Conveyor Drive Mechanical Work Solution

In automated logistics facilities, motorized conveyor systems pull cargo carts along linear tracks. Mechanical engineers calculate the mechanical work done by tension cables to size drive motors and optimize electrical energy consumption.

A motor-driven cable pulls a cargo pallet along a straight horizontal rail with displacement vector $\vec{d} = 12\hat{i}\text{ meters}$. Because of the overhead placement of the drive pulley, the cable pulls on the pallet with a constant force vector $\vec{F} = 350\hat{i} + 120\hat{j}\text{ Newtons}$.

1. Compute the mechanical work done by the cable on the pallet using the dot product.
2. Find the angle $\theta$ between the force vector and the direction of motion.
3. Determine what portion of the total force magnitude contributes directly to moving the pallet forward.

Step-by-Step Solution:

1. Compute Mechanical Work ($W$): Using the algebraic definition of the dot product:

   $$W = \vec{F} \cdot \vec{d} = (350)(12) + (120)(0) = 4200 + 0 = 4200\text{ Joules}$$

2. Calculate Angle ($\theta$): Find the magnitudes of $\vec{F}$ and $\vec{d}$:

   $$|\vec{F}| = \sqrt{350^2 + 120^2} = \sqrt{122500 + 14400} = \sqrt{136900} = 370\text{ N}$$
   $$|\vec{d}| = 12\text{ m}$$
   Apply the cosine formula:
   $$\cos\theta = \frac{\vec{F} \cdot \vec{d}}{|\vec{F}| |\vec{d}|} = \frac{4200}{(370)(12)} = \frac{4200}{4440} \approx 0.9459$$
   $$\theta = \cos^{-1}(0.9459) \approx 18.92^\circ$$

3. Forward Component of Force: The component of force parallel to the displacement is:

   $$F_{\parallel} = |\vec{F}|\cos\theta = 370(0.9459) = 350\text{ N}$$
   (Notice this matches the $x$-component $F_x = 350\text{ N}$).

Meaning of the Answer & Real-Life Application:

The total work performed on the cargo pallet over the 12-meter travel is $4200\text{ J}$ ($4.2\text{ kJ}$). The $120\text{ N}$ upward component of force ($y$-axis) acts at $90^\circ$ to the rail and therefore does exactly $0\text{ J}$ of work along the track, although it reduces normal force and friction. Mechanical engineers utilize this calculation to determine the required wattage of the conveyor motor ($P = W / \Delta t$).