Structural Safe-Load Limits for Cargo Cranes (Mechanical Engineering) Solution

Problem Overview

We analyze a motorized shipyard gantry crane hoisting a cargo container along a vertical track. The winch drum is offset horizontally by a fixed distance $x = 24.0\text{ m}$ . At a cable length $z = 30.0\text{ m}$ , the winch reels cable inward at $\frac{dz}{dt} = -2.0\text{ m/s}$ . We calculate the vertical lift velocity $\frac{dy}{dt}$ and verify adherence to dynamic hoist load safety limits ( $v_{\text{max}} \le 3.0\text{ m/s}$ ).

Step-by-Step Mathematical Solution

Step 1: Geometric Setup & Differentiation

The cable geometry forms a right triangle where:

Horizontal drum offset $x = 24.0\text{ m}$ is a static constant ( $\frac{dx}{dt} = 0$ ).
Vertical distance from drum level down to the cargo spreader is $y(t)$ (dynamic variable).
Active cable length is the hypotenuse $z(t)$ (dynamic variable).

The governing formula is the Pythagorean Theorem:

x^2 + y^2 = z^2

(This formula and its time derivative principles are defined in Related Rates.)

Substitute the static horizontal offset $x = 24.0$ before differentiating:

24.0^2 + y^2 = z^2 \implies 576.0 + y^2 = z^2

Differentiating both sides implicitly with respect to time $t$ using the Chain Rule:

0 + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \implies y \frac{dy}{dt} = z \frac{dz}{dt}

Isolating the vertical velocity $\frac{dy}{dt}$ :

\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}

Step 2: Snapshot Calculation of $y$ and $\frac{dy}{dt}$

Find the vertical position $y$ when $z = 30.0\text{ m}$ : $576.0 + y^2 = (30.0)^2 = 900.0$ $y^2 = 900.0 - 576.0 = 324.0 \implies y = \sqrt{324.0} = 18.0\text{ m}$
Compute the instantaneous vertical velocity $\frac{dy}{dt}$ : $\frac{dy}{dt} = \frac{30.0}{18.0} (-2.0) = \frac{5}{3} (-2.0) = -\frac{10}{3}\text{ m/s} \approx -3.33\text{ m/s}$ The negative sign indicates the vertical distance $y$ from the gantry drum down to the container is decreasing, meaning the container is moving upward at a speed of $3.33\text{ m/s}$ .

Step 3: Structural Safety Compliance Evaluation

Regulatory Threshold: Maximum vertical hoist velocity $v_{\text{safe}} \le 3.0\text{ m/s}$ .
Calculated Vertical Velocity: $|v_{\text{vertical}}| = 3.33\text{ m/s}$ .
Safety Verdict: NON-COMPLIANT / UNSAFE EXCURSION. The container's vertical ascent speed ( $3.33\text{ m/s}$ ) exceeds the $3.0\text{ m/s}$ structural safety limit by $11\%$ , despite the winch reel speed being only $2.0\text{ m/s}$ .

Engineering Conclusion

In mechanical hoisting and crane automation, the winch spool tachometer measures only the line reel-in rate ( $\frac{dz}{dt}$ ), not the vertical cargo velocity ( $\frac{dy}{dt}$ ).

Because $\frac{dy}{dt} = \frac{z}{y}\frac{dz}{dt}$ , the vertical velocity is amplified by the geometric factor $\frac{z}{y} = \frac{1}{\cos\phi}$ (where $\phi$ is the angle of the cable relative to the vertical track). As the container is hoisted higher, $y$ decreases while the fixed horizontal offset $x = 24\text{ m}$ remains constant. Consequently, the cable angle $\phi$ widens, and the ratio $\frac{z}{y}$ increases rapidly:

When $y = 32\text{ m}$ ( $z = 40\text{ m}$ ): $\frac{z}{y} = 1.25 \implies |v_y| = 2.5\text{ m/s}$ (safe).
When $y = 18\text{ m}$ ( $z = 30\text{ m}$ ): $\frac{z}{y} = 1.67 \implies |v_y| = 3.33\text{ m/s}$ (exceeds $3.0\text{ m/s}$ limit).
When $y = 10\text{ m}$ ( $z = 26\text{ m}$ ): $\frac{z}{y} = 2.60 \implies |v_y| = 5.20\text{ m/s}$ (extreme hazard).

If crane control microcontrollers do not execute this related rates calculation in real time, a winch operator maintaining a constant motor spool speed of $2.0\text{ m/s}$ will unknowingly subject the crane structure and cargo rigging to severe shock accelerations and dynamic pendulum swinging. Microcontrollers must automatically derate and throttle down winch motor RPM as $y$ decreases to enforce a strict cap on $\frac{dy}{dt}$ .