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Static

Exam

mode!

Answer the following questions 20 questions From lt o 12 choose the correct answer Q ( l) If the friction force equals 25 nektons and the coefficient of friction 5 equals then the resultant reaction equals

© 56

© 65

® 13

© 75

F = jjR *’ ‘ 2 5

=

^?R

**'

R

=

6°

R = R / l + (j' = 6 0 x J1+

=65

3 (2 ) The m om ent o f the force 100N about the p o in t (O) N.m

© ®

® - io o J z

100

©

- z o o jz

-200

-1 0 0 x 4 s in 4 5 ° = -2 0 0 jZ Q (3 ) The opposite fig ure represents a rod in equilibrium , then F =

0 28

Newton

(D 16 Newton

®

2 Newton

20N

T

® 4 Newton

-1 2 x 4 -F x 2 + 20x4 = 0

12N

1 4m

~7\2m 2m

/ . -48-2F + 80 = 0 .\F = 16

Q(4) Three forces are represented com pletely by the sides o f an equilateral triangle ABC in the same c yclic o rder w ith scale 2cm o r each 8 N. If the side length o f the triangle is 5 cm ,then the lo rm o f the m om ent o f the couple e q u a ls .......... gm . w t . cm o .........................................

©

50V3

®

®

50

©

4 x 2 x ix 5 x 5 x s in 60° = 50 V3 2

50V2 100V2

S ta tic

Exam

Q (5) ABCD is right trapezium The set of forces form a a ouple Then

F=

llcm J) ION

©

30N

™■i6

in.

\ a

m

3

3) 20N

©

12N

\

-\

B+ -------- TJ------------ ►C

17cm

10

F = 20N

8

Q (6) U niform density and thickne ss w ire suspended fre ely from B Then BC inclined to the vertical angle o f measure

3)

t a r f 1-^

©

3)

tan-1 3

®

3

tan"1-^-

a

9

M

tan

3

^ ■

24cm

tan"1!

Q (7)

In the opposite fig u re If the rod is about to silid e , then

(Ri,R2)= ©

( w ,- |w )

®

©

( w ,w )

f 2

© (W ,2 W )

(w,iw)

y \

N

Q (8 ) In the given fig ure Ft and F2 act at the points (1,0 i and (5,0) are tw o parallel forces th e ir resultant R acts the poi n tD th e point D is y *

D

3)

(7,0) (8,0)

®

(8,0)

©

(2,0)

IN ! = J? + (- I f = J5 21| = j ? + ( - 6 f = 3 ^ 5 Md = 0 /. ^5 X L = 3^5 (L - 4).\ L = 3(L - 4) L = 3L-12 2L = 12 . L = 6 Point D is (7,0)

«—

J

\\

C

Static

Exam

Q(9) The center of gravity of three equal masses each of 2Kg placed at the vertices of a right -angled triangle whose length of two legs 3cm and 4cm ©

M) N )

f4

0 + 4 + 0 3+ 0 + 0

—1

3’

Q(10) ABC is an equilateral triangle the length of its side equals 18cm forces of magnitudes 8,6,13 Newtons act along AB, BC, AC the sum of algebraic moments of the resultant of thes forces about the centroid = .cm

®

.cm AD = 9^3

®

3^3N .cm 4,/3N .cm

/.MD = 1 x 9 ^ 3 = 3^3

Mm = -8x3^ 3-6x3^ 3 + 13x3^3 = ^

.cm

Q ( l l ) ABCD is a uniform lamina of side length 50cm Suspended from nail at A an equilibrium under a couple Of magnitude 250gmwt.cm If AB is vertical then W =R =......

