Question

part ii 45 marks answer three questions from this part. all questions carry equal marks. 8. (a) (i) draw a pulley - system that has a velocity ratio of 4. (ii) a body of mass 25 kg is pulled over a rough surface with a 35 n force. if the object accelerated at a rate of 1.5 m/s², calculate the frictional force acting on the object and the surface. 5 marks (b) define the following terms as applied to machines: (i) velocity ratio; (ii) efficiency. 4 marks (c) (i) a screw - jack of pitch 2 mm is to be used to lift a car of mass 8000 kg. the length of the tommy bar of the jack is 25 cm. calculate the effort that would have been required to attain an efficiency of 85 %. (ii) explain why the efficiency of a machine is always less than 100 %. 6 marks

Explanation:

Step1: Find net - force on the body in part (a)(ii)

According to Newton's second law $F_{net}=ma$. Given $m = 25\ kg$ and $a=1.5\ m/s^{2}$, so $F_{net}=25\times1.5 = 37.5\ N$.

Step2: Calculate frictional force in part (a)(ii)

The applied force $F = 35\ N$. Since $F_{net}=F - f$ (where $f$ is the frictional force), then $f=F - F_{net}$. So $f = 35-37.5=- 2.5\ N$. The negative sign indicates that the frictional force acts in the opposite direction of the motion. The magnitude of the frictional force is $2.5\ N$.

Step3: Calculate the load in part (c)(i)

The load $L$ is the weight of the car. Using $L = mg$, with $m = 8000\ kg$ and $g = 9.8\ m/s^{2}$, we have $L=8000\times9.8 = 78400\ N$.

Step4: Calculate the velocity - ratio of the screw - jack in part (c)(i)

The velocity - ratio $VR$ of a screw - jack is given by $VR=\frac{2\pi r}{p}$, where $r$ is the length of the tommy - bar and $p$ is the pitch. Given $p = 2\ mm=0.002\ m$ and $r = 25\ cm = 0.25\ m$. Then $VR=\frac{2\pi\times0.25}{0.002}=\frac{0.5\pi}{0.002}=250\pi$.

Step5: Calculate the effort in part (c)(i)

The efficiency $\eta=\frac{MA}{VR}$, and $MA=\frac{L}{E}$ (where $E$ is the effort). So $\eta=\frac{L}{E\times VR}$. Rearranging for $E$, we get $E=\frac{L}{\eta\times VR}$. Substituting $L = 78400\ N$, $\eta = 0.85$ and $VR = 250\pi$, we have $E=\frac{78400}{0.85\times250\pi}=\frac{78400}{212.5\pi}\approx118.5\ N$.

Step6: Explain machine efficiency in part (c)(ii)

In a machine, some work is always done against friction and other non - ideal factors. The work input $W_{i}$ is used to do useful work $W_{u}$ and overcome losses $W_{l}$. So $W_{i}=W_{u}+W_{l}$. Efficiency $\eta=\frac{W_{u}}{W_{i}}$. Since $W_{l}>0$, then $W_{u}

Step7: Define velocity ratio in part (b)(i)

The velocity ratio of a machine is the ratio of the distance moved by the effort to the distance moved by the load in the same time interval. Mathematically, $VR=\frac{d_{E}}{d_{L}}$.

Step8: Define efficiency in part (b)(ii)

The efficiency of a machine is the ratio of the mechanical advantage ($MA=\frac{L}{E}$) to the velocity ratio ($VR$). It can also be defined as the ratio of the useful work output to the total work input. $\eta=\frac{MA}{VR}=\frac{W_{u}}{W_{i}}\times100\%$.

Answer:

(a)(ii) The frictional force is $2.5\ N$.
(b)(i) The velocity ratio of a machine is the ratio of the distance moved by the effort to the distance moved by the load in the same time interval.
(b)(ii) The efficiency of a machine is the ratio of the mechanical advantage to the velocity ratio or the ratio of the useful work output to the total work input.
(c)(i) The effort required is approximately $118.5\ N$.
(c)(ii) In a machine, some work is always done against friction and other non - ideal factors. Since the work input is used for useful work and to overcome losses, and the useful work is always less than the work input, the efficiency of a machine is always less than $100\%$.