Answer to Question #299487 in Mechanics | Relativity for Amos

Question

The car screw Jack is used to lift a car whose weight is 8800N.The bar for rotating the jack is 0.21m long and pitch of the screw is 2 mm. Calculate the effort required to operate the screw Jack assuming 50% efficient. If the bar is rotated 10 times how far will the car be raised? Hence, calculate the work done in raising the car through this distance.

Answer 1

(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%:

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