Regulation 433.3.3 gives six examples where th
1 answer:
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The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
The omitted part of the question shown in bold format
The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
Answer:
C = 0.967 in. lb
Explanation:
Given that:
The lead of the threaded shaft of the C-clamp is 0.05 in.
∴ the pitch of the screw = 0.05 in
the mean radius of the thread is r = 0.15 in.
Assuming:
(μs)= 0.18 which implies the coefficient of the static friction
(μk) = 0.16 (coefficient of kinetic friction)
Force = 30-lb
What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
To determine the Couple (C) that must be applied; we use the expression:
C = Fr tan ( +
)
where; F = force
r = mean radius
= angle of kinetic friction
= pitch angle
NOW, let's take then one after the other.
From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:
Angle of static friction ()
() =
() = 10.204°
Angle of kinetic friction ()
= 9.0903°
To determine the pitch angle(); we apply the expression:
() =
() =
() =
() = 3.0368°
Have gotten our parameters to solve for Couple (C); then we have:
C = Fr tan ( +
)
substituting our values; we have:
C = (30 × 0.15) tan ( 9.0903 + 3.0368)
C = 4.5 × tan ( 12.1271)
C = 4.5 × 0.2148761968
C = 0.96669428854 in.lb
C = 0.967 in. lb
Therefore, 0.967 in. lb couple must be applied to the shaft to exert a 30-lb force on the clamped object.