Question

force x has a magnitude of 1210 pounds, and force y has a magnitude of 1470 pounds. they act on a single point at an angle of 60° to each other. find the magnitude of the equilibrant and the angle between the equilibrant and the 1210 - pound force.
what is the magnitude of the equilibrant?
(round to the nearest tenth.)
□ pounds
what is the angle between the equilibrant and force x?
(round to the nearest tenth.)
□°

Answer:

Part (i): Magnitude of the Equilibrant

To find the magnitude of the equilibrant, we first need to find the magnitude of the resultant of Forces \( X \) and \( Y \). The equilibrant has the same magnitude as the resultant but opposite in direction.

For two forces \( F_1 \) and \( F_2 \) acting at an angle \( \theta \) between them, the magnitude of the resultant \( R \) is given by the law of cosines:

$$ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} $$

Here, \( F_1 = 1210 \) pounds, \( F_2 = 1470 \) pounds, and \( \theta = 60^\circ \).

Step 1: Calculate \( F_1^2 \), \( F_2^2 \), and \( 2F_1F_2\cos\theta \)

• \( F_1^2 = 1210^2 = 1464100 \)
• \( F_2^2 = 1470^2 = 2160900 \)
• \( 2F_1F_2\cos\theta = 2 \times 1210 \times 1470 \times \cos(60^\circ) \)

Since \( \cos(60^\circ) = 0.5 \):

$$ 2 \times 1210 \times 1470 \times 0.5 = 1210 \times 1470 = 1778700 $$

Step 2: Sum these values and take the square root

$$ R = \sqrt{1464100 + 2160900 + 1778700} = \sqrt{5403700} $$

Calculating the square root:

$$ \sqrt{5403700} \approx 2324.6 $$

So, the magnitude of the equilibrant is also approximately \( 2324.6 \) pounds (since the equilibrant has the same magnitude as the resultant).

Part (ii): Angle between the Equilibrant and Force \( X \)

To find the angle \( \alpha \) between the equilibrant (which is opposite to the resultant) and Force \( X \), we can use the law of sines or law of cosines. Let's use the law of sines on the triangle formed by Forces \( X \), \( Y \), and the resultant \( R \).

In the triangle with sides \( F_1 = 1210 \), \( F_2 = 1470 \), and \( R \approx 2324.6 \), the angle opposite \( F_2 \) (let's call it \( \beta \)) can be found using the law of sines:

$$ \frac{\sin\beta}{F_1} = \frac{\sin\theta}{R} $$

Wait, actually, the angle between the resultant and Force \( X \) can be found using the law of cosines or the formula for the angle in the parallelogram of forces.

Alternatively, using the law of cosines for the angle \( \alpha \) between the equilibrant (opposite to \( R \)) and \( F_1 \):

The equilibrant is opposite to \( R \), so the angle between the equilibrant and \( F_1 \) is \( 180^\circ - \beta \), where \( \beta \) is the angle between \( R \) and \( F_1 \).

First, find \( \beta \) (angle between \( R \) and \( F_1 \)) using the law of sines:

$$ \frac{\sin\beta}{F_2} = \frac{\sin\theta}{R} $$

$$ \sin\beta = \frac{F_2 \sin\theta}{R} $$

$$ \sin\beta = \frac{1470 \times \sin(60^\circ)}{2324.6} $$

$$ \sin(60^\circ) \approx 0.8660 $$

$$ \sin\beta = \frac{1470 \times 0.8660}{2324.6} \approx \frac{1273.02}{2324.6} \approx 0.5476 $$

$$ \beta \approx \arcsin(0.5476) \approx 33.2^\circ $$

Now, the angle between the equilibrant (opposite to \( R \)) and \( F_1 \) is \( 180^\circ - 33.2^\circ = 146.8^\circ \)? Wait, no, that can't be right. Wait, maybe I mixed up the angle.

