FrictionHard
Question
In the given figure the value(s) of mass m for which the 100 kg block remains in static equilibrium is
(g = 10 m/s2)

(g = 10 m/s2)

Options
A.35 kg
B.37 kg
C.83 kg
D.85 kg
Solution

There are two possibilities
(i) 100 kg block slides down the incline 100 kg
(ii) 100 kg block slides up the incline 100 kg
case-(i)

we get, 100 g sin 37 − m × (100 g) cos 37 − mg = 0
⇒ m =
Case - (ii)

mg = 100 g sin 37 + m g cos 37 × 100
⇒ m =
To remain in equilibrium, m ∈ [36, 84] kg
therefore, m can be 37 and 83 kg.
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