Question

The value of mass m for which the 100 kg block remains is static equilibrium is

Answer 1

you forgot to add figure, well i attached .

solution :- see figure, a block of mass 100kg is placed in rough inclined plane which is inclined at an angle of 37° with horizontal and coefficient of friction is 0.3 as shown in figure.

for static equilibrium, means, bodies do not move.

case 1 :- if we assume block of mass 100kg is moving downward then friction acts upward along plane.
for block of mass, 100kg :
100gsin37° = fr + T
but we know, friction = coefficient of friction × normal reaction
e.g., fr = 0.3 × 100gcos37°
now, 100gsin37° = 0.3 × 100gcos37° + T
100 × 10 × 3/5 = 30 × 10 × 4/5 + T
600 = 240 + T
T = 360 N .....(i)
for block of mass, m kg :
mg = T
m × 10 = 360 => m = 36 kg


case 2 :- when block of mass 100kg is moving upward then, friction acts downward along plane .

for block of mass, 100kg :
100gsin37° + fr = T
100gsin37° + u × 100gcos37° = T
100 × 10 × 3/5 + 0.3 × 100 × 10 × 4/5 = T
600 + 240 = T
T = 840 N

for block of mass , mkg :
mg = T
m × 10 = 840 => m = 84 kg


hence, mass of block either 36 kg or 84 kg

Answer 2

It's a similar question with different values.. It will be helpful