## Brake Force

Consider a braking system with a 10 m long brake arm and a rotational speed of 25 rev/s. If the total brake power is 50 kW, what is the applied force (N) at the end of the brake arm?

Expand Hint
Brake Power:
$$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon$$$where $$\tau$$ is torque, $$\upsilon$$ is rotational speed, $$F$$ is the force at the end of the brake arm, and $$R$$ is the brake arm’s length. Hint 2 No conversion is needed for rotational speed since the problem statement specifies rev/s. Brake Power: $$\dot{W}_b=2\pi \tau \upsilon =2\pi FR\upsilon$$$
where $$\tau$$ is torque in $$N \cdot m$$ , $$\upsilon$$ is rotational speed in $$rev/sec$$ , $$F$$ is the force at the end of the brake arm, and $$R$$ is the brake arm’s length. Thus,
$$F=\frac{\dot{W}_b}{2\pi R\upsilon }=\frac{50,000W}{2\pi (10m)(25rev/s)}$$$$$=\frac{50,000kg\cdot m^2}{2\pi (10m)s^3(25\frac{rev}{s})}=\frac{50,000kg\cdot m}{1,570s^2}=31.8\:N$$$
31.8 N
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