Question

Two blocks A and B are resting against the wall and floor as shown in the figure.Find the minimum value of P that will hold the system in equilibrium. Take μ=0.25 at the floor,μ=0.3 at the wall and μ=0.2 between the blocks.

Given : μ=0.25 at floor
             μ=0.2 between blocks
To find : Minimum value of force P

Answer 1

 

The impending motion of block A is to move down and that of block B is to move towards left
F_S1=μ1N₁=0.3N₁  
F_S2=μ₂N₂=0.2N₂
F_S3=μ₃N₃=0.25N₃ ……………..(1) 
Block A is under equilibrium
Applying conditions of equilibrium

ΣF_Y=0
-500+F_S1+F_S2sin60+N₂cos60=0 
0.3N₁+0.6732N₂=500 ………….(2)

Similarly, ΣF_X=0 

N₁+F_S2cos60-N₂sin60=0

N₁+0.2N₂ x 0.5 – N₂ x 0.866 = 0 (From 1)

N₁-0.766N₂=0 ……….(3)

Solving (2) and (3)

N₁=424.1417 N

N₂=553.71 N

Applying conditions of equilibrium on block B

ΣF_Y=0 
-1000+N3-F_S2sin60-N₂ cos60=0 
N₃-0.6732N2=1000 
N3=1372.7576 N
ΣF_X=0 
-P-F_S3-F_S2cos60+N₂sin60=0 
-0.25N₃-0.2N₂ x 0.5 +N₂ x 0.866 = P
P=80.9525 N
The minimum value of force P that will hold the system in equilibrium is 80.9525 N