Problem 72 - Algebra and Trigonometry

Question

The magnitude and direction exerted by two tugboats towing a ship are 4200 pounds, \(\mathrm{N} 65^{\circ} \mathrm{E},\) and 3000 pounds, \(\mathrm{S} 58^{\circ} \mathrm{E}\) respectively. Find the magnitude, to the nearest pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.

Step-by-Step Solution

Answer

Once calculations are performed according to the steps, we get the magnitude of the resultant force to the nearest pound and its direction to the nearest tenth of a degree.

Step 1: Resolve each vector into components

The first step is to decompose the given vectors force1 and force2 into their horizontal (x) and vertical (y) components. The x-component for a vector can be calculated as magnitude * cos(direction) and the y-component as magnitude * sin(direction). Considering north and east as positive, we get for the first force \(F1_x = 4200 \cos{65}\) and \(F1_y = 4200 \sin{65}\). Similarly for the second force \(F2_x = -3000 \cos{58}\) (as the direction is south) and \(F2_y = 3000 \sin{58}\).

Step 2: Calculate the components of the resultant vector

The x and y components of the resultant force (F) can simply be obtained by adding together the corresponding components of force1 (F1) and force2 (F2). Therefore \(F_x = F1_x + F2_x\) and \(F_y = F1_y + F2_y\).

Step 3: Find the magnitude of the resultant force

The formula to calculate the magnitude of a force (or a vector in general) is \(\sqrt{F_x^2 + F_y^2}\), where \(F_x\) is the x-component and \(F_y\) is the y-component.

Step 4: Find the direction of the resultant force

To find the direction of the resultant force, the tangent function can be used. The formula being \(theta = \arctan{(F_y/F_x)}\). For this problem, must consider the quadrant in which the resulting vector is. If \(F_x\) is negative and \(F_y\) is positive, the direction is South-East. Add 180 degrees to the result obtained from the \(\arctan\) function.