Question

express each force in cartesian vector form. diagram: f₁=5 kn, f₂=2 kn, angles 60°, 45°, x, y, z axes. multiple - choice options: f₁=(2.17i + 3.75j + 4.33k) kn, f₂ = -2j kn; f₁=(4.33i + 3.75j + 2.17k) kn, f₂ = -2j kn; f₁=(2.50i + 3.54j + 2.50k) kn, f₂ = -2j kn; f₁=(4.33i + 3.54j + 4.33k) kn, f₂ = -2j kn

Explanation:

Step1: Analyze \( \mathbf{F_1} \) components

First, find the magnitude of \( \mathbf{F_1} \) in the \( xy \)-plane. The angle with the \( y \)-axis is \( 45^\circ \), so the \( y \)-component in the \( xy \)-plane: \( F_{1,xy} \cos(45^\circ) \), \( x \)-component: \( F_{1,xy} \sin(45^\circ) \), where \( F_{1,xy} = F_1 \sin(60^\circ) \) (since \( F_1 \) makes \( 60^\circ \) with the \( z \)-axis, so the angle with \( xy \)-plane is \( 30^\circ \)? Wait, no: the angle between \( F_1 \) and \( z \)-axis is \( 60^\circ \), so the projection onto \( xy \)-plane is \( F_1 \sin(60^\circ) \), and the angle between this projection and \( y \)-axis is \( 45^\circ \). So:

- \( F_{1x} = F_1 \sin(60^\circ) \sin(45^\circ) \)
- \( F_{1y} = F_1 \sin(60^\circ) \cos(45^\circ) \)
- \( F_{1z} = F_1 \cos(60^\circ) \)

Given \( F_1 = 5 \, \text{kN} \):

\( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \), \( \cos(60^\circ) = 0.5 \), \( \sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.7071 \)

Step2: Calculate \( F_{1x} \)

\( F_{1x} = 5 \times 0.8660 \times 0.7071 \approx 5 \times 0.6124 \approx 3.062 \)? Wait, no, wait: maybe the angle with \( y \)-axis is \( 45^\circ \), and the angle with \( z \)-axis is \( 60^\circ \). Wait, maybe I mixed up. Let's re-express:

The force \( F_1 \) has:

- Angle with \( z \)-axis: \( 60^\circ \), so the component along \( z \): \( F_{1z} = F_1 \cos(60^\circ) = 5 \times 0.5 = 2.5 \, \text{kN} \)
- The projection onto \( xy \)-plane: \( F_{1,xy} = F_1 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} \approx 4.330 \, \text{kN} \)
- Now, in the \( xy \)-plane, the angle with \( y \)-axis is \( 45^\circ \), so:
- \( F_{1y} = F_{1,xy} \cos(45^\circ) = 4.330 \times \frac{\sqrt{2}}{2} \approx 4.330 \times 0.7071 \approx 3.061 \)? No, wait the options have \( 3.75 \) or \( 3.54 \). Wait, maybe the angle with \( x \)-axis? Wait, the diagram: the \( y \)-axis is vertical, \( x \) and \( z \) are horizontal. Wait, maybe the angle between \( F_1 \) and \( y \)-axis is \( 45^\circ \), and the angle between its projection on \( xz \)-plane? No, the diagram shows \( 60^\circ \) with \( z \)-axis and \( 45^\circ \) with \( y \)-axis? Wait, maybe I made a mistake. Let's check the options.

Wait, the options for \( F_1 \): let's compute each component:

Wait, maybe the angle with \( y \)-axis is \( 45^\circ \), so the angle between \( F_1 \) and \( y \)-axis is \( 45^\circ \), so the \( y \)-component is \( F_1 \cos(45^\circ) \), and the projection onto \( xz \)-plane is \( F_1 \sin(45^\circ) \). Then, the angle between this projection and \( z \)-axis is \( 60^\circ \), so:

- \( F_{1z} = F_1 \sin(45^\circ) \cos(60^\circ) \)
- \( F_{1x} = F_1 \sin(45^\circ) \sin(60^\circ) \)
- \( F_{1y} = F_1 \cos(45^\circ) \)

Wait, let's try this:

\( F_1 = 5 \, \text{kN} \)

\( F_{1y} = 5 \cos(45^\circ) \approx 5 \times 0.7071 \approx 3.535 \approx 3.54 \, \text{kN} \)

\( F_{1, xz} = 5 \sin(45^\circ) \approx 3.535 \, \text{kN} \)

\( F_{1z} = 3.535 \cos(60^\circ) \approx 3.535 \times 0.5 = 1.7675 \)? No, that's not matching. Wait, the options have \( F_1 = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN} \). Let's check:

If \( F_{1z} = 2.5 \, \text{kN} \), then \( F_1 \cos(\theta_z) = 2.5 \), so \( \cos(\theta_z) = 2.5 / 5 = 0.5 \), so \( \theta_z = 60^\circ \), which matches. Then \( F_{1,xy} = 5 \sin(60^\circ) \approx 4.330 \, \text{kN} \). Now, if the angle between \( F_{1,xy} \) and \( y \)-axis is \( 45^\circ \), then:

