Discussion
The force (F) acting on the block, as shown in the figure is 2 N. The stiffness of the spring (k) is 100 N/m.
Assume that the block is in static equilibrium and its mass is negligible.
If the static deflection of the spring \((\Delta)\) is 0.01 m, which one of the following is the corresponding angle \((\theta)\) in radians?
Answer: D
The angle is \(\theta = \frac{\pi}{6}\).
The force F is normal to the sloping face, which makes angle \(\theta\) with the horizontal base, so F makes the same angle \(\theta\) with the vertical. Its horizontal component pushes the block towards the wall and is resisted by the spring; the vertical component is taken by the floor.
\[ \begin{aligned} F\sin\theta &= k\Delta = 100 \times 0.01 = 1\ \text{N} \\ \sin\theta &= \frac{1}{2},\ \theta = \frac{\pi}{6} \end{aligned} \]
Using the cosine instead of the sine gives \(\pi/3\), the complementary angle. \(\pi/4\) and \(\pi/5\) give horizontal components of 1.41 N and 1.18 N, which do not match the spring force of 1 N.
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