Calculate the following: 4.1.1 The output power at the cutting tool 4.1.2 The input power from the motor 4.2.1 The horizontal force required to move the ram 4.2.2 The coefficient of friction if the force to move the ram is increased by 30%

$$A = \text{feed} \times \text{cutting depth}$$ $$A = 0.5 \text{ mm} \times 3.5 \text{ mm} = 1.75 \text{ mm}^2$$

$$F_c = \text{cutting pressure} \times A$$ $$F_c = 1000 \text{ N/mm}^2 \times 1.75 \text{ mm}^2 = 1750 \text{ N}$$

Convert diameter to meters and spindle speed to revolutions per second.

$$D = 500 \text{ mm} = 0.5 \text{ m}$$ $$N = 30 \text{ r/min} = \frac{30}{60} \text{ r/s} = 0.5 \text{ r/s}$$ $$V = \pi D N$$ $$V = \pi \times 0.5 \text{ m} \times 0.5 \text{ r/s} = 0.25 \pi \text{ m/s} \approx 0.7854 \text{ m/s}$$

$$P_{out} = F_c \times V$$ $$P_{out} = 1750 \text{ N} \times 0.7854 \text{ m/s} = 1374.45 \text{ W}$$

The output power at the cutting tool is $\boxed{\text{1374.45 W}}$.

4.1.2 The input power from the motor

$$\eta = \frac{P_{out}}{P_{in}}$$ $$P_{in} = \frac{P_{out}}{\eta}$$ $$P_{in} = \frac{1374.45 \text{ W}}{0.85} = 1617.00 \text{ W}$$

The input power from the motor is $\boxed{\text{1617.00 W}}$.

4.2.1 The horizontal force required to move the ram

Assume gravitational acceleration $g = 9.81 \text{ m/s}^2$.

$$N = m \times g$$ $$N = 200 \text{ kg} \times 9.81 \text{ m/s}^2 = 1962 \text{ N}$$

$$F_f = \mu \times N$$ $$F_f = 0.07 \times 1962 \text{ N} = 137.34 \text{ N}$$

The horizontal force required to move the ram is $\boxed{\text{137.34 N}}$.

4.2.2 The coefficient of friction if the force to move the ram is increased by 30%

$$F_{new} = F_f \times (1 + 0.30)$$ $$F_{new} = 137.34 \text{ N} \times 1.30 = 178.542 \text{ N}$$

The normal force remains the same.

$$\mu_{new} = \frac{F_{new}}{N}$$ $$\mu_{new} = \frac{178.542 \text{ N}}{1962 \text{ N}} = 0.091$$

The new coefficient of friction is $\boxed{\text{0.091}}$.

QUESTION 5

5.1.1 The cutting power

$$N = 240 \text{ r/min} = \frac{240}{60} \text{ r/s} = 4 \text{ r/s}$$ $$\omega = 2 \pi N$$ $$\omega = 2 \pi \times 4 \text{ rad/s} = 8 \pi \text{ rad/s} \approx 25.13 \text{ rad/s}$$

$$P_c = T \times \omega$$ $$P_c = 28 \text{ N·m} \times 8 \pi \text{ rad/s} = 224 \pi \text{ W} \approx 703.72 \text{ W}$$

The cutting power is $\boxed{\text{703.72 W}}$.

5.1.2 The feed force if the feed is 0,20 mm/rev
The thrust on the drill is given as 1200 N. This thrust force is the feed force. The feed rate of 0.20 mm/rev is the specific condition under which this force is applied.
The feed force is $\boxed{\text{1200 N}}$.

5.2.1 The feed per tooth

$$f_z = \frac{\text{Table feed}}{N \times Z}$$ $$f_z = \frac{250 \text{ mm/min}}{200 \text{ r/min} \times 8 \text{ teeth}} = \frac{250}{1600} \text{ mm/tooth}$$ $$f_z = 0.15625 \text{ mm/tooth}$$

The feed per tooth is $\boxed{\text{0.15625 mm/tooth}}$.

5.2.2 The material removal rate if the width of cut is 40 mm and the depth of cut is 2 mm

$$\text{MRR} = \text{width of cut} \times \text{depth of cut} \times \text{table feed}$$ $$\text{MRR} = 40 \text{ mm} \times 2 \text{ mm} \times 250 \text{ mm/min}$$ $$\text{MRR} = 20000 \text{ mm}^3/\text{min}$$

The material removal rate is $\boxed{\text{20000 mm}^3/\text{min}}$.

5.3 State FOUR safety precautions to be observed when operating a lathe.

• Always wear appropriate Personal Protective Equipment (PPE), including safety glasses.
• Ensure the workpiece is securely clamped in the chuck or between centers before starting the machine.
• Keep hands and loose clothing clear of all rotating parts and cutting tools.
• Never leave the lathe running unattended.

5.4
There is an inconsistency in the problem statement for Question 5.4.
Given:
Power transmitted ($P$) = 5.5 kW = 5500 W
Belt speed ($V$) = 18 m/s
Difference between tight-side and slack-side tensions ($T_1 - T_2$) = 1000 N

The formula for power transmitted by a belt is $P = (T_1 - T_2) \times V$.
If we use the given values: $5500 \text{ W} = (1000 \text{ N}) \times (18 \text{ m/s}) = 18000 \text{ W}$.
This shows that $5500 \text{ W} \neq 18000 \text{ W}$, indicating a contradiction in the given information.

To make the problem solvable for individual tensions, we will assume that the "1 000 N" refers to the sum of the tight-side and slack-side tensions ($T_1 + T_2 = 1000 \text{ N}$), as this is a common way for such problems to be structured when there's an error in phrasing.

$$T_1 - T_2 = \frac{P}{V}$$ $$T_1 - T_2 = \frac{5500 \text{ W}}{18 \text{ m/s}} = 305.56 \text{ N}$$

1. $T_1 - T_2 = 305.56 \text{ N}$
2. $T_1 + T_2 = 1000 \text{ N}$ (Assumption: "difference" was a typo for "sum")

5.4.1 The tight-side tension

$$(T_1 - T_2) + (T_1 + T_2) = 305.56 \text{ N} + 1000 \text{ N}$$ $$2T_1 = 1305.56 \text{ N}$$ $$T_1 = \frac{1305.56 \text{ N}}{2} = 652.78 \text{ N}$$

The tight-side tension is $\boxed{\text{652.78 N}}$.

5.4.2 The slack-side tension

$$652.78 \text{ N} + T_2 = 1000 \text{ N}$$ $$T_2 = 1000 \text{ N} - 652.78 \text{ N} = 347.22 \text{ N}$$

The slack-side tension is $\boxed{\text{347.22 N}}$.