Question

The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Drill Bit? This problem has been solved! See the answer. The high speed industrial drill with diameter of 98 cm develops 5.85hp at 1900 Rpm. What torque and force is applied to the drill bit?

Answer 1

Answer:

1) Torque = 21.934 Nm

2) force = 44.76 N

Explanation:

Power = 5.88 hp

1 hp = 746 W

Power = 5.88 x 746 = 4364.1 W

Angular speed in Rpm = 1900 rpm

But angular speed w = (2¶N)/60 rad/s

= (2 x 3.142 x 1900)/60 = 198.968 rad/s

From,

1) Power P = T x w

Where T = torque

T = P/W = 4364.1/198.968 = 21.934 Nm

2) diameter of drill = 98 cm

Radius = 98/2 = 49 cm = 49x10^-2 m

From torque T = Force x radius

Force = Torque /radius

F = 21.934/49x10^-2 = 44.76 N

Answer 2

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

                         T = 5252*P / N

                         T = 5252*5.85 / 1900

                         T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

                          T = F*r

Where,   r = d / 2

                          F = 2T / d

                          F = 2*16.171 / 0.98

                          F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.