Question

part a express force f as a cartesian vector, then determine its direction angles. f = 3 kn. diagram with x, y, z axes, distances (1m, 2m, 3m, 4m), points a/b, and force f. multiple - choice options: f = (-2i + j + 2k) kn, α = 131.8°, β = 70.5°, γ = 48.2°; f = (-4i + j + 4k) kn, α = 48.2°, β = 70.5°, γ = 48.2°; f = (-2i + j + 2k) kn, α = 48.2°, β = 70.5°, γ = 48.2°; f = (-4i + 2j + 4k) kn, α = 131.8°, β = 70.5°, γ = 48.2°.

Explanation:

Step1: Find the position vector of AB

First, we determine the coordinates of points A and B. From the diagram, the displacement in x: \( 1 - 2 = -1 \) m? Wait, no, let's check the distances. Wait, the horizontal (x), y, z: Let's see the vector from B to A. The x-component: from the x-direction, the difference is \( 1 - 2 = -1 \)? Wait, no, looking at the diagram, the x-distance: 2m and 1m? Wait, maybe the position vector of AB: let's see the coordinates. Let's assume B is at (0,0,0)? Wait, no, the y-direction: 2m, z-direction: 3m, and the x-direction: 2m and 1m? Wait, maybe the vector from A to B? Wait, the force is from A to B? Wait, the force F is along AB. Let's find the vector AB. Let's get the components:

From the diagram, the x-component: \( 1 - 2 = -1 \)? Wait, no, maybe the differences: x: \( 1 - 2 = -1 \)? Wait, the horizontal (x) direction: the two points have x-coordinates 2m and 1m? So \( \Delta x = 1 - 2 = -1 \) m? Wait, no, maybe the vector is from B to A? Wait, the length of AB: let's calculate the magnitude of AB. The components: \( \Delta x = -1 \) m? Wait, no, looking at the diagram, the x-distance: 2m (from the left) and 1m? Wait, maybe the x-component is \( 1 - 2 = -1 \), y-component: \( 4 - 0 = 4 \)? No, the y-direction: 2m? Wait, the diagram has y-axis, z-axis, x-axis. Let's re-examine:

The coordinates: Let's say point B is at (2, 0, 3) and point A is at (1, 4, 1)? No, maybe better to find the vector AB. The differences:

x: \( 1 - 2 = -1 \) m? Wait, the x-direction: the two parallel lines have 1m and 2m, so \( \Delta x = 1 - 2 = -1 \) m?

y: \( 4 - 0 = 4 \) m? Wait, the y-direction has a 4m length? Wait, the diagram shows a 4m segment.

z: \( 1 - 3 = -2 \) m? Wait, no, the z-direction: 3m and 1m? Wait, maybe I got the coordinates wrong. Let's look at the vertical (z) direction: 3m, and the x-direction: 2m and 1m (so \( \Delta x = 1 - 2 = -1 \)), y-direction: 2m? Wait, no, the diagram has a 4m line. Wait, maybe the vector AB has components: \( \Delta x = -1 \) m, \( \Delta y = 4 \) m, \( \Delta z = -2 \) m? No, that doesn't make sense. Wait, the correct way: the position vector from A to B (or B to A) – let's calculate the vector AB. Let's see the coordinates:

From the diagram, the x-component: 2m (right) and 1m (left), so \( \Delta x = 1 - 2 = -1 \) m? Wait, no, maybe the vector is from B to A: so A is at (1, 4, 1) and B is at (2, 0, 3)? Then \( \vec{AB} = (1 - 2, 4 - 0, 1 - 3) = (-1, 4, -2) \)? No, that magnitude would be \( \sqrt{(-1)^2 + 4^2 + (-2)^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \approx 4.58 \), but the force is 3 kN. Wait, maybe I messed up the components. Let's look again:

Wait, the diagram has:

- x-direction: 2m and 1m (so difference is \( 1 - 2 = -1 \) m? Or \( 2 - 1 = 1 \) m? Wait, the arrow is from A to B? Wait, the force F is along AB, so the vector of F should be in the direction of AB. Let's find the correct components.

Wait, the vertical (z) direction: 3m (from B up) and 1m (from A up), so \( \Delta z = 1 - 3 = -2 \) m?

y-direction: 2m (from B in y) and 4m (from A in y), so \( \Delta y = 4 - 2 = 2 \) m? Wait, no, the diagram shows a 4m line in y? Wait, maybe the vector AB has components:

x: \( 1 - 2 = -1 \) m,

y: \( 4 - 0 = 4 \) m? No, the y-axis is horizontal. Wait, maybe the correct components are:

x: \( 1 - 2 = -1 \) m,

y: \( 2 - 0 = 2 \) m? No, the diagram has a 4m segment. Wait, I think I made a mistake. Let's calculate the vector AB correctly.

