The question is incomplete. We have to calculate :
a). the cutting force
b). volumetric metal removal rate, MRR
c). the horsepower required at the cut
d). if the power efficiency of the machine tool is 90%, determine the motor horsepower
Solution :
Given :
Cutting velocity (v) = 500 ft/min
= 500 x 12 in/min
= 6000 in/min
Feed , f = 0.025 in/rev
Depth of cut, d = 0.2 in
b). Volumetric material removal rate, MRR = v.f.d
= 6000 x 0.025 x 0.2
= 30 
c). Horsepower required = MRR x unit horsepower
= 30 x 1.6
= 48 hp
a). Cutting force,
(1 hp = 550 ft lbf /sec)
= 3168 lbf
d). Machine HP required
= 53.33 HP
Answer:
Explanation:
Given data:
mass = 2.00 kg
slope angle = 38.0
From figure
balancing force
.....1
Balancing torque
......2
for pure rolling
from 1 and 2nd equation
N =normal force 
solving for coefficent of friction we get
Answer:
dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)
Explanation:
d/dx (x − 4y) = d/dx (e^(2x + 3y − 1))
1 − 4 dy/dx = e^(2x + 3y − 1) (2 + 3 dy/dx)
Since x − 4y = e^(2x + 3y − 1):
1 − 4 dy/dx = (x − 4y) (2 + 3 dy/dx)
1 − 4 dy/dx = 2 (x − 4y) + 3 (x − 4y) dy/dx
1 − 4 dy/dx = 2x − 8y + (3x − 12y) dy/dx
1 − 2x + 8y = (4 + 3x − 12y) dy/dx
dy/dx = (1 − 2x + 8y) / (4 + 3x − 12y)
Answer:
The magnitude of work input is 178.79 kJ
Explanation:
The process is an isothermal compression process because the temperature inside the cylinder is constant.
W = nRT ln(P1/P2)
n is the number of moles of air in the cylinder = mass/MW = 1.3/29 = 0.045 kgmol
R is gas constant = 8.314 kJ/kgmol.K
T is temperature of air = 24 °C = 24+273 = 297 K
P1 is intial pressure of air = 120 kPa
P2 is final pressure of air = 600 kPa
W = 0.045×8.314×297 ln(120/600) = 111.11661 ln 0.2 = 111.11661 × -1.609 = -178.79 kJ
