Question

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 mus = 0.18. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?

Answer 1

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 ($\theta_k$ + $\alpha$)

where; F = force

r =  mean radius

$\theta_k$= angle of kinetic friction

$\alpha$ = 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 ($\theta_s$)  $= tan^{-1}(U_s)$

($\theta_s$) = $tan^{-1}(0.18)$

($\theta_s$) = 10.204°

Angle of kinetic friction ($\theta_k$)  $= tan^{-1}(U_k)$

$\theta_k$ $= tan^{-1}(0.16)$

$\theta_k$ = 9.0903°

To determine the pitch angle($\alpha$); we apply the expression:

($\alpha$) = $tan^{-1}(\frac{p}{2 \pi r} )$

($\alpha$) = $tan^{-1}(\frac{0.05}{2 \pi 0.15} )$

($\alpha$) = $tan^{-1}(0.0530516 )$

($\alpha$) = 3.0368°

Have gotten our parameters to solve for Couple (C); then we have:

C = Fr  tan ($\theta_k$ + $\alpha$)

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.