Looking at the diagram:

- The x-component: the distance between the two parallel lines is 1m (from 2m to 1m), so \( \Delta x = 1 - 2 = -1 \) m?

- The y-component: the length of the slanted line is 4m? Wait, no, the horizontal (y) direction: 2m (from B) and 4m? Wait, maybe the vector AB has components:

\( \Delta x = -1 \) m,

\( \Delta y = 4 \) m,

\( \Delta z = -2 \) m? No, that can't be. Wait, let's calculate the magnitude of AB. Let's see the coordinates:

Point A: (1, 4, 1)

Point B: (2, 0, 3)

Then \( \vec{AB} = (2 - 1, 0 - 4, 3 - 1) = (1, -4, 2) \)

Magnitude of AB: \( \sqrt{1^2 + (-4)^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \approx 4.583 \) m

But the force F is 3 kN, so the unit vector in the direction of AB is \( \frac{\vec{AB}}{|\vec{AB}|} = \frac{(1, -4, 2)}{\sqrt{21}} \)

Then F vector would be \( F \times \) unit vector: \( 3 \times \frac{(1, -4, 2)}{\sqrt{21}} \approx 3 \times (0.218, -0.873, 0.436) \approx (0.654, -2.619, 1.308) \) kN. But this doesn't match the options. So I must have messed up the components.

Wait, maybe the vector is from A to B:

Point A: (2, 0, 1)

Point B: (1, 4, 3)

Then \( \vec{AB} = (1 - 2, 4 - 0, 3 - 1) = (-1, 4, 2) \)

Magnitude: \( \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21} \approx 4.583 \) m

Force F is 3 kN, so F vector: \( 3 \times \frac{(-1, 4, 2)}{\sqrt{21}} \approx 3 \times (-0.218, 0.873, 0.436) \approx (-0.654, 2.619, 1.308) \) kN. Still not matching.

Wait, the options have components like -2i, +j, +2k. Let's check the magnitude of the vector in the options. Let's take the first option: F = (-2i + j + 2k) kN. Magnitude: \( \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \) kN. Oh! That's the magnitude of F (3 kN). So the unit vector must have components (-2/3, 1/3, 2/3), because \( \sqrt{(-2/3)^2 + (1/3)^2 + (2/3)^2} = \sqrt{4/9 + 1/9 + 4/9} = \sqrt{9/9} = 1 \). So the vector AB must have components proportional to (-2, 1, 2). Let's check the direction angles.

Direction angle \( \alpha \) (with x-axis): \( \cos \alpha = \frac{F_x}{|F|} = \frac{-2}{3} \), so \( \alpha = \arccos(-2/3) \approx 131.8^\circ \)

Direction angle \( \beta \) (with y-axis): \( \cos \beta = \frac{F_y}{|F|} = \frac{1}{3} \), so \( \beta = \arccos(1/3) \approx 70.5^\circ \)

Direction angle \( \gamma \) (with z-axis): \( \cos \gamma = \frac{F_z}{|F|} = \frac{2}{3} \), so \( \gamma = \arccos(2/3) \approx 48.2^\circ \)

Which matches the first option: F = (-2i + j + 2k) kN, \( \alpha = 131.8^\circ \), \( \beta = 70.5^\circ \), \( \gamma = 48.2^\circ \)

Step2: Verify the components and direction angles

- The vector components: F_x = -2, F_y = 1, F_z = 2. Magnitude: \( \sqrt{(-2)^2 + 1^2 + 2^2} = 3 \) kN, which matches F = 3 kN.
- Direction angle \( \alpha \): \( \cos \alpha = -2/3 \), so \( \alpha = \arccos(-2/3) \approx 131.8^\circ \)
- Direction angle \( \beta \): \( \cos \beta = 1/3 \), so \( \beta = \arccos(1/3) \approx 70.5^\circ \)
- Direction angle \( \gamma \): \( \cos \gamma = 2/3 \), so \( \gamma = \arccos(2/3) \approx 48.2^\circ \)

This matches the first option.

Answer:

F = (-2i + j + 2k) kN, \( \alpha = 131.8^\circ \), \( \beta = 70.5^\circ \), \( \gamma = 48.2^\circ \) (corresponding to the first option: F = (-2i + j + 2k) kN, \( \alpha = 131.8^\circ \), \( \beta = 70.5^\circ \), \( \gamma = 48.2^\circ \))