Answer:
Explanation:
For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.
X axis
-Aₓ - f_e +T = 0 (1)
Y axis
N₁ - W_y = 0 ( 2)
let's use trigonometry for the weight components
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
We write the diagram for the second body.
Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards
W₂ -T = 0 (3)
we add the equations is 1 and 3
- W₁ sin θ - μ N₁ + W₂ = 0
from equation 2
N₁ = W₁ cos θ
we substitute
-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0
W₂ = m₁ g (without ea - very expensive)
This is the smallest value that supports the equilibrium system
