Question

A vertical screw with single start square threads of 50 mm mean diameter and 12.5 mm pitch is raised against a load of 10 kn by means of a hand-wheel, the boss of which is threaded to act as a nut. The axial load is taken up by a thrust collar which supports the wheel boss and has mean diameter of 60 mm. The coefficient of friction is 0.15 for the screw and 0.18 for the collar. If the tangential force applied by each hand to the wheel is 100 n, find suitable diameter of the hand-wheel.

Answer 1

Answer: 1.122m

Explanation: Check the enclosed photo

Answer 2

Concept introduction:

Because thrust is a mechanical force, it must be produced by the propulsion system making intimate contact with a working fluid. The most common way to generate thrust is to accelerate a mass of gas.

Given:

Given that

$d=50mm\\P=12.5\\W=10Kn\\D=60mm\\R=30mm$

$tan$∅=$0.5$

To find:

We have to find  suitable diameter of the hand-wheel.

Solution:

$tan\alpha =\frac{p}{\pi d} =\frac{125}{\pi *50} \\=0.08$

and the tangent force required

$P=10*10^{3} (\frac{0.08+0.15}{1.008*0.15} )\\=2328N$

We also know that the total force required tp turn the wheel

$T=2328*\frac{50}{2}+0.18*10*10^{3} *30\\=58200+54000\\=112200$

We know that the forced applied by the hand wheel

$=100D_{1} Nnm$

Then by calculating two equation

$D_{1} =11220/100\\=1122mm\\=1.122m$

Final answer:

Hence, the suitable diameter of the hand wheel will be $1.122m$.

#SPJ2