Problem 62 - Algebra and Trigonometry

Question

A force of 4 pounds acts in the direction of \(50^{\circ}\) to the horizontal. The force moves an object along a straight line from the point \((3,7)\) to the point \((8,10),\) with distance measured in feet. Find the work done by the force.

Step-by-Step Solution

Answer

The work done is given as either \(5 \cos (50^{\circ})\) foot-pounds or \(3 \sin (50^{\circ})\) foot-pounds. The significance of these terms will depend on the preference of the units used for work done in the problem, but both answers are identical and correct.

Step 1: Find the Displacement Vector

To calculate the displacement vector, subtract the final point from the initial point. This would be calculated as \((8-3, 10-7)\) and found to be \((5,3)\). This means that the object moved 5 feet in the x-direction (horizontal) and 3 feet in y-direction (vertical).

Step 2: Resolve the Force

To resolve the force into its horizontal component and vertical component, the student can use the trigonometric functions of the angle \(50^{\circ}\). The horizontal component of the force (\(F_x\)) is \[F \cos (50^{\circ}) = 4 \cos (50^{\circ})\] and the vertical component of the force (\(F_y\)) is \[F \sin (50^{\circ}) = 4 \sin (50^{\circ})\]. This separates the force into components that line up with the displacement (horizontal) and perpendicular to the displacement (vertical).

Step 3: Calculate the Work Done

Now the work done can be computed as the product of the horizontal force and the horizontal displacement, or equivalently, the vertical force and the vertical displacement. This can be calculated as \(F_x \times X\) or \(F_y \times Y\). Solving this will give \[4 \cos (50^{\circ}) \times 5 = 5\cos (50^{\circ})\] or \[4 \sin (50^{\circ}) \times 3 = 3 \sin (50^{\circ})\]