©

2

© 10

5

© 25

W=R

W x 25 = 250

.\W = 10cm

Q(12) The algebraic sum of the moments of this system of forces about C equals ©

(D

570

©

580 15cm C

500

©

600

Mc = 12x 15 + zjz x 15sin45° + 8x4 5 + 6 x 0 = 570Kg.wt.cm

O

-HU

Static

Exam

Q (13) A body of mass SOOgm.wt is placed on a rough plane which is inclined to the horizontal at an angle of measure 6 where tane =-|> the body is connected by a string passing over a smooth pulley at the top of the plane from the other end of the string a scale pan of weight 25gm.wt if the least weight should be put in the scale pan so that the body in equilibrium is 175gm find the coefficient of friction then prove that the greatest weight which should by in the scale pan with keeping equilibrium is 575gm.wt In case of least force : The body is about to move down T = 200

R = W cos 0 = 500 X1 = 300 5 T + |jR = 5OOsin0 200 + jj x 300 = 500 x £ 5 300|a = 200 ,.M = f In case of the greatest weight: The body is about to move upward T = R = 500 cos 0 = 500 x 35 = 300

W

T = 5OOsin0 + |jR W = 5 0 0 x 4 + - | x 300 = 600 5 3 greatest weight: 600-25=75

Q (15) A force F at the xy-plane acts on the triangle A O B. If the algebraic measure o f the m om ent o f F about point O is equal to 84 Newton . m , the algebraic measure o f its m om ent about p oint A is equal to -1 0 0 Newton . m , and the algebraic measure o f its m om ent at point B is equal to zero, determ ine F

Let the force (L,M) M e =[(3,0)-(0,0) x(L,/W)=84 3/W= 84 ,\M = 28 = [(3,0)-(0,4) x (L ,M ) = -1 0 0 L = 46 3/W+ 4L = -100 \ 4L = -184 F = V282+462 = 10V29 m

a

Static

Exam

Q (16) Find the m om ent o f the ofrce about (0)

i

k

0

j 0

10

16

12

-2 0

= ~ 120 i = 160 j

x

8tn

&

Q (17) (The resultant of two parallel forces is 75Newtons , one of the two forces equals 50 Newton’s and acts at a distance 25 cm from the line action of the resultant find the second force and also the distance between the two forces if the given force and the resultant act in the same direction

F2 = R - F 2 = 7 5 - 5 0 = 25 50 x 25 = 25 x AC AC = 50cm F, = 50

R = 75

F2 = ?

Q (18) ABCD is a rectangle , AB=9cm , BC=24cm , H and O are the mid-points of BC and AD , the forces 27,36 and 45N act along Afe,§fi and HA find the two forces act along B?> and

to be the system in equilibrium

The forces 27,36,45 in cyclic order 27 _ 36 _ 45 _ « 9 12 15 Norm of couple momemt = 3 x : j- jx 9 x 12j = -324 Fx12sin0 = 324

/. F = 45

Static

Exam

Q (19) A 10 Kg.wt weight is placed on a rough plane which is inclined to the horizontal. A force is applied to the weight parallel to the line of greatest slope upwards . If the limit force lies which the applied , so as to make the weight abou to move upwards is 6Kg.wt ,the limit force lies which the applied , so as to make the weight about to move . downwards is 4Kg.wt find (1) the angle of inclined of the horizontal plane (2) the friction coefficient between the body and the plane About to move upwards About to move downwards 1Osin0 + pR = 6 -> (l) dfm 1Osin0 = 4 + |jR -» (2) |jR = 6 - 1 0 s in 6 -»■ (3) R = 10 cos 0 —» (4) S ub stitu ting from 3 in 2 /. 1Osin0 = 4 + 6-1O sin0

2Osin0 = 1 0 si n0 = =

.*.0 = 30°

/. R = 10x cos30° = 5Vs substituting in (3)/. px5^3 = 6 -1 0 sin 3 0 |i = 1 sjz\x = 1 ------

Q(20) ABC is a fine lamina of uniform thickness and density in the form of a right - angled triangle at B where AB = 12 cm, BC = 20 cm and X , Y and Z are the midpoints of AB , BC and CA respectively. The triangle CYZ is cut off and coincided with the triangle Y B X. If the system is freely suspended from point B. Find the tangent of the angle of inclination of BC to the vertical in the equilibrium position.

