Calculate each of the following: 4.1.1 The output power at the cutting tool 4.1.2 The input power from the motor

The diameter $D = 500 \text{ mm} = 0.5 \text{ m}$.
The spindle speed $N = 30 \text{ r/min} = \frac{30}{60} \text{ r/s} = 0.5 \text{ r/s}$.

The cutting depth $d = 3.5 \text{ mm}$.
The feed $f = 0.5 \text{ mm/rev}$.

The cutting pressure (specific cutting energy) $P_c = 1000 \text{ N/mm}^2$.

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

4.1.2 The input power from the motor

The drive efficiency $\eta = 85\% = 0.85$.

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

Question 4.2

b) A shaping machine has a ram of mass $200 \text{ kg}$. The ram moves in slides with a coefficient of friction of 0.07 between the ram and the slides.

4.2.1 The horizontal force required to move the ram

The mass of the ram $m = 200 \text{ kg}$. Assume acceleration due to gravity $g = 9.81 \text{ m/s}^2$.

The coefficient of friction $\mu = 0.07$.

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%

The force is increased by $30\%$, so $F_{h,new} = F_h \times (1 + 0.30)$.

The normal force $N$ remains the same ($1962 \text{ N}$).

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

Question 5.1

a) A drilling machine has a spindle speed of $240 \text{ r/min}$ and a drill of $20 \text{ mm}$ diameter. The thrust on the drill is $1200 \text{ N}$ and the cutting torque is $28 \text{ N} \cdot \text{m}$.

5.1.1 The cutting power

The spindle speed $N = 240 \text{ r/min} = \frac{240}{60} \text{ r/s} = 4 \text{ r/s}$.

The cutting torque $T_c = 28 \text{ N} \cdot \text{m}$.

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

5.1.2 The feed force if the feed is 0.20 mm/rev

The feed force is the thrust force acting on the drill. The problem states that "The thrust on the drill is $1200 \text{ N}$". The value of the feed ($0.20 \text{ mm/rev}$) is a characteristic of the operation but does not change the given thrust force.
The feed force is $\boxed{\text{1200 N}}$.

Question 5.2

b) A horizontal milling machine uses a cutter of $150 \text{ mm}$ diameter. The cutter has 8 teeth and is running at $200 \text{ r/min}$. The table feed is $250 \text{ mm/min}$.

5.2.1 The feed per tooth

The table feed $V_f = 250 \text{ mm/min}$.
The number of teeth $Z = 8$.
The running speed $N = 200 \text{ r/min}$.

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

The width of cut $w = 40 \text{ mm}$.
The depth of cut $d = 2 \text{ mm}$.
The table feed $V_f = 250 \text{ mm/min}$.

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

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

• Always wear appropriate Personal Protective Equipment (PPE), including safety glasses or a face shield, and avoid loose clothing, jewelry, or long hair that could get caught in rotating parts.
• Ensure the workpiece is securely clamped in the chuck or between centers before starting the machine to prevent it from being thrown out.
• Never leave the chuck key in the chuck after tightening, as it can become a dangerous projectile when the machine is started.
• Keep hands and fingers clear of all rotating parts and cutting tools during operation, and use a brush or hook to remove chips, never bare hands.

Question 5.4

d) A belt drive transmits $5.5 \text{ kW}$ at a belt speed of $18 \text{ m/s}$. The difference between tight-side and slack-side tensions is $1000 \text{ N}$.

There is an inconsistency in the problem statement.
The power transmitted by a belt drive is given by the formula:

Where $P$ is the power, $T_1$ is the tight-side tension, $T_2$ is the slack-side tension, and $V$ is the belt speed.

Given:
$P = 5.5 \text{ kW} = 5500 \text{ W}$
$V = 18 \text{ m/s}$
$T_1 - T_2 = 1000 \text{ N}$

If we use the given power and belt speed to calculate the tension difference:

This calculated difference ($305.56 \text{ N}$) contradicts the explicitly stated difference of $1000 \text{ N}$.
Since the problem provides contradictory information, it is impossible to uniquely determine $T_1$ and $T_2$ individually. To find $T_1$ and $T_2$, a second independent equation (e.g., the ratio of tensions $T_1/T_2$) would be required, which is not provided.

Therefore, the tight-side tension and slack-side tension cannot be calculated from the given information.