WBJEE · Physics · Laws of Motion
Block \(B\) lying on a table weighs 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
Let weight of \(A\) is \(W^{\prime}\). From the free body diagram. For equilibrium of the system, \[ \begin{aligned} T \cos \theta &=\mu N=\mu W \\ T \sin \theta &=W^{\prime} \end{aligned} \] where \(T=\) tension in the thread lying between knot and the support. On divindg, we…
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