Answer to Question #94579 in Mechanics | Relativity for Wisdom

Question

A screw jack whose pitch is 22mm is used to raise a motor car of mass 900kg through a height of 20.0cm the lenght of the tommy bar of the jack is 60% efficient calculate the velocity ratio,mechanical advantage of the Jack, effort required, wordone by

Answer 1

The condition seems to be a bit incomplete. The full problem is:

A screw jack whose pitch is 2 mm is used to raise a motor car of mass 900 kg through a height of 20.0 cm. The length of the tommy bar of the jack is 40 cm. If the jack is 60% efficient. Calculate:

(i)      The velocity ratio

(ii)     Mechanical advantage of the jack

(iii)     Effort required

(iv)     Work done by the effort (g = 10 ms^-2)

Solution:

(i) The velocity ratio is the length of the circle formed by the tommy bar over the pitch:

$$\text{V.R.}=\frac{2\pi R}{p}=\frac{2\cdot3.14 \cdot0.4}{0.002}=1256.637.$$

(ii) The mechanical advantage:

$$\text{M.A.}=\eta\text{V.R.}=0.6\cdot1256.637=753.982.$$

(iii) The effort required is:

$$E=\frac{\text{load}}{\text{M.A.}}=\frac{mg}{\text{M.A.}}=\frac{900\cdot10}{753.982}=11.94\text{ N}.$$

(iv) Work done by the effort. Keep in mind that here we should include the loss of efficiency, or we should only lift the car for 20 cm, but also apply additional effort to overcome friction and other factor included in the efficiency of 60%:

$$W=\frac{\text{work to lift the car}}{\eta}=\frac{mgh}{\eta}=3000\text{ J}.$$