Answer:
transmission, driveshafts, differential and axles
Explanation:
The powertrain consists of the prime mover (e.g. an internal combustion engine and/or one or more traction motors) and the drivetrain - all of the components that convert the prime mover's power into movement of the vehicle (e.g. the transmission, driveshafts, differential and axles); whereas the drivetrain does not.
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
<h3>The static equilibrium is given as:</h3>
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area =
d = 1.5
= 1.767 inches²
<h3>Solve for inertia</h3>
= 0.2485inches⁴
Solve for the tensile force from here
30x10³ =
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
Read more on tensile force here: brainly.com/question/25748369
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2