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Chapter 17: Problem 3

Determine the magnitude of the moment about the \(y\)-axis of the force \(F=500\) \(\left(F_{x}=300, F_{y}=200, F_{z}=\right.\) ?) acting at \((4,-6,4)\). a) 186 c) 2580 b) 1385 d) 3185

Short Answer

Expert verified
Answer: 1000

Step by step solution

01

Find the force vector

First, we need to determine the force vector in Cartesian coordinates \((F_x, F_y, F_z)\). The x and y components are given, so we need to find the z component. We use the magnitude of the force and its components provided: \(F = \sqrt{F_x^2 + F_y^2 + F_z^2}\) Plug in the given values of F, \(F_x\), and \(F_y\): \(500 = \sqrt{(300)^2 + (200)^2 + F_z^2}\) Solving for \(F_z\), we get: \(F_z \approx 100\) So, the force vector is \((300, 200, 100)\).
02

Find the position vector

The position vector is determined from the point where the force is applied. It is given by the coordinates: \(\vec{r} = (4, -6, 4)\)
03

Find the moment vector

Now, we will find the moment vector by taking the cross product of the position vector and the force vector: \(\vec{M} = \vec{r} \times \vec{F}\) \(\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -6 & 4 \\ 300 & 200 & 100 \end{vmatrix}\) Calculating the cross product, we get: \(\vec{M} = (400, 1000, -2200)\)
04

Find the magnitude of the moment about the y-axis

The magnitude of the moment about the \(y\)-axis is the y-component of the moment vector: \(M_y = 1000\) Thus, the magnitude of the moment about the y-axis is 1000, which is not in the given options. The answer seems to be wrong, double-check the given options and the problem statement.

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