Question

two forces act at a point in the plane. the angle between the two forces is given. find the magnitude of the resultant force.
forces of 135 and 150 lb, forming an angle of 112°24
the magnitude of the resultant force is □ lb.
(round to the nearest pound as needed.)

Explanation:

Step1: Convert angle to decimal degrees

First, convert \( 112^\circ 24' \) to decimal degrees. Since \( 1^\circ = 60' \), then \( 24'=\frac{24}{60}=0.4^\circ \). So the angle \( \theta = 112.4^\circ \).

Step2: Apply the law of cosines

The formula for the magnitude of the resultant force \( R \) when two forces \( F_1 \) and \( F_2 \) act at an angle \( \theta \) between them is \( R=\sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \)? Wait, no, actually, the correct formula from the law of cosines for the resultant of two vectors (forces) is \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos(180^\circ - \theta)} \)? Wait, no, let's recall: when two forces \( F_1 \) and \( F_2 \) are at an angle \( \alpha \) between them, the magnitude of the resultant \( R \) is given by \( R=\sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\alpha} \)? Wait, no, the law of cosines for the triangle formed by the two forces and the resultant: if the angle between \( F_1 \) and \( F_2 \) is \( \alpha \), then the angle opposite the resultant in the parallelogram (or triangle) is \( 180^\circ - \alpha \)? Wait, no, let's think again. The correct formula is \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\alpha} \) when \( \alpha \) is the angle between the two forces? Wait, no, actually, the law of cosines for the resultant: if you have two vectors \( \vec{F_1} \) and \( \vec{F_2} \) with an angle \( \theta \) between them, then the magnitude of the resultant \( \vec{R}=\vec{F_1}+\vec{F_2} \) is \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \)? Wait, no, that's when the angle between them is \( \theta \), but actually, the correct formula is \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \) where \( \theta \) is the angle between the two vectors. Wait, let's check with an example: if \( \theta = 0^\circ \), then \( R = F_1 + F_2 \), which matches \( \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos0^\circ}=\sqrt{(F_1 + F_2)^2}=F_1 + F_2 \). If \( \theta = 180^\circ \), then \( R = |F_1 - F_2| \), which matches \( \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos180^\circ}=\sqrt{(F_1 - F_2)^2}=|F_1 - F_2| \). So the formula is correct: \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \), where \( \theta \) is the angle between the two forces.

Given \( F_1 = 135 \) lb, \( F_2 = 150 \) lb, and \( \theta = 112.4^\circ \).

First, calculate \( \cos(112.4^\circ) \). Let's compute that: \( \cos(112.4^\circ) \approx \cos(112^\circ 24') \approx -0.3827 \) (using calculator: \( 112.4^\circ \), cosine of that is approximately \( \cos(112.4) \approx -0.382683 \)).

Now, compute \( F_1^2 = 135^2 = 18225 \)

\( F_2^2 = 150^2 = 22500 \)

\( 2F_1F_2 = 2 \times 135 \times 150 = 40500 \)

Now, plug into the formula:

\( R = \sqrt{18225 + 22500 + 40500 \times (-0.382683)} \)

First, compute the term inside the square root:

\( 18225 + 22500 = 40725 \)

\( 40500 \times (-0.382683) \approx -40500 \times 0.382683 \approx -15500.66 \)

Now, add to 40725: \( 40725 - 15500.66 \approx 25224.34 \)

Then, take the square root: \( R \approx \sqrt{25224.34} \approx 158.82 \approx 159 \) (rounded to the nearest pound)

