ sr giving a coefficient of
3
i.
Example 7
Write down and simplify the first three tenns in the expansioils (in ascending powers
ofx) of(a) (1 - 3; )' and (b) (2 + x)5
•
Hence find the coefficient ofx' in the expansion of(2 - 2x- 3;' )'.
(a) (1 _ 3x)s; 1 + 5(-;l,t,) + l0(-;l,t,)2 ; 1- 15x + 45x'
2 2 2 ... 2 2
(b) (2 + x)5 ; 25 + 5(24)(x) + 10(2')(x') ... ; 32 +BOx+ BOx'
We notice that (2- 2x- 3;' )5 is the product of (a) and (b)
; [(1 - 3x )(2 + x)]'
2
; [1 _ 15x + 45x' ...][32 +BOx+ BOx'...]
2 2 .
The term in x' will be the sum of the products linked together, so the coefficient of x'
is BO- ().2 X BO) + ( ~ X 32); 2002 2 .
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307.
308. Example 11
Jff(x) = x'- Jx' +x +2, solve the equationf(x) = 0.
x- I is not a factor of f(x).
Verify that x + I is also not a factor.
Try x- 2 and verify that this is a factor.
Then f(x) = (x- 2)(x'- x- 1).
The roots of f(x) = 0 are x = 2 and the roots of x'- x- I = 0,
. 1±-15 •
I.e. x =- 2
- =1.62 or --0.62.
Example 12
Given thotf(x) = x' -2x' +2x, solve the equationf(x) = 4.
f(x) = 4 gives x' - 2x' + 2x - 4 = 0. To solve this equation, we first factorise the
polynomial x' - 2x' + 2x - 4.
Check that x + I and x- I are not factors. Now try x- 2.
Divide the polynomial by x - 2 to obtain the other factor.
The equation is (x·- 2)(x' + 2) = 0 and the only root is x = 2 as x' + 2 = 0 has no real
roots.
Example 13
Find the nature and x-coordinates ofthe turning points on the curve
y = 3X' +4x' -6x'- 12x +I.
::;: = 12x' + !2x'- 12x- 12 = 12(x' +.i' -x- I)
::;: =Owhenx'+.i'-x-1=0.
x- I is a factor of the left hand side of this equation.
Thenx' +.i'-x-1 = (x-I)(x' + 2x+ I)= (x-I)(x+ 1)2
and so::;: = 0 whenx= I
or x =-I.
d' •
J = 12(3.i' + 2x- I)
Wh I d'y 0 th' . .. .en x = , d.xl > so IS IS a mmnnum pomt.
When x =-I, ~ = 0 so we use the sign test on ::;: = (x- l)(x + 1)2
•
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320.
321.
322.
323.
324.
325. Example 18
The first term ofa GP exceeds the second term by 4 and the sum of the 2nd and 3rd
terms is 2j_ Find the first three terms.
a -ar = 4
and ar + ar' = ~
a- ar 8 3
Divide (i) by (ii) to eliminate a: ar + ar' =4 + 3 =2
1- r 3
Then r+r' = 2
which gives 2 - 2r =3r + 3r' or 3r' + 5r- 2 =0.
Hence (3r- l)(r + 2) = 0 giving r = t or -2.
From (i), when r =f, a= 6 and when r =-2, a= 1.
. 2 4 8 16
The first three terms are therefore etther 6, 2, 3 or 3 , -3 , 3.
Example 19
(i)
(ii)
A store finds that it is selling 10% less ofan article each week. In the first week it sold
500. In which week will it be first selling less than 200?
The number of articles sold forms a GP with a = 500 and r = 0.9.
[If a= 500 then r, is 10% less i.e. r, = 0.9a]
r. = 500(0.9)~1
and we require the least value of n for which 500(0.9)"-' < 200
i.e. 0.9"- 1
< 0.4. Then 0.9" < 0.4 x 0.9 = 0.36.
This can be found quickly using the x' key of a calculator and testing values of 0.9"
for say n = 5, 6, ... and stopping when the result is first< 0.36. This will be for n = 10.
In the lOth week less than 200 are sold. (An alternative method using logarithms is
shown in Chapter 15).
