Question
Download Solution PDFA brake-block shown below has a face width of 300 mm.
The mean coefficient of friction is 0.25. For an activating force of 400 N, the braking torque in Nm is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
Free body diagram of lever and brake dram are
For lever moment about O, ∑Mo = 0
⇒ (400 N) × (400 + 200) mm = N × 200 mm
\(\Rightarrow Normal\;force,\;N = \frac{{400 \times 600}}{{200}} = 1200\;N\)
Tangential force, Ft = μN = 0.25 × 1200 = 300 N
Braking torque = Tb = Ft × r = 300 × 0.15 = 45 N.m
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