INDIAN INSTITUTE OF TECHNOLOGY BOMBAY DEPARTMENT OF CIVIL ENGINEERING

MID SEMESTER EXAMINATION CE 102 - ENGINEERING MECHANICS

Date: 23-02-2017 Total Marks: 30 Time: 03:00 pm – 5:00 pm

Read the questions carefully. Draw clear FBD wherever necessary, show the steps clearly and mark the final answers. Assume suitable additional data, if required, and state the same clearly. Answer all four questions.

1. The pipe ACDE is supported by ball-and-socket joints at A and E and by the wire DF. Determine the tension in the wire when a 640 N load is applied at B as shown in Figure 1.

[8 Marks]

Figure 1

2. The elements of a rear suspension for a front-wheel-drive car are shown in Figure 2. Determine the magnitude of the force in the suspension FE and link CD if the normal force F exerted on the tire has a magnitude 3600 N.

[8 Marks]

Figure 2

3. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If x = 450 mm and, the mass of the counterweight S is 2 kg, determine the mass of the load L, required to maintain the balance. [7 Marks]

Figure 3

4. A triangular shaped uniform gate placed at the end of a freshwater channel is supported by a pin and bracket at A and rests on a frictionless support at B. The pin is located at a distance h = 0.10 m below the centre of gravity C of the gate. Determine the height of water d in the channel that would open the gate.

[7 Marks]

!

Figure 4

24/2/17�2�

28060+220

130 C FBD -1

60mm D

FDC L= -2802+-602245

⇐MB=O = 205205A E

. B

⇒Fbcx

6=0×(260+90)FBX

260 FBY

+ FDCX 28=0×24590165

F= 3600N

- 3600×165=0 ⇒ Fact 1898N

9 EFY =

FDcx6g +3600 +FBy=O

=> FBY = -3998N

F

130 FBD -2

+245 FEF+60435 B FB×

A E.

3998

Li = -202nF.

60 200 90 = 5769

( EMA = FEFX 4<3,1×260 -3998×(601200+90)

⇒ FEFE5923N

�3�

r

FBD -1

100 FED

C

150+350 A

4505 = 2kg

9 EMC = Sx (150+350+450) - FEDXIOO

= O

⇒ FED =l9zS = 19kg

FBD - 2

250 150

E G.

F

19kg FGB

{ EMF = -19×250+150 XFGB =O

=> FGB = 95kg- BD -3

FGB

_AB

150 350L

9 EMA = 4150 - FGBX(150+350)=0⇒

L=9g5f~~ 106kg

�4�

Neglect atmospheric pressure contributionsince It will almost add up to zero

.

h= 0.1-

cost = 075F .

×-

n w 0.75 0.85d . a = d-

x1B@5

-

x ,a) A

h0.25

COSL 3

0.15b = 0.8/3

23×0.4.

C = 0.15

0.4N

th should pass through A

Fhe acosx - C = e - �1�

a-

4 A Sind

)#z- x

asinxtecosx=D - �2�

CD= 0.68N

b

Alternate Approach

CEMA = Fhcosxfeacostn - c)

- Fhsindfb - asinx ]=o⇒ ( -0.125+0.083 d) Fh + (-0.132

+0.294 d) th

=o=> D= 0.68W

@

I MAE = 0

in T I A

⇒ a=

{480/160,240}- {0/0,240}

D A

Feb = { 0,490,0} - { 480,160,240 }^ F DXFID =

'121 { -16,11 ,-8 }

TEIA = {480,160/0} - {0/0,240}^ E A

XEIA =' 17 {6/2/-3}

to = { 0, -640,0 } 's

RBTA = {200/0,240} - {0/0,240}EMAE =

TDIAXTF# ftp.T/EtatrB1AXT.%fb

= - IGOTFD + 54857=0

⇒ TFD = 3428N