1. First : statics

2. Two forces of magnitude F , 2F
Newton act at particle ,
magnitude of their resultant is R
st
If the mag. of the 1 - is
increased by 4 N
& the mag. of the 2 nd is doubled,
–

3. then the measure of
the angle between
the resultant
st
& the 1 – force does
not change ,
Find the value of F, 2F

4. f R = 4 √ 7 N , Find the measure o
the angle between the two forces
he
let ∝ be the angle between
the 2 force ,

5. Solution

6. The angle between
st
the
Resultant & the 1 -
force
does not change

7. st nd
Tan θ for 1 - = Tan θ for 2 -
∴
∴
∴ 2F sin ∝ 4 F sin ∝
=
F + 2F cos ∝ F + 4 + 4F cos
∝
∴ 2F + 4F cos ∝ = F + 4 + 4Fcos
∴ 2F = F + 4
∴ F = 4

8. 2 2 2
R = F + 4 F + 2 F (2F) Cos ∝
∴
∴
2
∴ (4 √ 7) = 16 + 64 + 64 Cos ∝
64 Cos ∝ = 32
∴ Cos ∝ = 1
2
°
∴ ∝ = 60

9. A B C D H O is a regular
Hexagon of side length 10 cm.
forces of mag. 3 , 2 √ 3 , 4 , 3 √ 3 ,
2 N act at the vertex D
n directions DC , DB , DA , DO,DH
respectively

10. Find the magnitude & dir. o
the resultant & the distance
from B to the line
of action of the resultant .

11. Solution

12. F 1 ( 3 , 0 º )
=
F 2= (2 √ 3 ,30) y
F O A
3=( 4 , 60 º)
F 4= ( 3 √ 3 , 90 º ) 3√3
4
F 5 ( 2 , 120 º
= )
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

13. X = 3 + 2 √ 3 Cos 30 º + 4 Cos 60 º
+ 3 √ 3 Cos 90º + 2 Cos 120º = 7
y
O A
3√3
4
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

14. Y = 3 Sin 0 º + 2 √ 3 sin
30º
+ 4 sin 60º + 3 √ 3 sin 90 º
y
+ 2 sin 120º = A √ 3
7
O
3√3
4
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

15. Notice that both X , Y are
ositive then θ lies in the 1- quan
st
y
O A
3√3
4
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

16. 2 2 2
∴R = X + y = 49 + 147 = 196
∴ R = 14 N ,
Y
Tan θ = X = √ 3
y
O A ∴θ = 60º
3√3
4
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

17. ∴ R acts along DA ,
Rup = 14 sin 60º = 7 √ 3 cm.
y
O A
3√3
4
H B
2 √3
30 º 30 º
2 30 º
30 º
C X
D 3

18. A particle moves from point
A ( -3 ,1) to point B ( 3 , 9 )
under the action of the force
^ ^
F = 3 i + 4 j .
Find the alg. Component &
the vector component of F
in dir . Of AB

19. Solution

20. AB = B – A = ( 6 , 8 )
alg . comp. = F .
AB
|| AB ||
(3,4)= . (6,8)
√ 36 + 64
= 18 + 32 =
50 = 5
units
10 10

21. .
Vector comp. = ( F
|| AB ||
)
AB AB
2
50 ^ ^
= (6 i + 8 j
100
^ ^
= 3 i + 4 j

22. F ′ ,1 F ′ 2, F ′3 are three coplanar
^ ^
forces where F1 = 3 i – 5 j ,
^ ^ ^ ^ j act res
= - 7i + 2 j , F 3 = 4 i + 3
2
n A ( 2, 41 ) , B ( 5 , -1 ) , C ( -2 , 1

23. Prove that the system is
equivalent to a couple & find
the norm of its moment .

24. Solution

25. Firstly Find the resultant of
F 2 F 3 for example
,

26. R ′ = F ′2 + F ′
3
^ ^ ^ +^
= (-7 i + 2 j)+ (4i 3j)
^ ^
∴R ′ = - 3 i + 5j ∴ R ′ = -1 F ′
∴ The system is equivalent
to a couple .

27. ∴M = MA = AB × F + AC
^ × F^ ^ ^
= ( 3i – 5j) × (-7i + ^
^ ^ ^ 2j)
^ ^
+ (- 4i – 3j) × ( 4i + 3j)
= - 29 k + O = - 29 K
∴ || M || = 29 units

28. AB is a uniform rod of length
60 cm , its weight is 400 gm. wt
rests on a horizontal position on
a support 20 cm from A &
is kept in equilibrium by means
of a vertical light string
at its end B .

29. Find :
 The tension in the string &
The reaction of the support .

30. Find :
The mag. of the weight that
should be suspended at A
so that the tension in
the string is about to be
vanished .

31. Solution

32. ∴
∴ T + R = 400
, M B = 0 ∴ 40 R – 400 × 30 = 0
∴R = 300 gm. wt
∴ T = 100 gm . wt
T R
C
B A
30cm 10cm 20cm
400 W

33. When T = 0 ∴ MC = 0
∴400 × 10 – 20 w = 0
∴ W = 200 gm . Wt.
T R
C
B A
30cm 10cm 20cm
400 W

34. A B C D is a rectangle in
Which AB = 6 cm , BC = 8 cm.
Forces of mag.15 , 20 , 3 , 4 , F
Newton act along BA , CB , DC
AD , AC resp. If the system is
Equivalent to a couple ,
find F & the moment
norm of the couple

35. Solution