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Through a point not on a line, one and only one liParallel lines are cut by a transversal such that the alternate interior angle
1 answer:
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Answer:
a. θ = 30°
b. μ = √3 / 15 ≈ 0.115
Step-by-step explanation:
Draw a free body diagram for each scenario (see attached figure). The body has four forces acting on it:
- Weight pulling down
- Normal force perpendicular to the incline
- Applied force parallel up the incline
- Friction force parallel to the incline
Remember that friction opposes the direction of motion. So when the body is sliding up, friction points down the incline. And when the body is sliding down, friction points up the incline.
Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.
For sliding up, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Sum of the forces parallel to the incline:
∑F = ma
P₁ − f − mg sin θ = 0
P₁ − Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₁ − mgμ cos θ − mg sin θ = 0
Now for sliding down, sum of the forces normal to the incline:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
And sum of the forces parallel to the incline:
∑F = ma
P₂ + f − mg sin θ = 0
P₂ + Nμ − mg sin θ = 0
Substituting the expression for normal force:
P₂ + mgμ cos θ − mg sin θ = 0
We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.
So we have two equations and two unknowns (μ and θ):
P₁ − mgμ cos θ − mg sin θ = 0
P₂ + mgμ cos θ − mg sin θ = 0
Let's start by adding the equations together:
P₁ + P₂ − 2 mg sin θ = 0
P₁ + P₂ = 2 mg sin θ
sin θ = (P₁ + P₂) / (2 mg)
Plugging in the values:
sin θ = (6 + 4) / (2 × 10)
sin θ = 1/2
θ = 30°
Now we can plug this into either equation and find μ.
P₁ − mgμ cos θ − mg sin θ = 0
6 − (10 cos 30°) μ − 10 sin 30° = 0
6 − 5√3 μ − 5 = 0
1 = 5√3 μ
μ = √3 / 15
μ ≈ 0.115
