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?

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

it is a negative acceleration

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle