A particle is constrained to move at a constant speed on an inclined plane (ABCD) along the curved path shown in the figure. Edges AD and BC are parallel to the y axis. The inclined plane makes an angle $\theta$ with the xy-plane. The velocity vector of the particle makes an angle $\phi$ with the dotted line which is parallel to edge AB. If the speed of the particle is $2$ m/s, $\phi = 30^\circ$, and $\theta = 40^\circ$, then the z-component of the velocity of the particle in m/s is _____________.
The particle moves at a constant speed on the inclined plane. We are required to find the z-component of the velocity vector. Given:
- Speed of the particle, \(v = 2 \, \text{m/s}\)
- Angle \(\phi = 30^\circ\)
- Inclined plane angle \(\theta = 40^\circ\)
The velocity vector makes an angle \(\phi\) with the dashed line parallel to edge AB.
The z-component of the velocity vector can be derived using the inclination angle \(\theta\). The relation for the z-component of velocity \(v_z\) is:
\({v_z} = v \cdot \sin\theta \cdot \cos\phi\)
Substituting the given values:
\({v_z} = 2 \cdot \sin(40^\circ) \cdot \cos(30^\circ)\)
Calculating each term:
- \(\sin(40^\circ) \approx 0.6428\)
- \(\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660\)
Thus,
\({v_z} = 2 \cdot 0.6428 \cdot 0.8660 = 1.114\)
The calculated value is approximately: \({v_z} \approx 1.11 \, \text{m/s}\) in the negative z-direction, as inferred from the diagram.
Thus, the z-component of the velocity of the particle is -1.11 m/s.
Important Questions from Vector Algebra
Vector A̅ = ŷ.3 + ẑ.2 and B̅ = x̂.5 + ŷ.8 extend from the origin. Find A̅.B̅ Choose the correct answer.
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If non - zero a, b, c are such that a + b + c = 0, then the value of \(\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab}\) is
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Two forces F1 and F2 are used to pull a car, which met an accident. The angle between the two force is θ. Find the value of θ for the resultant force is equal to \(\sqrt{(F_1^2 + F_2^2)}\)
