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

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.