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.
Create a free account to view solution
View Solution FreeMore Friction Questions
A block A (5 kg) rests over another block B (3 kg) placed over a smooth horizontal surface. There is friction between A ...A contact force exerted by one body on horizontal surface is equal to the normal force (≠ 0) between them. It can ...(i) From the figure shown, find out acceleration of 3 kg block....A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 ...Co-efficient of friction between blocks and surface is 0.1. If at some instant the acceleration of 10 kg mass is 12 m/s2...