Problem 509 | Friction

Neglecting friction
 

 

$T_A = 3000 \sin \theta = 3000(\frac{3}{5})$

$T_A = 1800 \, \text{ N}$
 

$T_B = 2000 \sin \alpha = 2000(\frac{4}{5})$

$T_B = 1600 \, \text{ N}$
 

2T_A is greater than T_B, thus, the system will move to the left if contact surfaces are frictionless.
 

Considering friction
The angle of static friction at A, ɸ_A = arctan 0.30 = 16.70°, is not enough to hold the block from sliding the incline of angle θ = arctan (3/4) = 36.87° from horizontal. If T_B is insufficient to hold 2T_A statically the system will move to the left, otherwise, the system is stationary.
 

 

Assume the blocks are stationary (use μ_s)
$N_A = 3000 \cos \theta = 3000(\frac{4}{5}) = 2400 \, \text{ N}$

$f_A = 0.30N_A = 0.30(2400) = 720 \, \text{ N}$

$T_A = 3000 \sin \theta - f_A = 3000(\frac{3}{5}) - 720 = 1080 \, \text{ N}$

$2T_A = 2160 \, \text{ N}$
 

$N_B = 2000 \cos \alpha = 2000(\frac{3}{5}) = 1200 \, \text{ N}$

$f_B = 0.40N_B = 0.40(1200) = 480 \, \text{ N}$

$T_B = 2000 \sin \alpha + f_B = 2000(\frac{4}{5}) + 480 = 2080 \, \text{ N}$
 

T_B < 2T_A. T_B is insufficient to hold the system in static equilibrium, thus, the blocks are moving to the left.
 

Blocks are moving to the left (Use μ_k)
$f_A = 0.20N_A = 0.20(2400) = 480 \, \text{ N}$           answer

$f_B = 0.30N_B = 0.30(1200) = 360 \, \text{ N}$           answer