AP EAMCET · PHYSICS · Laws of Motion
A block \(B\), lying on a table, weights \(w\). The coefficient of static friction between the block and the table is \(\mu\). Assume that, the cord between \(B\) and the knot is horizontal. The maximum weight of the block \(A\) for which the system will be stationary is
- A \(\frac{w \tan \theta}{\mu}\)
- B \(\mu w \tan \theta\)
- C \(\mu w \sqrt{1+\tan ^2 \theta}\)
- D \(\mu w \sin \theta\)
Answer & Solution
Correct Answer
(B) \(\mu w \tan \theta\)
Step-by-step Solution
Detailed explanation
Given that, weight of block \(B=w\) Coefficient of friction between block and table \(=\mu\) Let the maximum weight of block \(A=w_A\) To keep the cord between block \(B\) and knot horizontal, block must remains in equilibrium. Now, from FBD of knot and block \(B\), we get…
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