Block B of weight 'W' is lying on a table . The coefficient of static friction between the block and table is μ. 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
Wtanθμ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
μWtanθ
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
μW√1+tan2θ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
μWsinθ
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is DμWtanθ Let weight of A is W′. From the free body diagram For equilibrium of the system, Tcosθ=μN=μW .................... (i) Tsinθ=W′ ................... (ii) where, T= tension in the thread lying between knot and the support. On dividing Eq. (ii) by Eq. (i), we get TsinθTcosθ=W′μW ⇒tanθ=W′μW ⇒W′=μWtanθ