Answer:
a = 2.3 m/s²
T = 45 N
Explanation:
Draw a free body diagram for each mass.
For the mass on the incline, there are four forces:
Weight force mg pulling down.
Normal force N perpendicular to the incline.
Friction force Nμ pushing down the incline.
Tension force T pulling up the incline.
For the hanging mass, there are two forces:
Weight force Mg pulling down.
Tension force T pulling up.
Sum of the forces on the hanging mass in the -y direction:
∑F = ma
Mg − T = Ma
T = Mg − Ma
Sum of the forces on the sliding mass in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces on the sliding mass in the parallel direction:
∑F = ma
T − mg sin θ − Nμ = ma
Substitute:
Mg − Ma − mg sin θ − mgμ cos θ = ma
Mg − mg (sin θ + μ cos θ) = ma + Ma
Mg − mg (sin θ + μ cos θ) = (m + M) a
a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)
Plug in values:
a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)
a = 2.3 m/s²
Now find tension:
T = Mg − Ma
T = 6.0×9.8 − 6.0×2.3
T = 45 N