Wait, wait, maybe I made a mistake in the formula. Wait, actually, the correct formula for the resultant of two forces using the law of cosines is \( R = \sqrt{F_1^2 + F_2^2 - 2F_1F_2\cos(180^\circ - \theta)} \)? No, no, let's recall the parallelogram law: the resultant of two vectors is the diagonal of the parallelogram, so the triangle formed has sides \( F_1 \), \( F_2 \), and \( R \), with the angle between \( F_1 \) and \( F_2 \) being \( \theta \), so the law of cosines is \( R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta \) when \( \theta \) is the angle between the two vectors (because in the parallelogram, the angle between the vectors is \( \theta \), and the triangle has angle \( 180^\circ - \theta \)? Wait, no, I think I confused the angle. Let's draw the parallelogram: two vectors \( \vec{F_1} \) and \( \vec{F_2} \) with angle \( \theta \) between them. The resultant \( \vec{R} = \vec{F_1} + \vec{F_2} \) is the diagonal. So the triangle formed by \( \vec{F_1} \), \( \vec{F_2} \), and \( \vec{R} \) has sides \( F_1 \), \( F_2 \), and \( R \), with the angle between \( F_1 \) and \( F_2 \) being \( \theta \). Wait, no, actually, in the triangle, the angle between \( F_1 \) and \( F_2 \) is \( 180^\circ - \theta \)? No, that's not right. Let's use the formula correctly. Let's check with \( \theta = 90^\circ \): then \( R = \sqrt{F_1^2 + F_2^2} \), which matches \( \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos90^\circ} = \sqrt{F_1^2 + F_2^2} \), since \( \cos90^\circ = 0 \). So that works. So when \( \theta = 90^\circ \), the formula gives the correct result. So my initial formula was correct.

Wait, but when I calculated \( \cos(112.4^\circ) \), that's correct. Let's recalculate:

\( F_1 = 135 \), \( F_2 = 150 \), \( \theta = 112.4^\circ \)

\( R^2 = 135^2 + 150^2 + 2 \times 135 \times 150 \times \cos(112.4^\circ) \)

Calculate each term:

\( 135^2 = 18225 \)

\( 150^2 = 22500 \)

\( 2 \times 135 \times 150 = 40500 \)

\( \cos(112.4^\circ) \approx \cos(112 + 24/60) = \cos(112.4) \approx -0.382683 \)

So \( 40500 \times (-0.382683) \approx -40500 \times 0.382683 \approx -15500.66 \)

Then, \( R^2 = 18225 + 22500 - 15500.66 = 40725 - 15500.66 = 25224.34 \)

Then, \( R = \sqrt{25224.34} \approx 158.82 \approx 159 \)

Wait, but let's check with another approach. Let's use the formula for the resultant of two forces: \( R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} \), where \( \theta \) is the angle between them. Wait, but when \( \theta \) is the angle between the forces, the formula is correct. Let's verify with \( F_1 = 135 \), \( F_2 = 150 \), \( \theta = 112.4^\circ \).

Alternatively, maybe I made a mistake in the sign. Wait, \( \cos(112.4^\circ) \) is negative, so \( 2F_1F_2\cos\theta \) is negative, so \( R^2 = F_1^2 + F_2^2 + 2F_1F_2\cos\theta = 135^2 + 150^2 + 2 \times 135 \times 150 \times \cos(112.4^\circ) \). Since \( \cos(112.4^\circ) \) is negative, this is \( 18225 + 22500 - 40500 \times 0.382683 \), which is what I did. So that's correct.

Calculating \( \sqrt{25224.34} \approx 158.82 \), which rounds to 159 lb.

Wait, but let's use a calculator for more precision. Let's compute \( \cos(112^\circ 24') \). First, 24' is 0.4 degrees, so 112.4 degrees. \( \cos(112.4) \approx \cos(180 - 67.6) = -\cos(67.6^\circ) \). \( \cos(67.6^\circ) \approx 0.3827 \), so \( \cos(112.4^\circ) \approx -0.3827 \). Then,

\( R^2 = 135^2 + 150^2 + 2 \times 135 \times 150 \times (-0.3827) \)

\( = 18225 + 22500 - 40500 \times 0.3827 \)

\( = 40725 - 15500.35 \)

\( = 25224.65 \)

\( R = \sqrt{25224.65} \approx 158.82 \), which is approximately 159 lb when rounded to the nearest pound.

Answer:

\( \boxed{159} \)