## 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**

**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**

**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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