A flat belt fits around a pulley which has a diameter of 45cm. The belt has a speed of 17,6m/s and transmits 9N per mm belt. The effective pull force in the belt is 423N and the belt has a width of 18N. Calculate:

Question

a) The power transmitted by the belt in KW
b) The tight side of the force
c) pulley speed in r/s

Answer 1

a) The power transmitted by the belt can be calculated using the formula:

P = FV

Where:
P = power transmitted (in watts)
F = force transmitted (in newtons)
V = belt speed (in m/s)

Plugging in the values, we get:

P = 423N * 17.6m/s
P = 7469.8 watts

Converting to kilowatts:

P = 7469.8 watts / 1000
P = 7.4698 kW

Therefore, the power transmitted by the belt is 7.4698 kW.

b) The tight side force can be calculated using the formula:

T = F1 + (m * a)

Where:
T = tight side force (in newtons)
F1 = effective pull force (in newtons)
m = mass per unit length of belt (in kg/m)
a = acceleration of the belt (in m/s^2)

First, we need to calculate the mass per unit length of the belt:

m = (9N/mm) * (18mm) / 1000
m = 0.162 kg/m

Acceleration of the belt can be calculated using the formula:

a = V^2 / r

Where:
V = belt speed (in m/s)
r = pulley radius (in m)

Plugging in the values, we get:

a = (17.6m/s)^2 / (0.45m / 2)
a = 154.56 m/s^2

Now we can calculate the tight side force:

T = 423N + (0.162kg/m * 154.56 m/s^2)
T = 423N + 26.15N
T = 449.15N

Therefore, the tight side force is 449.15N.

c) The pulley speed in r/s can be calculated using the formula:

ω = V / r

Where:
ω = pulley speed (in r/s)
V = belt speed (in m/s)
r = pulley radius (in m)

Plugging in the values, we get:

ω = 17.6m/s / 0.45m
ω = 39.11 r/s

Therefore, the pulley speed in r/s is 39.11.