Question No 1
A pulley system uses a flat belt of cross section area 1000 mm square and density 1150 kg/m^3. The angle of the lap on the smaller wheel is 130, the coefficient of friction is 0.32 and the maximum force allowed in the belt is 550N and velocity is 9 m/sec. Calculate
 The maximum power when centrifugal force is not included
 The maximum power when centrifugal force is included
 The initial belt tensions
Solution
 a. Maximum Power without centrifugal force
Power= P=(T_1 T_2 )v
T_1/e^uθ = T_2
550/e^(0.32×3.14) = T_2
T_2=201.36 N
P=(550 201.36)9
P= 3137.71 WAns
 b. Maximum Power with centrifugal force
Power= P=(T T_c ) v ×C
C=1 1/e^uθ
C=1 1/e^(0.32*3.14)
C=0.63
T_c=mv^2
m=area ×length ×density
m=0.001×1×1150=1.15 kg
T_c= 1.15×9^2=93.15 N
P=(550 93.15) 9 ×0.63=2590.33 WAns
 c. initial belt tensions
Without Centrifugal force
T_o=((T_1+ T_2 ))/2
T_o=(550+201.36)/2=375.68 N
With centrifugal Force
T_o=((T_1+ T_2+2T_c ))/2
T_o=(550+201.36+2(93.15))/2 =468.83 N Ans
Question 2
The following data is for a conical clutch
Inside diameter 35 mm
Outside diameter 150 mm
Coefficient of friction 0.35
Included angle 120
Speed 2500 rev/min
Calculate the axial forced needed to allow the transmission of 1000 watts without slipping using
 The uniform pressure theory
 The uniform wear theory
 If the clutch is replaced by a multiplate clutch with 3 contact surface for the same axial force what power can be transmitted by
 The uniform wear theory
 Uniform pressure theory
Solution
Power= P= τω
P/ω= τ
τ=1000/((2 ×3.14 ×2500)/60)=3.82 Nm Ans
Uniform pressure theory
T_r=2 π u p_n cosec α ((〖ro〗^3 〖ri〗^3)/3)
3.82=2 ×3.14×0.35 × p_n× cosec 60 ×((〖0.075〗^3 〖0.0175〗^3)/3)^
p_n=3.82/0.000152=25035 Pa
Axial Force = W = 2 π p_n r dr
W= 2 π ×25035×((0.075+ 0.0175)/2)×(0.0750.0175)
W = 418.10 N Ans
Uniform Wear theory
T_r=2 π u C cosec α ((〖ro〗^2 〖ri〗^2)/2)
3.82=2 ×3.14×0.35 × C× cosec 60 ×((〖0.075〗^2 〖0.0175〗^2)/2)^
C = 3.82/0.002922 = 1307.03
Axial Force = W =2 π C dr
W = 2 π ×1307.03× (0.0750.0175)
W = 471.96 N Ans
Multi disc clutch
Uniform pressure
Torque= T=uW/3 ×(Do^3Di^3)/(DoDi) ×n
T= (0.35×418)/3 ×(〖0.15〗^3〖0.035〗^3)/(0.1500.035) ×3
T=48 ×0.029×3=4.17 Nm
Power=P=((2×3.14×N))/60×T
P=1091.15 WAns
Uniform wear
Torque= T=uW/3 ×(Do+Di) ×n
T=(0.35×471.96 )/4 ×(0.15+0.035)×3
T=22.91 N
Power=P=((2×3.14×N))/60×T
P=5994.78 WAns
Question 3
3) The following gear train has an efficiency of 75% and the drum rotates at 75 rpm. If the torque input from the motor is 8Nm.Calculate
 The number of teeth on the gear X
 The speed at which the load is moving up as the rope on the drum
 The power input by the motor
 The power output by the gear train
 The load on the drum
Solution
a. Number of Teeth
Gear ratio= 50/x × 100/30 × 80/25=75/2000
50/x ×3.34 ×3.2=0.0375
50/x=0.0375/10.688
50/x=0.003508
x=50/0.00350=14222.22=14223 Teeths Ans
b. Speed
v=r × ω
v=0.2/2 × (2 ×3.14 ×75)/60
v= 0.785 m/sec Ans
c. Power of motor
P= τω
P=8 ×(2 ×3.14 ×2000)/60
P = 1674.66 W Ans
d. Power output
As efficiency of the train is 75% so
P_out=0.75 × P_motor
P_out=0.75 ×1674.66
P_out= 1256 W Ans
e. load torque
A power at the drum is the power of gear train so
P_out= τω
τ=P_out/ω
τ=1256/((2 ×3.14 ×75)/60)
τ=160 Nm Ans
Question 4
4) The diagram below shows two masses on two rotors in
planes B and C. Determine the masses to be added to the rotor in plane A and D
at a radius of 50 mm which will produce static and dynamic balance.
Solution
M mass kg

R distance mm

M*r

X distance from A

M*r*x


A

Ma

50

50Ma

0

0

B

5

20

100

80

8000

C

3

50

150

150

22500

D

Md

50

50Md

250

12500Md

From the polygon
for M*r*x we have
12500Md = 25000
Md = 2 Kg Ans
Mr of point D = 50*2
= 100 at an angle of 58 degree Ans
From the polygon
for M*r we have
50 Ma = 105
Ma = 2.1 Kg at
an angle of 35 degree Ans
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