A CNC vertical milling is used for cutting a straight line slot in the x-y plane. The cutter is located at point P. The slope \( \frac{dy}{dx} \) of the straight line created by the cutter is 1.25. The feed rate of x-axis is 120 mm/min. The new position of the cutter after 20 seconds is:
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When solving problems with motion along an axis and a given slope, use the relationship \( \Delta y = {slope} \times \Delta x \) to find the change in the y-coordinate.
We are given the following information:
- Initial position: \( P(10, 5) \)
- Slope of the line: \( \frac{dy}{dx} = 1.25 \)
- Feed rate of x-axis: 120 mm/min
- Time: 20 seconds
First, calculate the distance the cutter travels along the x-axis in 20 seconds:
\[
{Distance along x-axis} = {Feed rate} \times {Time} = 120 \, {mm/min} \times \frac{20}{60} \, {min} = 40 \, {mm}.
\]
So, the cutter moves 40 mm along the x-axis. The new x-coordinate of the cutter is:
\[
x_{{new}} = 10 + 40 = 50 \, {mm}.
\]
Next, use the slope \( \frac{dy}{dx} = 1.25 \) to calculate the change in the y-coordinate. The slope tells us that for every 1 mm movement along the x-axis, the y-coordinate changes by 1.25 mm. Since the cutter moves 40 mm along the x-axis, the change in y is:
\[
\Delta y = 1.25 \times 40 = 50 \, {mm}.
\]
Thus, the new y-coordinate of the cutter is:
\[
y_{{new}} = 5 + 50 = 55 \, {mm}.
\]
So, the new position of the cutter after 20 seconds is \( (50, 55) \).