\( F_{1y} = F_{1,xy} \cos(45^\circ) \approx 4.330 \times 0.7071 \approx 3.06 \), no. Wait, maybe the angle between \( F_{1,xy} \) and \( x \)-axis is \( 45^\circ \)? Then \( F_{1x} = F_{1,xy} \cos(45^\circ) \), \( F_{1y} = F_{1,xy} \sin(45^\circ) \). Wait, no, \( y \)-axis is vertical. Wait, maybe the diagram has \( F_1 \) in the \( yz \)-plane? No, the \( x \)-axis is also there. Wait, the correct approach:

For a force in 3D, the components are:

\( F_x = F \sin\theta \sin\phi \)

\( F_y = F \cos\theta \)

\( F_z = F \sin\theta \cos\phi \)

Where \( \theta \) is the angle with \( y \)-axis, and \( \phi \) is the angle with \( z \)-axis in the \( xz \)-plane. Wait, maybe \( \theta = 45^\circ \) (angle with \( y \)-axis), \( \phi = 60^\circ \) (angle with \( z \)-axis). Then:

\( F_{1x} = 5 \sin(45^\circ) \sin(60^\circ) \)

\( F_{1y} = 5 \cos(45^\circ) \)

\( F_{1z} = 5 \sin(45^\circ) \cos(60^\circ) \)

Wait, \( \sin(45^\circ) = \sqrt{2}/2 \approx 0.7071 \), \( \sin(60^\circ) = \sqrt{3}/2 \approx 0.8660 \), \( \cos(60^\circ) = 0.5 \)

\( F_{1x} = 5 \times 0.7071 \times 0.8660 \approx 5 \times 0.6124 \approx 3.062 \) – no. Wait, the option \( F_1 = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN} \):

Check \( F_{1z} = 2.5 \), so \( 5 \cos\theta_z = 2.5 \implies \theta_z = 60^\circ \), correct.

\( F_{1y} = 3.54 \), so \( 5 \cos\theta_y = 3.54 \implies \cos\theta_y = 3.54/5 \approx 0.708 \implies \theta_y \approx 45^\circ \), correct (since \( \cos(45^\circ) \approx 0.7071 \)).

\( F_{1x} = 2.5 \), so \( 5 \sin\theta_y \sin\theta_z = 2.5 \). Wait, \( \sin\theta_y = \sin(45^\circ) \approx 0.7071 \), \( \sin\theta_z = \sin(60^\circ) \approx 0.8660 \), so \( 5 \times 0.7071 \times 0.8660 \approx 3.06 \), no. Wait, maybe \( \theta_z = 45^\circ \) and \( \theta_y = 60^\circ \)? No, the diagram shows \( 60^\circ \) with \( z \)-axis.

Wait, maybe the force \( F_1 \) has:

- Angle with \( y \)-axis: \( 45^\circ \), so \( F_{1y} = 5 \cos(45^\circ) \approx 3.535 \approx 3.54 \)
- Angle with \( z \)-axis: \( 60^\circ \), so \( F_{1z} = 5 \cos(60^\circ) = 2.5 \)
- Then, the \( x \)-component: since in 3D, \( F_x^2 + F_y^2 + F_z^2 = F_1^2 \)

So \( F_x^2 + (3.54)^2 + (2.5)^2 = 5^2 \)

\( F_x^2 + 12.53 + 6.25 = 25 \)

\( F_x^2 = 25 - 18.78 = 6.22 \implies F_x \approx 2.49 \approx 2.50 \)

Ah! So that's the key. So:

\( F_{1x} = \sqrt{F_1^2 - F_{1y}^2 - F_{1z}^2} \)

Wait, no, because the angle with \( y \)-axis is \( 45^\circ \), so \( F_{1y} = F_1 \cos(45^\circ) \), angle with \( z \)-axis is \( 60^\circ \), so \( F_{1z} = F_1 \cos(60^\circ) \). Then \( F_{1x} \) is the remaining component:

\( F_{1x} = \sqrt{F_1^2 - F_{1y}^2 - F_{1z}^2} \)

Compute:

\( F_{1y} = 5 \cos(45^\circ) \approx 3.535 \)

\( F_{1z} = 5 \cos(60^\circ) = 2.5 \)

\( F_{1x}^2 = 5^2 - (3.535)^2 - (2.5)^2 = 25 - 12.53 - 6.25 = 6.22 \implies F_{1x} \approx 2.49 \approx 2.50 \)

So \( F_1 = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN} \)

For \( F_2 \), it's along the negative \( y \)-axis, so \( F_2 = -2\mathbf{j} \, \text{kN} \)

So the correct option is the one with \( F_1 = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN} \), \( F_2 = -2\mathbf{j} \, \text{kN} \)

Answer:

\( \mathbf{F_1} = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN} \), \( \mathbf{F_2} = -2\mathbf{j} \, \text{kN} \) (corresponding to the option \( \mathbf{F_1} = (2.50\mathbf{i} + 3.54\mathbf{j} + 2.50\mathbf{k}) \, \text{kN}, \mathbf{F_2} = -2\mathbf{j} \, \text{kN} \))