Question

use the parallelogram rule to find the magnitude of the resultant for the two forces shown in the figure.

the magnitude of the resultant force is \boxed{} lb.
(round to the nearest tenth as needed.)

Explanation:

Step1: Recall the Law of Cosines

The parallelogram rule for vector addition (here, forces are vectors) forms a triangle (or parallelogram) where the resultant is the diagonal. Using the Law of Cosines: \( c^{2}=a^{2}+b^{2}-2ab\cos(C) \), but wait, the angle between the two forces: the given angle is \( 60^\circ \), but in the parallelogram, the angle between the sides (forces) and the resultant - actually, when using the parallelogram, the angle between the two forces is \( 60^\circ \), so the triangle formed has sides \( a = 15 \), \( b = 4 \), and the included angle \( C = 180^\circ - 60^\circ= 120^\circ \)? Wait no, wait: the parallelogram rule: if we have two vectors \( \vec{F_1}=15\) lb at \( 60^\circ \) from the horizontal, and \( \vec{F_2}=4\) lb horizontal. Wait, actually, the angle between the two vectors is \( 60^\circ \). Wait, no, let's visualize: the 4 lb force is horizontal (along x - axis), the 15 lb force makes \( 60^\circ \) with the horizontal (the 4 lb force). So the angle between the two vectors is \( 60^\circ \). Then, the resultant \( R \) can be found using the Law of Cosines where \( R^{2}=15^{2}+4^{2}-2\times15\times4\times\cos(180^\circ - 60^\circ) \)? Wait, no, wait: the Law of Cosines formula for the magnitude of the resultant when two vectors \( \vec{A} \) and \( \vec{B} \) with angle \( \theta \) between them is \( R=\sqrt{A^{2}+B^{2}+2AB\cos(\theta)} \). Wait, yes! Because if the angle between the two vectors is \( \theta \), then the formula is \( R^{2}=A^{2}+B^{2}+2AB\cos(\theta) \). Wait, let's correct: when you have two vectors \( \vec{A} \) and \( \vec{B} \), the magnitude of their sum \( \vec{R}=\vec{A}+\vec{B} \) is given by \( |\vec{R}|=\sqrt{|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos(\theta)} \), where \( \theta \) is the angle between \( \vec{A} \) and \( \vec{B} \). In this case, \( |\vec{A}| = 15 \) lb, \( |\vec{B}| = 4 \) lb, and \( \theta = 60^\circ \) (since the 15 lb force makes \( 60^\circ \) with the 4 lb force (horizontal)).

So \( R^{2}=15^{2}+4^{2}+2\times15\times4\times\cos(60^\circ) \)

Step2: Calculate each term

First, \( 15^{2}=225 \), \( 4^{2}=16 \), \( \cos(60^\circ)=0.5 \)

Then, \( 2\times15\times4\times0.5 = 60 \)

So \( R^{2}=225 + 16+60=301 \)

Wait, no, wait: \( 225+16 = 241 \), then \( 241 + 60=301 \)? Wait, no: \( 225+16=241 \), \( 2\times15\times4=120 \), \( 120\times\cos(60^\circ)=120\times0.5 = 60 \). So \( 241+60 = 301 \). Then \( R=\sqrt{301}\approx17.3 \)? Wait, that can't be right. Wait, maybe I messed up the angle. Wait, the angle between the two forces: if the 15 lb force is at \( 60^\circ \) above the horizontal, and the 4 lb force is along the horizontal, then the angle between them is \( 60^\circ \), so the formula is \( R^{2}=A^{2}+B^{2}+2AB\cos(\theta) \), where \( \theta \) is the angle between the vectors. Wait, but let's check with another approach. Let's resolve into components.

Horizontal component of 15 lb force: \( 15\cos(60^\circ)=15\times0.5 = 7.5 \) lb

Vertical component of 15 lb force: \( 15\sin(60^\circ)=15\times\frac{\sqrt{3}}{2}\approx12.99 \) lb

Horizontal component of 4 lb force: 4 lb (since it's along x - axis)

Total horizontal component \( F_x=7.5 + 4=11.5 \) lb

Total vertical component \( F_y=12.99 \) lb

Then, magnitude of resultant \( R=\sqrt{F_x^{2}+F_y^{2}}=\sqrt{(11.5)^{2}+(12.99)^{2}} \)

Calculate \( 11.5^{2}=132.25 \), \( 12.99^{2}\approx168.74 \)

Sum: \( 132.25+168.74 = 300.99\approx301 \), so \( R=\sqrt{301}\approx17.3 \) lb. Wait, but let's check the Law of Cosines again. Wait, maybe the angle in the Law of Cosines is \( 120^\circ \). Wait, no: when you have two vectors, the angle between them is \( \theta \), and the resultant is the diagonal of the parallelogram. The triangle formed by the two vectors and the resultant has sides \( A \), \( B \), and \( R \), with the angle opposite \( R \) being \( 180^\circ-\theta \)? No, no. Wait, the parallelogram: if you have vectors \( \vec{A} \) and \( \vec{B} \), the parallelogram has sides \( \vec{A} \) and \( \vec{B} \), and the diagonal is \( \vec{A}+\vec{B} \). The triangle formed by \( \vec{A} \), \( \vec{B} \), and \( \vec{A}+\vec{B} \) has sides \( |\vec{A}| \), \( |\vec{B}| \), and \( |\vec{A}+\vec{B}| \), with the angle between \( \vec{A} \) and \( \vec{B} \) being \( \theta \). So the Law of Cosines should be \( |\vec{A}+\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}+2|\vec{A}||\vec{B}|\cos(\theta) \), because the angle between \( \vec{A} \) and \( \vec{B} \) is \( \theta \), so in the triangle, the included angle is \( \theta \), not \( 180^\circ-\theta \). Wait, maybe my initial component method was wrong. Wait, no: the horizontal component of 15 lb is \( 15\cos(60^\circ)=7.5 \), plus 4 lb is 11.5. Vertical component is \( 15\sin(60^\circ)\approx12.99 \). Then \( R=\sqrt{11.5^{2}+12.99^{2}}=\sqrt{132.25 + 168.74}=\sqrt{300.99}\approx17.3 \). Wait, but let's recalculate the Law of Cosines correctly. Let's take \( a = 15 \), \( b = 4 \), angle between them \( \theta = 60^\circ \). Then \( c^{2}=15^{2}+4^{2}+2\times15\times4\times\cos(60^\circ) \). \( 15^{2}=225 \), \( 4^{2}=16 \), \( 2\times15\times4 = 120 \), \( \cos(60^\circ)=0.5 \), so \( 120\times0.5 = 60 \). Then \( c^{2}=225 + 16+60 = 301 \), so \( c=\sqrt{301}\approx17.3 \). Yes, that's correct.

Step2: Compute the square root

Calculate \( \sqrt{301}\approx17.3 \) (rounded to the nearest tenth)

Answer:

\( 17.3 \)