Wait, actually, the equilibrant is opposite to the resultant, so the angle between the equilibrant and Force \( X \) is \( 180^\circ - \) the angle between the resultant and Force \( X \).

Alternatively, let's use the law of cosines on the triangle formed by Force \( X \), the equilibrant \( E \) (magnitude \( R \)), and the angle between them \( \alpha \). The angle between Force \( X \) and Force \( Y \) is \( 60^\circ \), so the angle between Force \( X \) and the equilibrant (which is opposite to the resultant) can be found by considering the parallelogram.

Wait, maybe a better approach: The equilibrant is equal in magnitude to the resultant and opposite in direction. So, the angle between the equilibrant and Force \( X \) is \( 180^\circ - \) the angle between the resultant and Force \( X \).

To find the angle between the resultant \( R \) and Force \( X \) (let's call it \( \beta \)):

Using the law of cosines:

$$ \cos\beta = \frac{F_1^2 + R^2 - F_2^2}{2 \times F_1 \times R} $$

$$ \cos\beta = \frac{1210^2 + 2324.6^2 - 1470^2}{2 \times 1210 \times 2324.6} $$

Calculate numerator:

$$ 1210^2 = 1464100 $$

$$ 2324.6^2 \approx 5403700 $$

$$ 1470^2 = 2160900 $$

$$ \text{Numerator} = 1464100 + 5403700 - 2160900 = 4706900 $$

Denominator:

$$ 2 \times 1210 \times 2324.6 \approx 2 \times 1210 \times 2324.6 \approx 5606772 $$

$$ \cos\beta = \frac{4706900}{5606772} \approx 0.8395 $$

$$ \beta \approx \arccos(0.8395) \approx 32.9^\circ $$

Therefore, the angle between the equilibrant (opposite to \( R \)) and Force \( X \) is \( 180^\circ - 32.9^\circ = 147.1^\circ \)? Wait, that doesn't seem right. Wait, maybe I made a mistake in the direction.

Wait, actually, the equilibrant is a force that balances the resultant, so it is equal and opposite to the resultant. So, the angle between the equilibrant and Force \( X \) is \( 180^\circ - \) the angle between the resultant and Force \( X \). But let's check with the law of sines again.

Wait, maybe the angle between the equilibrant and Force \( X \) is \( 180^\circ - 60^\circ - \beta \)? No, that might not be correct.

Wait, let's start over. The two forces \( X \) and \( Y \) act at \( 60^\circ \) to each other. The resultant \( R \) is the diagonal of the parallelogram formed by \( X \) and \( Y \). The equilibrant \( E \) is equal in magnitude to \( R \) and opposite in direction. So, the angle between \( E \) and \( X \) is \( 180^\circ - \) the angle between \( R \) and \( X \).

To find the angle between \( R \) and \( X \), we can use the formula:

$$ \tan\beta = \frac{F_2 \sin\theta}{F_1 + F_2 \cos\theta} $$

Here, \( F_1 = 1210 \), \( F_2 = 1470 \), \( \theta = 60^\circ \)

$$ \tan\beta = \frac{1470 \times \sin(60^\circ)}{1210 + 1470 \times \cos(60^\circ)} $$

$$ \sin(60^\circ) \approx 0.8660, \cos(60^\circ) = 0.5 $$

$$ \tan\beta = \frac{1470 \times 0.8660}{1210 + 1470 \times 0.5} = \frac{1273.02}{1210 + 735} = \frac{1273.02}{1945} \approx 0.6545 $$

$$ \beta \approx \arctan(0.6545) \approx 33.2^\circ $$

So, the angle between \( R \) and \( X \) is approximately \( 33.2^\circ \). Therefore, the angle between \( E \) (opposite to \( R \)) and \( X \) is \( 180^\circ - 33.2^\circ = 146.8^\circ \). But this seems large. Wait, maybe the angle between the equilibrant and Force \( X \) is \( 180^\circ - 60^\circ - 33.2^\circ = 86.8^\circ \)? No, that doesn't make sense.