ABC -» ' 0 + 0 + 20 0 + 12 + 0 3 5 3 0 +0+10 0 +6 +0 XYB —»j 3 ’ 3 -P 10 + 10 + 20 0 + 6 + 0 ZYC —»f

M

Mass ABC XYB ZYC mass 4K

K

K

X

20 3

10 3

40 3

y

4

2

2

x=

f4K x# +K x 4 ? -K x ^ 3 3 3 4K + K - K

tan0 = 4 , # = | i 6 25

10cm

25 6

Y

( 4Kx 4 + K x 2 - K x 2 ^ L

4K + K - K

J

y 10cm (10,0)

(20 ,0)

Static

Exam

Answer the following questions From lt o 12 choose the correct answer Q ( l) If the ratio between the resultant reaction force and the normal eaction force Equals i j z : 3 then the angle of friction is equal © 45°

® 30'

CD 90°

© 60‘ J3

cos0 = z j%

2

3 (2 ) A body o f w eight 3Newtons is placed on a horizontal rough

Diane whose c o e ffic ie n t w ith th e body is

then th e fric tio n fo rc e

® [1,»[

© [0 ,1]

< [4 i [04]

Q (3 ) In the opposite figure : A BCD is a parallelogram m(ZA)= 60c X , Y , Z and L are mid points AB, BC, CDand AD given that the orces are in equilibrium then F =.... N

-6 x -2 ^ L + F x ^ L = 0

F= 3

Q (4 ) I f a se t o f fo rc e s are in equilibrium then :

®

R=0

(D

R

5=0

0

© R=1 ® otherw ise

Static

Exam

Q(5) The resultant of two parallel forces is 350Newtons , one of the two forces equals 500 Newton’s and acts at a distance 51cm then distance between the two force

3) 30cm 3) 170cm

® 150cm

© 51cm

c

3

51cm

> «

-Q

v 350

o

* = Fi -F 2 F, = 5 0 0 -3 5 0 = 150 500x51 = 150 x AC /.AC 170cm

Q (6)The force F = 12i + mj acts at the p o in t A = ( - 5,3) and its line o f action bisect BCWhere B = (1,2), C = (1,9) use the m om ent to the value o f m D

©

S>

©

B = (V )

(- 6,l)x (12,m) = -(- 6,-6) x(l 2,m).-. -6m -1 2 = -(- 6m + 72); 6m -1 2 = 6m - 72

.\12m = 60 .\m = 5

c -*w>

Q (7 ) if a driver of a car wants to ascend an plane inclined to the horizontal by an ingle of measure 60° ,then the static friction coefficient between the wheels of the :are and the inclined plane must be not less than...............

3) 1 3) o

® ©

J3 Jz

tan 60° = J3

Q (8 ) The sum o f m om ent o f the forces about B Equals

3) 3)

©

©

- 250 x i x 110 + 200cos 30° x 150-150 x 60

B

1■

K r 60cm

SOcm

40cm 200N

Static

Exam

Q(9) AB is a uniform rod of length 90 cm and mass 5 kg. C and D are its two trisection points from end A. Masses of magnitudes 1 , 2 , 3 and 4 kg are placed at A , B , C and D respectively. Then distance from the center of i

. , ( 0 ,0 )

m gl a

(3 0 ,0 ) Q

30cm

'

( A C

3 Kg

n'i

|5Kg|

(90,0)

( 6 0 > °) D

4Kg

30cm

—mB 2 Kg

r

©

48cm

©

49cm

(D

7cm

©

50cm

xr_( 1x0 + 3x30 + 4x60 + 2x90 + 5x 45^ _ 49cm “ I 1 +3 +4 + 2 + 5 )

Q(10) The m om ent o f the force 6N about the p o in t O equals

©

14

®

16

®

18

©

19

X

2

4

3

:.X

= |

M 0 = 6 x |

= 1 6

Q (1 1) ABC is a triangle in which AB=AC =26cm , BC= this set of forces is equivalent to a couple the magnitude of its moment

© -3360

-1680 ©

-3365

A D =

J

z

6

2

-

102 = 2 4 c m

K =

= 7 x 2 x f -jx 20 x 24^ = -3360

^

do

= ^

r =

20

7

-240