Geometric Means
If a, b and c are consecutive terms of a GP, then b is the geometric mean of a and c.
~ = ~ so b2
= ac or b = &. For example, the geometric mean of 2 and 32 is 8 as
,J2 x 32 = 8.
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484. Example 4
Rain is falling vertically at 5 km h-1• A man is sitting by the window of a train
travelling at 40 km hc1
• In what direction do the raindrops appear to cross the windows
ofthe train?
Fig. 21.17
The vectors are shown in Fig. 21.17.
40
(train)
The velocity of the rain relative to the train= r - t.
So tan 9 = J0 giving 9 ~ 7.1°..
5
r (rain)
Thus the rain drops appear to cross the windows at 7.I0
to the horizontal.
Example 5
A ship is sailing due north at a constant speedof12 knots. A destroyer sailing at 36
knots·is 30 nautical miles due east ofthe sliip. At-this moment, the destroyer is ordered
to intercept the ship. Find
(a) the course which the destroyer should take,.
(b) the velocity of the destroyer relative to the ship,
(c) the time taken for the destroyer to reach the ship.
[It is assumed that both the ship and the destroyer do not change their velocities.].
c
~"T --- - v I
Fig. 21.18·
t-sI
I
30n';;, ------ t-s
-.._--J
Fig.21.18 shows the.·positions of the sltip. S and the destroyer D. We reduceS' torest
by. introducing a.velocity of'l2 knots due south to both Sand D. Then the course of
-> . ->
Dis DC and; ttrinterceptS; its track (DT)·must lie along DS.
474
485. By the cosine rule,
IoWI'= 402
+ 102 -2 x 40 X 10 x cos 30° giving IoW I~ 31.7.
AI sine sin30° .. e 910so, 10 = 3IT g1vmg ~ . .
Hence the true velocity of the wind is 31.7 km h-1
towards the direction (60°- 9.1°)
= 50.9° or from the direction 230.9°.
2 By calculation (using vectors)
[By vectors, take i along oE and j along oN.
Qt = (40 sin 60°)i + (40 cos 60°)j = (20 -J3)i + 20j
WiG= -!Oi
OW= OC + Wi(; = (20..J3- IO)i + 20j
-> . r;; ->
IOW /2
= (20 '13 -10)2
+ 202
giving IOW I~ 31.7.
AI 2o,f3-w . . 509oso, tan ~ = 20
glV!ng ~ ~ . .
Thus we obtain the same results as before.
3 First draw a sketch and label it with all the information given (it should be a rough
version of Fig. 21.19). The actual drawing must be done carefully. Choose a
suitable scale to ensure reasonably accurate results, say I em for 4 km h-1.
Draw a north line ON as in Fig. 21.19.
From 0, draw OC 10 em long with LNOC = 60°.
From C, draw CW 2.5 em long parallel to OE.
Join OW.
Measure OW (and convert to km h-1) and LNOW.
Compare with the calculated values above.
Exercise 21.2 (Answers on page 649.)
1 Aeroplane A is flying due Nat !50 km h-1
• Aeroplane B is flying due Eat 200 km h-1
•
Find the velocity of B relative to A.
2 Two cars A and B are travelling on roads which cross at right angles. Car A is
travelling due east at 60 km h-1, car B is travelling at 40 km h-1 due north, both going
towards the crossing. Find the velocity of B relative to A. [The magnitude and
direction must be given].
3 A passenger is on the deck of a ship sailing due east at 25 km h-1• The wind is blowing
from the north-east at 10 km h-1• What is the velocity of the wind relative to the
passenger?
4 A road (running north-south) crosses a railway line at right angles. A passenger in a
car travelling north at 60 km h-1 and 600 m south of the bridge, sees a train, travelling
west at 90 km h-1, which is 800 m east of the bridge. Find the velocity of the train
relative to the car.
476
486.
487.
488.
489.
490.
491.
492.
493.
494.
495.
496.
497.
498.
499.
500.
501.
502.
503.
504.
505.
506.
507.
508.
509.
510.
511. (a) We use unit vectors, taking i along the horizontal direction and j the vertical
direction as shown in Fig. 23.12.
