Question

two forces of 17.3 pounds and 42.5 pounds act on a body with an angle of 21.9 degrees between them. the force of 42.5 pounds acts on the body in the horizontal direction. choose the correct approximation for the direction angle the resultant vector makes with the horizontal force. 6.3° 21.9° 15.7° 10.9°

Explanation:

Step1: Decompose the non - horizontal force

The non - horizontal force \(F = 17.3\) pounds. The horizontal component \(F_x=17.3\cos(21.9^{\circ})\) and the vertical component \(F_y = 17.3\sin(21.9^{\circ})\).
Using a calculator, \(F_x\approx17.3\times0.928 = 16.05\) and \(F_y\approx17.3\times0.373=6.45\).

Step2: Calculate the total horizontal force and use the tangent formula

The total horizontal force \(F_{totalx}=42.5 + 16.05=58.55\).
Let \(\theta\) be the angle of the resultant with the horizontal force. Then \(\tan\theta=\frac{F_y}{F_{totalx}}\).
Substitute \(F_y = 6.45\) and \(F_{totalx}=58.55\), so \(\tan\theta=\frac{6.45}{58.55}\approx0.11\).

Step3: Find the angle

\(\theta=\arctan(0.11)\approx6.3^{\circ}\) (using a calculator to find the arctangent value).

Answer:

$6.3^{\circ}$