Wait, maybe I confused the angle. Let's use the law of cosines for the angle between \( E \) and \( X \). The triangle formed by \( X \), \( E \), and the equilibrant? No, the equilibrant is opposite to \( R \), so the angle between \( E \) and \( X \) is the angle between \( -R \) and \( X \). So, if \( R \) makes an angle \( \beta \) with \( X \), then \( -R \) makes an angle \( 180^\circ - \beta \) with \( X \).

But let's calculate \( \beta \) again using the law of cosines:

$$ \cos\beta = \frac{F_1^2 + R^2 - F_2^2}{2 \times F_1 \times R} $$

We have \( F_1 = 1210 \), \( F_2 = 1470 \), \( R \approx 2324.6 \)

$$ \cos\beta = \frac{1210^2 + 2324.6^2 - 1470^2}{2 \times 1210 \times 2324.6} $$

$$ 1210^2 = 1464100 $$

$$ 2324.6^2 \approx 5403700 $$

$$ 1470^2 = 2160900 $$

$$ \text{Numerator} = 1464100 + 5403700 - 2160900 = 4706900 $$

$$ \text{Denominator} = 2 \times 1210 \times 2324.6 \approx 5606772 $$

$$ \cos\beta = \frac{4706900}{5606772} \approx 0.8395 $$

$$ \beta \approx \arccos(0.8395) \approx 32.9^\circ $$

So, the angle between \( R \) and \( X \) is \( 32.9^\circ \), so the angle between \( E \) (opposite to \( R \)) and \( X \) is \( 180^\circ - 32.9^\circ = 147.1^\circ \). But this seems too large. Wait, maybe the angle between the equilibrant and Force \( X \) is \( 180^\circ - 60^\circ - 32.9^\circ = 87.1^\circ \)? No, that doesn't fit.

Wait, perhaps the correct approach is to use the law of cosines for the angle between the equilibrant and Force \( X \). The equilibrant \( E \) has magnitude \( R \approx 2324.6 \), Force \( X \) has magnitude \( 1210 \), and the angle between them can be found using the law of cosines in the triangle formed by \( X \), \( E \), and the resultant? No, that's not the right triangle.

Wait, actually, the equilibrant is a force that, when added to the resultant, gives zero. So, \( E + R = 0 \), so \( E = -R \). Therefore, the angle between \( E \) and \( X \) is the angle between \( -R \) and \( X \), which is \( 180^\circ - \) the angle between \( R \) and \( X \).

To find the angle between \( R \) and \( X \), we can use the formula for the angle in the parallelogram:

$$ \tan\beta = \frac{F_2 \sin\theta}{F_1 + F_2 \cos\theta} $$

We have:

$$ \tan\beta = \frac{1470 \times \sin(60^\circ)}{1210 + 1470 \times \cos(60^\circ)} \approx \frac{1470 \times 0.8660}{1210 + 1470 \times 0.5} \approx \frac{1273.02}{1210 + 735} \approx \frac{1273.02}{1945} \approx 0.6545 $$

$$ \beta \approx \arctan(0.6545) \approx 33.2^\circ $$

So, the angle between \( R \) and \( X \) is \( 33.2^\circ \), so the angle between \( E \) (opposite to \( R \)) and \( X \) is \( 180^\circ - 33.2^\circ = 146.8^\circ \), which rounds to \( 146.8^\circ \) or \( 146.9^\circ \) when rounded to the nearest tenth.

Final Answers

(i) Magnitude of the Equilibrant:

\( \boxed{2324.6} \) pounds (rounded to the nearest tenth)

(ii) Angle between the Equilibrant and Force \( X \):

\( \boxed{146.9} \) degrees (rounded to the nearest tenth) (Note: The exact value may vary slightly due to rounding during calculations, but this is the approximate result.)