Question

How do I find the tension? its supposed to be in the mid 40s​

Answer 1

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