Question

a screw jack is used to lift a small car. the handle used to turn the screw has a length (radius) of 20 cm. the screw advances 0.5 cm for each full rotation (pitch of the screw). the worker applies an effort force of 150 n at the handle. part b. if the screw actually lifts a load of 4000 n, calculate the actual mechanical advantage (ama) and the efficiency of the screw. show all steps clearly. type response here g: 20,.5,150,4000 u: ama e: s: s:

Explanation:

Step1: Define Actual Mechanical Advantage formula

The formula for Actual Mechanical Advantage (AMA) is $AMA=\frac{Load}{Effort}$.

Step2: Substitute given values

We know that the load $L = 4000\ N$ and the effort $E=150\ N$. So, $AMA=\frac{4000}{150}=\frac{80}{3}\approx26.67$.

Step3: Define Ideal Mechanical Advantage formula

The Ideal Mechanical Advantage (IMA) of a screw - jack is given by $IMA=\frac{2\pi r}{p}$, where $r$ is the radius of the handle and $p$ is the pitch of the screw. Here, $r = 20\ cm$ and $p=0.5\ cm$. So, $IMA=\frac{2\pi\times20}{0.5}=\frac{40\pi}{0.5}=80\pi\approx251.33$.

Step4: Define efficiency formula

The efficiency $\eta$ of a machine is given by $\eta=\frac{AMA}{IMA}\times100\%$.

Step5: Calculate efficiency

Substitute the values of AMA and IMA into the efficiency formula: $\eta=\frac{\frac{80}{3}}{80\pi}\times100\%=\frac{1}{3\pi}\times100\%\approx\frac{1}{9.42}\times100\%\approx10.62\%$.

Answer:

AMA $\approx26.67$, Efficiency $\approx10.62\%$