Sample Problem of Pulley – Belt - Friction

Sample Problem of Pulley – Belt - Friction
The tension in the pulley is 110 N when stationary. Calculate the tension in each side and the power transmitted when the belt
Is on the point of slipping on the smaller wheel.
The wheel is 240 mm diameter and the coefficient of friction is 0.32. The angle of lap is 165º. The wheel speed is 1500 rev/min
a)
b)
c)
d)
F1 = 140 N ,
F1 = 157 N ,
F1 = 130 N ,
F1 = 120 N ,
F2= 60 N , P= 1,500 W
F2= 62 N , P= 1,786 W
F2= 70 N , P= 1,632 W
F2= 50 N , P= 1,720 W
Solution
Belt Velocity = 𝜋 ∗ 𝑛 ∗ 𝐷 = 3.14 ∗ 1500 ∗ 0.24 ÷ 60 = 18.85 𝑚/𝑠
Lap angle = (165/180) ∗ 𝜋 = 2.88 𝑅𝑎𝑑𝑖𝑎𝑛𝑠
(Initial Tension)
Carlos Batista M. Sc. P. Eng. CCEA
CBA Engineering Support Services
www.cbaengineer.com
F1 = F2
When Stationary
F1 + F2 = 220 N
Tension in belts
F1/F2 = 𝒆𝒖 𝜽 = 𝒆𝟎.𝟑𝟐 𝟐.𝟖𝟖 = 2.513 N
F1 = 2.513 × F2
=
2.513 x (220 – F1)
=
552.9 – 2.513 x F1
F1 + 2.513 x F1 = 552.9
F1 = 157.4 N
3.513 x F1 = 552.9
When we obtained the first tension F1, in the FE exam is not necessary to
continue solving F2 and Power (P) , because if we see the answers,
we can identify rapidly the correct answer
Carlos Batista M. Sc. P. Eng. CCEA
CBA Engineering Support Services
www.cbaengineer.com
2
However, Is wise idea to continue a little more to be sure that our
Previous calculation is correct.
F2 = 157 ÷ 𝟐. 𝟓𝟏𝟑 = 𝟔𝟐. 𝟔 𝐍
𝐅𝟐 = F1 ÷ 2.513
Excellent, We are more
Sure that answer b is the
Correct answer
P = V ( F1 – F2) = 18.85 m/s x ( 157.4 – 62.6) N =
Be careful with units conversion
Carlos Batista M. Sc. P. Eng. CCEA
CBA Engineering Support Services
www.cbaengineer.com
3
1,786 W