Respuesta :
Answer:
a = g (m₂ -m₁ sin θ) / (m₁ + m₂)
Explanation:
To solve this we must locate a reference system in the inclined plane, in this case the axis eg. it is parallel to the plane and the axis and is perpendicular to the plane, we take the direction to the right as positive.
Let's write Newton's second law for block 1 on the plane
X axis
T - Wₓ = m₁ a
Y Axis
N –[tex]W_{y}[/tex] = 0
We use trigonometry to find the components of the weight
sin θ = Wₓ / W
Wₓ = W sin θ
cos θ = W_{y} / W
W_{y}= W cos θ
We substitute
T - m₁ g sin θ = m₁ a
N = m₁ g cos θ
We write Newton's equations for block 2 that is hanging.
Note that if block 1 goes up, block 2 must go down, therefore for this block the positive direction is down.
W₂ - T = m₂ a
Let's write the system of equations
T - m₁ g sin θ = m₁ a
m₂ g - T = m₂ a
Let's add
m₂ g - m₁ g sin θ = (m₂ + m₁) a
a = g (m₂ -m₁ sin θ) / (m₁ + m₂)
Acceleration is the same for both blocks as they are connected by a rope
