Question

Two four-wheelers are trying to drag a third four-wheeler out of the mud. The first four-wheeler

Answer 1

The magnitude of the resultant force on the third four-wheeler is 385.75 N in the direction of N 74.2⁰ E.

The given parameters:

• Applied force on the first wheel, F₁ = 300 lb at N 30 E
• Applied force on the second wheel, F₂ = 270 lb at N 55 W

The X-component and Y-component of the applied forces is calculated as follows;

$F_x = F \times cos(\theta)\\\\F_y = F \times sin(\theta)$

$X-component : F_x = \ \ 300 \times cos(30)\ -\ 270 \times cos(55) = 104.94 \ N \\\\Y - component: F_y = 300 \times sin(30) \ + \ 270 \times sin(55) = 371.2 \ N$

The magnitude of the resultant force on the third four-wheeler is calculated as follows;

$F = \sqrt{F_x^2 + F_y^2} \\\\F = \sqrt{104.94^2 \ + \ 371.2^2} \\\\F = 385.75 \ N$

The direction of the resultant force is calculated as follows;

$\theta = tan^{1} (\frac{F_y}{F_x} )\\\\\theta = tan^{-1} (\frac{371.2}{104.94} )\\\\\theta = 74.2 \ ^0\\\\\theta = N \ 74.2^0 \ E$

The complete question is below:

Two four-wheelers are trying to drag a third four-wheeler out of the mud. The first four-wheeler is traveling in the direction of N 30 E and the second four-wheeler is traveling in the direction of N 55 W If the four-wheelers are pulling with forces of 300 pounds and 270 pounds. respectively; find the magnitude and directions (as a quadrant bearing) of the resultant force on the third four-wheeler.

Learn more about resultant forces here: https://brainly.com/question/25239010