TS EAMCET · Physics · Laws of Motion
A wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force \(\mathbf{F}\) applied perpendicular to the incline. If the mass of the box is \(1 \mathrm{~kg}\), the angle of inclination is \(30^{\circ}\) and the coefficient of static friction between the box and the inclined plane is 0.2 , the minimum magnitude of \(\mathbf{F}\) is (Use \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(0 \mathrm{~N}\), as \(30^{\circ}\) is less than angle of repose
- B \(\geq 1 \mathrm{~N}\)
- C \(\geq 3.3 \mathrm{~N}\)
- D \(\geq 16.3 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(\geq 16.3 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
According to question, we can draw the following diagram. \( \begin{aligned} m g \sin \theta & =\mu(F+m g \cos \theta) \\ F & =\frac{m g \sin \theta}{\mu}-m g \cos \theta \\ & =m g\left[\frac{\sin \theta}{\mu}-\cos \theta\right] \end{aligned} \) Here,…
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