Can i have answer of this question please?
1 answer:
You might be interested in
check photo solve
check photo solve
check photo solve
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.