In case band and block there are 10 blocks. The coefficient of friction is 0.4. Each Block subtends an angle 16 with the centre of the wheel. The brake absorbs 225 Kw At 300rpm. The effective diameter is 80 cm. Find the force applied at the end of lever 25 cm long if band is attached at a distance of 15 cm and 3 cm either side of the fulcrum. -

n = 10

2θ = 16 $\hspace{3.7cm}$ so θ= 8

De = 80cm = 0.8m $\hspace{2cm}$so Re = 0.4m

L = 25cm = 0.25 m

A = 15cm = 0.15m

B = 3cm = 0.03m

. μ= 0.4

Formula

$TB/TA = ((1+μ \tanθ)/ (1-μ \tanθ))^n$

(Tb – Ta)Re = T

Sol:

Torque $= T = P/w = P \times 60/ 2πN = 225000 \times 60 / 2 π \times 300 \\ = 7162 Nm$

$(Tb – Ta)Re = T \\ Tb – Ta = 7162/0.4 = 17.9 kN …i$

$TB/TA = ((1+μ \tanθ)/ (1-μ \tanθ))^n$

TB/TA $= ((1+0.4 \tan8)/ (1-0.4 \tan8))^{10} \\ = 3 …ii$

Solving i and ii

Ta = 8.95kN

Tb = 26.85 kN

Taking moment about hinge O

$Ta \times A – Tb \times B – F \times L = 0 \\ 8.95 \times 0.15 – 26.85 \times 0.03 – F \times 0.25 = 0 \\ F = 2.15 kN$