Fig. 23.12
+I
I
I
I
I
20 N
50°______.....
i
The horizontal component is (20 cos 50°)i z 12.9i.
The vertical component is (20 sin 50°)j z 15.3j.
Hence the horizontal and vertical components are 12.9 Nand 15.3 N respectively.
(b) We take i along the plane andj normal to the plane as shown in Fig.23.l3.
Fig. 23.13
---
j
I
I
I
I
I
I
--
20 N
The component along the plane is (20 cos 30°)i z 17.3i.
The component normal to the plane is (20 sin 30°)j = 10j.
Hence the components along and normal to the plane are 17.3 Nand 10 N respec-
tively.
COPlANAR FORCES ACTING ON A PARTICLE
Coplanar forces are forces acting in the same plane. We can find the resultant of any
number of coplanar forces acting on a particle by resolving the forces in 2 perpendicular
directions of our choice and then recombining the components to obtain the resultant.
502
512. Example 3
Forces ofmagnitudes 1 N, 2 N, 3 N, 4 Nand 5 N act along the lines OA, OB, OC, OD
and OE respectively, where OABCDE is a regular hexagon. Find the resultant ofthe
. 'forces.
Li
Fig. 23.14 A B
The forces acting are shown in Fig.23.14.
Take 1and j along and perpendicular to OC respectively.
Sum of components along OC
= (I cos 60° + 2 cos 30° + 3 + 4 cos 30° + 5 cos 60°)1 = 11.21
Sum of components perpendicular to OC
= (-1 sin 60°- 2 sin 30° + 0 + 4 sin 30° + 5 sin 60°)j = 4.5j
Hence the resultant forcer= 11.21 + 4.5j (Fig.23.15).
Fig. 23.15
4.5j r I
C2JI
e
0 11.21
I r I'= 11.22
+ 4.52
giving I r I= 12.1.
The angle 9 which the resultant makes with OC is given by
e 45
· · e 21 9°tan = !1.2 gtvmg = . .
Hence the resultant is 12.1 N-acting at 21.9° to OC.
We can also obtain the resultant graphically.
c
Let the forces along OA, OB, OC, OD, OE be a, b, c, d, e, respectively. A suitable
scale is chosen, say 1em: 1 N. We s(arl by drawing a line to represent a in magnitude
and direction. (We can start with any force, the order is immaterial). Then we add the
other forces, one at a time, each force starting from the end point of the one before.
503
513.
514.
515. 4 Two forces P Nand Q N include an angle of 120° and their resultant is {19 N. If
the included angle between the forces were 60°, their resultant would be 7 N. Find
p~~ ~
5 Find the magnitude and direction of the resultant of the following coplanar forces
acting at a point 0: 10 N in direction 000°; 5 N in direction 090°; 20 N in direction
135°; 10 N in direction 225°.
6 The resultant of two forces X N and 3 N is 7 N. If the 3 N force is reversed, the
resultant is {19 N. Find the value of X and the angle between the two forces;
7 Two forces of 13 N and 5 N act at a point. Find the angle between the forces when
their resultant makes the largest possible angle with the 13 N force. Find also the
magnitude of the resultant when the angle between the forces has this value. (C)
8 The resultant of a force 2F N in a direction 090° and a force F N in a direction 330°
is a force of 12 N. Calculate the value of F.
It is required to add a third force X N in a direction 270° so that the resultant of the
system is in a direction 000°. Calculate the value ofX. (C)
9 Two concurrent forces of equal magnitude have a resultant of 12 N. When one force
is reversed the resultant becomes 6 N. Calculate the magnitude of each force and the
angle between them. (C)
10 Fig.23.20 shows four forces in a plane. Given that cos 8 = ~ and that the resultant of
the forces is 6--/2 N in a direction 225°, calculate the values of P and Q. (C)
2QN
Fig.23.20
2PN
11 Four horizontal forces of magnitudes 1 N, 2 N, 3 N, and 4 N act at a point in the
directions whose bearings are 000°, 060°, 120° and 270° respectively. Calculate the
magnitude of their resultant.
A fifth horizontal force of magnitude 3 N now acts at the same point so that the
resultant of all five forces has a bearing of 090°. Find the bearing of this fifth force.
(C)
506
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