'a mechanical jack is used to lift a load of 3KN. the mean diameter of the screw thread is 60mm and the pitch is 12mm. if the coefficient of friction is 0.26 calculate the torque required to lfit he load and the efficiency of the jack'
Use "conservation of energy" together with E= fd where E is the energy required to apply force f over distance d. In order to lift an object of weight 3 KN a distance d meters, you must use 3000d Joules of energy. With pitch 12 mm, to move upward d meters the screw must turn d/0.012 meters. To get 3000d Joules of energy over that distance you must apply force (3000d)/(d/0.012)= 3000(0.012)= 36 Newtons. With coefficient of friction 0.26, that is a total of (1.26)(36)= 45.36 Newtons. Finding the torque seems impossible here since you would need to know, not the "mean diameter of the screw thread", but the length of the arm over which you are applying the force.
refer to the diagram ...
$F = $ applied force
$N = W\cos{\theta} + F\sin{\theta}$
$f = \mu N$
$W =$ axial weight
$p =$ pitch distance
$d =$ screw diameter
parallel to the incline ...
$F\cos{\theta} = W\sin{\theta} + \mu W \cos{\theta} + \mu F\sin{\theta}$
$F = W\tan{\theta} + \mu W + \mu F \tan{\theta}$
$F - \mu F \tan{\theta} = W\tan{\theta} + \mu W$
$F(1 - \mu \tan{\theta}) = W(\tan{\theta} + \mu)$
note ... $\tan{\theta} = \dfrac{p}{2\pi r}$
$F\left(1 - \dfrac{\mu p}{2\pi r}\right) = W\left(\dfrac{p}{2\pi r} + \mu \right)$
$F\left(\dfrac{2\pi r - \mu p}{2\pi r}\right) = W\left(\dfrac{p+2\mu \pi r}{2\pi r} \right)$
$F = W\left(\dfrac{p+2\mu \pi r}{2\pi r - \mu p} \right)$
$\tau = Fr = Wr\left(\dfrac{p+2\mu \pi r}{2\pi r - \mu p} \right)$
efficiency = (work out)/(work in) = $\dfrac{Wp}{\tau \cdot 2\pi}$
... substitute the given values for $W$, $p$, $\mu$, and $r = \dfrac{d}{2}$