MEC420 – TEST NO. 1 Fakulti Kejuruteraan Mekanikal Universiti Teknologi MARA

Name:

UiTM No.:

Score:

Instruction: There are two (2) questions in this test. Answer all questions.

Question 1 (60 Marks)

a) Consider a system which consists of several masses connected by an inextensible rope as shown in Figure Q1. If the system is released from rest where the masses of the pulleys are negligible and mA = 2mB = 50 kg,

(i) draw the free-body and kinetic diagrams and also state all your assumptions clearly, (10 Marks)

(ii) determine the acceleration of block A and block B, and the tension in the rope. (35 marks)

Soln: Figure Q1

FBD and KD of each block:

Assumptions:

1) Rope is inextensible

2) Mass of all pulleys is negligible

3) No friction in pulleys and btw the rope

4) No air resistance

Kinematics: 4yA + yB = l (1)

Differentiating with respect to time yields

4 Ay + By = 0 or 4vA + vB = 0 (2)

4 Ay + By = 0 or 4aA + aB = 0 (3)

Kinetics: Block A,

[Fx = max]: WA – 4T = mAaA mAg – 4T = mAaA

or aA = g – 4T/mA (4)

Block B,

[Fy = may]: WB – T = mBaB mBg – T = mBaB

or aB = g – T/mB (5)

Substituting (4) and (5) into (3), yields T = )16(

)(5

BA

BA

mm

mmg

=

)16(2

)(25

BB

BB

mm

mmg

=

18

10 gmB

Therefore,

aA = – 1.09 m/s2, aB = 4.36 m/s

2 and T = 136.25 N [ans]

b) Based on your answers in part a), what is the relative velocity of block A to block B 2 s after they are released? [10 Marks]

Soln: vA/B = vA – vB where vA = uA + aA t = 0 + (– 1.09)(2) = – 2.18 m/s or 2.18 m/s

vB = – 4vA = – 4(– 2.18) = 8.72 m/s

vA/B = – 2.18j – (8.72j) = – 10.90j m/s2 or 10.90 m/s [ans]

c) Explain the difference in the answer for part a) if the friction in all the pulleys were to be

included. [5 Marks]

Soln: The system will experience energy losses, which will result in reduced acceleration

for both blocks.

B

AmA

mBB

AmA

mB

B

A

T T T T

WA = 50g

WB = 25g

x

y

T

mAaA

mBaB

[+4]

[+10]

[+3]

[+2]

[+3]

[+4]

[+2]

[+3]

[+3]

[+3]

[+2]

[+10]

[+3] [+3]

[+3]

[+2]

[+2]

[+2]

Question 2 (40 Marks)

a) Ball bearings leave the horizontal trough with a velocity of magnitude u and fall through

the 100-mm-diameter hole as shown in Figure Q2. Determine (i) the maximum value of

u if the ball bearings were to enter the hole BC, and (ii) the corresponding time of flight.

[35 Marks]

Figure Q2

Soln: Maximum u corresponds to when the ball bearings hit end C. Choosing the x-y

coordinate with the origin at A, we have at position C, we have x = 0.4 m, y = – 0.180 m.

For the motion in the x direction, we have

vx = vocos = u

and x = xo + vocos t = 0 + ut (1)

and for the motion in the y direction, we have

vy,o = vosin = 0

y = yo + (vosint – 21 gt2 = 0 + 0t – 2

1 gt2 (2)

Substituting the corresponding values at C into eq(1) through (2), we obtain

from (1) u = 0.4/t (4)

from (2) – 21 gt2 = – 0.180 (5)

Solving eq(4) and (5) gives

u = 2.09 m/s and t = 0.1916 s [Ans]

b) If the ball is now fired at an angle = 30o counter clockwise from the horizontal with the

same initial velocity u found in part a), what is your comment? Explain. [5 Marks]

Soln: The ball bearings will travel farther than before. i.e. the horizontal distance

covered will be more than 400 mm and they will not drop into the hole. Therefore, the

initial velocity u must be reduced if the ball bearings were to drop into the hole.

A u

B

300 mm 100 mm

180 mm

C

[+5]

[+5] [+5]

[+2]

[+3]

[+2]

[+3]

[+3]

[+2]

[+3] [+3]

[+2]

[+2]