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Problem 504

# A particle is at the origin of the $$\mathrm{x}, \mathrm{y}$$ and $$\mathrm{z}$$ axis in the standard coordinate system in $$\mathrm{R}^{3}$$. The particle is subject to the following forces: $$\mathrm{F}_{1}$$ : force of 5 units acting along $$0 \mathrm{x}$$ $$\mathrm{F}_{2}$$ : force of 3 units acting along $$z 0$$ $$\mathrm{F}_{3}:$$ force of 2 units acting along $$0 \mathrm{y}$$ $$\mathrm{F}_{4}$$ : force of $2 \sqrt{2}$$units acting towards 0 at an angle of$$\pi / 4$$to the$$\mathrm{x}$ and $$\mathrm{y}$$ axes in the xy plane. Find the resultant force on the particle.

Expert verified
The resultant force on the particle has a magnitude of $$\sqrt{74}$$ units and direction angles $$\alpha, \beta, \gamma$$ with respect to the x, y, and z axes respectively, where $$\alpha = \cos^{-1} \left(\frac{7}{\sqrt{74}}\right), \beta = \cos^{-1} \left(\frac{4}{\sqrt{74}}\right)$$, and $$\gamma = \cos^{-1} \left(\frac{3}{\sqrt{74}}\right)$$.
See the step by step solution

## Step 1: 1. Convert the Forces to Force Vectors

We are given 4 forces acting on the particle in different directions. Let's convert these forces into force vectors. Forces acting along the x, y, and z axes are easy to convert into vectors: - For $$\mathrm{F}_{1}$$, it acts along the x-axis (5 units): $$\vec{\mathrm{F}}_{1} = 5\hat{\mathrm{i}}$$ - For $$\mathrm{F}_{2}$$, it acts along the z-axis (3 units): $$\vec{\mathrm{F}}_{2} = 3\hat{\mathrm{k}}$$ - For $$\mathrm{F}_{3}$$, it acts along the y-axis (2 units): $$\vec{\mathrm{F}}_{3} = 2\hat{\mathrm{j}}$$ Now, for $$\mathrm{F}_{4}$$, we need to find the x and y components of the force vector since it acts at an angle of $$\frac{\pi}{4}$$ to both x and y axes: - The x-component of $$\mathrm{F}_{4}$$ is given by: $$2\sqrt{2} \cdot \cos(\frac{\pi}{4}) = 2$$ - The y-component of $$\mathrm{F}_{4}$$ is given by: $$2\sqrt{2} \cdot \sin(\frac{\pi}{4}) = 2$$ Therefore, the force vector of $$\mathrm{F}_{4}$$ is: $$\vec{\mathrm{F}}_{4} = 2\hat{\mathrm{i}} + 2\hat{\mathrm{j}}$$

## Step 2: 2. Find the Resultant Force Vector

Now, let's add all the force vectors to find the resultant force vector: $\vec{\mathrm{F}}_\mathrm{R} = \vec{\mathrm{F}}_{1} + \vec{\mathrm{F}}_{2} + \vec{\mathrm{F}}_{3} + \vec{\mathrm{F}}_{4}$ $\vec{\mathrm{F}}_\mathrm{R} = 5\hat{\mathrm{i}} + 3\hat{\mathrm{k}} + 2\hat{\mathrm{j}} + 2\hat{\mathrm{i}} + 2\hat{\mathrm{j}}$ Combining the components, we get: $\vec{\mathrm{F}}_\mathrm{ R} = (5+2)\hat{\mathrm{i}} + (2+2)\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$ $\vec{\mathrm{F}}_\mathrm{R} = 7\hat{\mathrm{i}} + 4\hat{\mathrm{j}} + 3\hat{\mathrm{k}}$

## Step 3: 3. Determine the Magnitude and Direction of the Resultant Force Vector

Now, we find the magnitude of the resultant force vector: $\Vert \vec{\mathrm{F}}_\mathrm{R} \Vert = \sqrt{(7^2)+(4^2)+(3^2)} = \sqrt{74}$ So, the magnitude of the resultant force is $$\sqrt{74}$$ units. To find the direction of the resultant force vector, we can calculate the angles with respect to the x, y, and z axes: $\alpha = \cos^{-1} \left(\frac{7}{\sqrt{74}}\right)$ $\beta = \cos^{-1} \left(\frac{4}{\sqrt{74}}\right)$ $\gamma = \cos^{-1} \left(\frac{3}{\sqrt{74}}\right)$ Hence, the resultant force on the particle has a magnitude of $$\sqrt{74}$$ units and direction angles $$\alpha, \beta, \gamma$$ with respect to the x, y, and z axes respectively.

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