Using the assembly-line balancing procedure, which of the following is the required cycle time in minutes per unit if the daily
1 answer:
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Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
x = Displacement
t = time
With the data we have we can find the time as well
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
Answer:
2.25in³
Explanation:
For a 12 awg conductor the minimum volume allowance as stated by the NEC is 2.25in³
See attached Table 314.16(B) from NEC 2011
