Question

part a express the force f₂ in cartesian vector form f₁=400 lb f₂=600 lb diagram with x, y, z axes, angles 45°, 60°, 30° options: f₂=(212i+212j−519k) lb, f₂=(155i+155j+300k) lb, f₂=(155i+155j−300k) lb, f₂=(367i+367j−300k) lb

Explanation:

Step1: Analyze the angles for \( F_2 \)

For \( F_2 = 600 \) lb, we need to find the components along \( x \), \( y \), and \( z \) axes. First, consider the angle with the \( y \)-axis (the vertical axis in the diagram). The angle between \( F_2 \) and the \( y \)-axis in the \( y - x \) plane? Wait, looking at the diagram, \( F_2 \) has angles: 45° with the \( x \)-axis (in the \( x - y \) plane? Wait, no, the dashed lines: there's a 30° angle with the \( y \)-axis? Wait, let's re-examine. The diagram shows for \( F_2 \): in the \( x - y \) plane, the angle with the \( x \)-axis is 45°? Wait, no, the angle between the dashed line (projection on \( x - y \) plane) and \( x \) is 45°, and the angle between \( F_2 \) and its projection (on \( x - y \) plane) is 30°? Wait, no, maybe the angles are: the angle with the \( y \)-axis in the \( y - z \)? No, let's use the standard method for 3D force components.

Wait, the force \( F_2 \): let's find the direction cosines. First, the angle with the \( y \)-axis? Wait, looking at the diagram, \( F_2 \) is in a plane where it makes 45° with the \( x \)-axis (in the \( x - y \) plane) and 30° with the \( y \)-axis? Wait, no, maybe the angles are: the angle between \( F_2 \) and the \( x \)-axis is 45°, the angle between its projection on \( x - y \) plane and \( y \)-axis is 30°? Wait, perhaps better to break into components:

First, find the component in the \( x - y \) plane (let's call this \( F_{xy} \)) and then the \( z \)-component? Wait, no, the diagram shows that \( F_2 \) has a 30° angle with the \( y \)-axis? Wait, maybe the angles are: the angle between \( F_2 \) and the \( y \)-axis is 60°? No, let's look at the options. The options have \( i \), \( j \), \( k \) components. Let's calculate each component:

For \( F_2 = 600 \) lb:

- The angle with the \( x \)-axis: 45° (in the \( x - y \) plane? Wait, the dashed line from \( F_2 \) to \( x \) has 45°, and the angle between \( F_2 \) and its projection (on \( x - y \) plane) is 30°? Wait, no, maybe the angle between \( F_2 \) and the \( y \)-axis is 60°? Wait, let's think of the 3D angles.

Wait, the correct approach: to find the \( i \) (x), \( j \) (y), and \( k \) (z) components.

First, the angle with the \( x \)-axis: 45°, the angle with the \( y \)-axis: let's see, the projection on \( x - y \) plane makes 30° with \( y \)? Wait, no, maybe the angle between \( F_2 \) and the \( y \)-axis is 60°? Wait, the diagram shows a 30° angle between \( F_2 \) and the dashed line (projection on \( x - y \) plane), and the dashed line makes 45° with \( x \). Wait, perhaps:

The component along \( x \): \( F_2 \sin(30°) \cos(45°) \)? No, wait, let's use the angles given. Wait, the force \( F_2 \): in the diagram, it's going into the \( x - y \) plane? No, the \( k \) component: if the force is below the \( x - y \) plane, the \( k \) component would be negative.

Wait, let's calculate each component:

- \( F_x \): \( F_2 \sin(30°) \cos(45°) \)? Wait, no, maybe the angle with the \( y \)-axis is 60°, so the angle with the \( x - y \) plane is 30° (since 90° - 60° = 30°). Wait, the standard formula for 3D force: \( F_x = F \sin(\theta) \cos(\phi) \), \( F_y = F \cos(\theta) \), \( F_z = -F \sin(\theta) \sin(\phi) \), where \( \theta \) is the angle with the \( y \)-axis, and \( \phi \) is the angle in the \( x - y \) plane with the \( x \)-axis.

Looking at the diagram, for \( F_2 \):

- The angle with the \( y \)-axis ( \( \theta \) ) is 60°? Wait, no, the diagram shows a 30° angle between \( F_2 \) and the dashed line (projection on \( x - y \) plane), so \( \theta = 30° \) (angle between force and \( x - y \) plane), and \( \phi = 45° \) (angle between projection and \( x \)-axis). Then:

\( F_x = F_2 \sin(\theta) \cos(\phi) \)

\( F_y = F_2 \cos(\theta) \)

\( F_z = -F_2 \sin(\theta) \sin(\phi) \)

Wait, let's check:

\( \theta = 30° \) (angle between \( F_2 \) and \( x - y \) plane), so the component perpendicular to \( x - y \) plane is \( F_2 \sin(\theta) \), but direction: if it's below the plane, \( z \) component is negative.

