To solve the problem of finding the resultant displacement both graphically and algebraically, we need to break the given displacements into their vector components. We'll handle this step by step:

Step 1: List the displacements and their directions.

1. 60 mm in the +y-direction:

   • This is a straightforward displacement: $\vec{D}_1 = (0, 60)$.

2. 30 mm in the -x-direction:

   • This is a displacement along the negative x-axis: $\vec{D}_2 = (-30, 0)$.

3. 40 mm at 150°:

   • We break this into components:
     • $x_3 = 40 \cos(150^\circ)$
     • $y_3 = 40 \sin(150^\circ)$ Using the values of cosine and sine for 150°:
     • $\cos(150^\circ) = -\frac{\sqrt{3}}{2}$
     • $\sin(150^\circ) = \frac{1}{2}$ So,
     • $x_3 = 40 \times \left(-\frac{\sqrt{3}}{2}\right) = -20\sqrt{3} \approx -34.64$
     • $y_3 = 40 \times \frac{1}{2} = 20$ Therefore, $\vec{D}_3 = (-34.64, 20)$.

4. 50 mm at 240°:

   • Again, we break this into components:
     • $x_4 = 50 \cos(240^\circ)$
     • $y_4 = 50 \sin(240^\circ)$ Using the values of cosine and sine for 240°:
     • $\cos(240^\circ) = -\frac{1}{2}$
     • $\sin(240^\circ) = -\frac{\sqrt{3}}{2}$ So,
     • $x_4 = 50 \times \left(-\frac{1}{2}\right) = -25$
     • $y_4 = 50 \times \left(-\frac{\sqrt{3}}{2}\right) = -25\sqrt{3} \approx -43.30$ Therefore, $\vec{D}_4 = (-25, -43.30)$.

Step 2: Add the components of the displacements algebraically.

Now, let's sum the x and y components separately:

• Total $x$-component: $x_{\text{total}} = 0 + (-30) + (-34.64) + (-25) = -89.64 \, \text{mm}$

• Total $y$-component: $y_{\text{total}} = 60 + 0 + 20 + (-43.30) = 36.70 \, \text{mm}$

Step 3: Find the resultant displacement magnitude and direction.

1. Magnitude: $R = \sqrt{x_{\text{total}}^2 + y_{\text{total}}^2}$ $R = \sqrt{(-89.64)^2 + (36.70)^2} \approx \sqrt{8034.58 + 1346.49} = \sqrt{9381.07} \approx 96.87 \, \text{mm}$

2. Direction (angle): $\theta = \tan^{-1} \left( \frac{y_{\text{total}}}{x_{\text{total}}} \right)$ $\theta = \tan^{-1} \left( \frac{36.70}{-89.64} \right) \approx \tan^{-1} (-0.409) = -22.3^\circ$

Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. Thus, the final angle is $180^\circ - 22.3^\circ = 157.7^\circ$.

Final Result:

• Magnitude of the resultant displacement: $96.87 \, \text{mm}$
• Direction of the resultant displacement: $157.7^\circ$

---

Graphical Solution:

To solve graphically, plot each vector according to its magnitude and direction, then measure the resultant vector's length and angle with respect to the x-axis. The algebraic solution should match the graphical result.

Would you like further details or any questions answered?

Here are five related questions:

1. How are vector components useful in solving displacement problems?
2. How can trigonometric functions help break vectors into components?
3. What is the significance of the direction angle in vector addition?
4. How would the displacement change if a third dimension (z-axis) was introduced?
5. How does the graphical method compare to the algebraic method in accuracy?

Tip: Always check the quadrant when determining the angle of a vector, especially if the x-component is negative!