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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.

Short Answer

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 \(\gamma = \cos^{-1} \left(\frac{3}{\sqrt{74}}\right)\).
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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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