This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the questions: Question 4.1 a) A lathe has a drive efficiency of 85\%. It is used to machine a workpiece of 500 mm diameter at a spindle speed of 30 r/min. The cutting depth of the tool is 3.5 mm and the feed of the tool is 0.5 mm per revolution of the workpiece. The cutting pressure on the cutting tool is 1000 N/mm^2. 4.1.1 The output power at the cutting tool Step 1: Calculate the cutting speed (V_c). The diameter D = 500 mm = 0.5 m. The spindle speed N = 30 r/min = (30)/(60) r/s = 0.5 r/s. V_c = D N = × 0.5 m × 0.5 r/s = 0.25 m/s ≈ 0.7854 m/s Step 2: Calculate the cross-sectional area of the chip (A_c). The cutting depth d = 3.5 mm. The feed f = 0.5 mm/rev. A_c = d × f = 3.5 mm × 0.5 mm = 1.75 mm^2 Step 3: Calculate the cutting force (F_c). The cutting pressure (specific cutting energy) P_c = 1000 N/mm^2. F_c = P_c × A_c = 1000 N/mm^2 × 1.75 mm^2 = 1750 N Step 4: Calculate the output power (P_out). P_out = F_c × V_c = 1750 N × 0.7854 m/s = 1374.45 W The output power at the cutting tool is 1374.45 W. 4.1.2 The input power from the motor Step 1: Use the drive efficiency () to find the input power (P_in). The drive efficiency = 85\% = 0.85. P_in = P_out = 1374.45 W0.85 = 1617.00 W The input power from the motor is 1617.00 W. Question 4.2 b) A shaping machine has a ram of mass 200 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 Step 1: Calculate the normal force (N). The mass of the ram m = 200 kg. Assume acceleration due to gravity g = 9.81 m/s^2. N = m × g = 200 kg × 9.81 m/s^2 = 1962 N Step 2: Calculate the horizontal force (F_h) required to overcome friction. The coefficient of friction = 0.07. F_h = × N = 0.07 × 1962 N = 137.34 N The horizontal force required to move the ram is 137.34 N. 4.2.2 The coefficient of friction if the force to move the ram is increased by 30% Step 1: Calculate the new horizontal force (F_h,new). The force is increased by 30\%, so F_h,new = F_h × (1 + 0.30). F_h,new = 137.34 N × 1.30 = 178.542 N Step 2: Calculate the new coefficient of friction (_new). The normal force N remains the same (1962 N). _new = F_h,newN = 178.542 N1962 N = 0.091 The new coefficient of friction is 0.091. Question 5.1 a) A drilling machine has a spindle speed of 240 r/min and a drill of 20 mm diameter. The thrust on the drill is 1200 N and the cutting torque is 28 N · m. 5.1.1 The cutting power Step 1: Convert the spindle speed to revolutions per second. The spindle speed N = 240 r/min = (240)/(60) r/s = 4 r/s. Step 2: Calculate the cutting power (P_c) using the cutting torque. The cutting torque T_c = 28 N · m. P_c = 2 N T_c = 2 × 4 r/s × 28 N · m = 224 W ≈ 703.72 W The cutting power is 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 N". The value of the feed (0.20 mm/rev) is a characteristic of the operation but does not change the given thrust force. The feed force is 1200 N. Question 5.2 b) A horizontal milling machine uses a cutter of 150 mm diameter. The cutter has 8 teeth and is running at 200 r/min. The table feed is 250 mm/min. 5.2.1 The feed per tooth Step 1: Calculate the feed per tooth (f_t). The table feed V_f = 250 mm/min. The number of teeth Z = 8. The running speed N = 200 r/min. f_t = (V_f)/(Z × N) = 250 mm/min8 teeth × 200 r/min = (250)/(1600) mm/tooth = 0.15625 mm/tooth The feed per tooth is 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 Step 1: Calculate the material removal rate (MRR). The width of cut w = 40 mm. The depth of cut d = 2 mm. The table feed V_f = 250 mm/min. MRR = w × d × V_f = 40 mm × 2 mm × 250 mm/min = 20000 mm^3/min The material removal rate is 20000 mm^3/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 kW at a belt speed of 18 m/s. The difference between tight-side and slack-side tensions is 1000 N. There is an inconsistency in the problem statement. The power transmitted by a belt drive is given by the formula: P = (T_1 - T_2) × V 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 kW = 5500 W V = 18 m/s T_1 - T_2 = 1000 N If we use the given power and belt speed to calculate the tension difference: T_1 - T_2 = (P)/(V) = 5500 W18 m/s ≈ 305.56 N This calculated difference (305.56 N) contradicts the explicitly stated difference of 1000 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. Send me the next one 📸