\( \phi = 45° \) (angle between projection on \( x - y \) plane and \( x \)-axis).

So:

\( F_x = 600 \sin(30°) \cos(45°) \)

\( \sin(30°) = 0.5 \), \( \cos(45°) = \frac{\sqrt{2}}{2} \approx 0.7071 \)

\( F_x = 600 * 0.5 * 0.7071 \approx 600 * 0.35355 \approx 212.13 \)? Wait, no, that's not matching. Wait, maybe the angle with the \( y \)-axis is 60°, so \( \theta = 60° \) (angle between \( F_2 \) and \( y \)-axis), then the angle with \( x - y \) plane is \( 90° - 60° = 30° \), same as before.

Wait, \( F_y = F_2 \cos(60°) = 600 * 0.5 = 300 \)? No, the options have \( j \) component as 155, 212, 367, etc. Wait, maybe I got the angles wrong.

Wait, looking at the diagram, \( F_2 \) has a 45° angle with the \( x \)-axis (in the \( x - y \) plane) and a 30° angle with the \( y \)-axis? No, the dashed line from \( F_2 \) to \( x \) is 45°, and the angle between \( F_2 \) and the dashed line (projection) is 30°, so the angle between \( F_2 \) and \( x \)-axis is 45°, and the angle between \( F_2 \) and its projection (on \( x - y \) plane) is 30°, so the angle with \( z \)-axis? No, 3D force components:

Let’s define the angles:

- The angle between \( F_2 \) and the \( x \)-axis: \( \alpha = 45° \)

- The angle between \( F_2 \) and the \( y \)-axis: \( \beta = 60° \) (since 90° - 30° = 60°? Wait, the diagram shows 30° between \( F_2 \) and the dashed line (projection on \( x - y \) plane), so the angle with \( y \)-axis is 60°?

Wait, the direction cosines: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \), where \( \gamma \) is the angle with \( z \)-axis.

But maybe the force \( F_2 \) is in a plane where it makes 45° with \( x \), 30° with \( y \), and the \( z \) component is negative.

Wait, let's calculate each component:

\( F_x = F_2 \cos(45°) \sin(30°) \)

\( F_y = F_2 \cos(30°) \)

\( F_z = -F_2 \cos(45°) \cos(30°) \)? No, this is getting confusing. Let's look at the options. The options for \( F_2 \) are:

1. \( (212i + 212j - 519k) \) lb

2. \( (155i + 155j + 300k) \) lb

3. \( (155i + 155j - 300k) \) lb

4. \( (367i + 367j - 300k) \) lb

Wait, let's compute \( F_x \), \( F_y \), \( F_z \) correctly.

Looking at the diagram, \( F_2 \) is in a direction where:

- The angle between \( F_2 \) and the \( x \)-axis (in the \( x - y \) plane) is 45°, so the projection on \( x - y \) plane is \( F_2 \cos(30°) \) (since the angle between \( F_2 \) and its projection is 30°), then:

Projection on \( x - y \) plane: \( F_{xy} = F_2 \cos(30°) \)

Then, \( F_x = F_{xy} \cos(45°) \)

\( F_y = F_{xy} \sin(45°) \)

\( F_z = -F_2 \sin(30°) \) (negative because it's below the \( x - y \) plane)

Let's compute:

\( F_2 = 600 \) lb

\( \cos(30°) = \frac{\sqrt{3}}{2} \approx 0.8660 \)

\( F_{xy} = 600 * 0.8660 \approx 519.6 \) lb

\( \cos(45°) = \sin(45°) = \frac{\sqrt{2}}{2} \approx 0.7071 \)

\( F_x = 519.6 * 0.7071 \approx 367 \) lb

\( F_y = 519.6 * 0.7071 \approx 367 \) lb

\( \sin(30°) = 0.5 \)

\( F_z = -600 * 0.5 = -300 \) lb

So \( F_2 = 367i + 367j - 300k \) lb, which is the fourth option.

Step2: Verify the calculations

- \( F_{xy} = 600 \cos(30°) \approx 600 * 0.8660 = 519.6 \)
- \( F_x = 519.6 \cos(45°) \approx 519.6 * 0.7071 \approx 367 \)
- \( F_y = 519.6 \sin(45°) \approx 519.6 * 0.7071 \approx 367 \)
- \( F_z = -600 \sin(30°) = -600 * 0.5 = -300 \)

This matches the fourth option: \( F_2 = (367i + 367j - 300k) \) lb.

Answer:

\( \boldsymbol{F_2 = (367i + 367j - 300k) \text{ lb}} \) (corresponding to the option \( \boldsymbol{F_2 = (367i + 367j - 300k) \text{ lb}} \))