McKenzie has enough paint to paint 108 square feet. She wants to paint her garage door, which has a height of 12 feet. The garag
2 answers:
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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
Answer:
First Case:
Second Case:
Step-by-step explanation:
We know that the first three terms of an arithmetic series are:
Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.
Therefore, we can write the second term as;
And, likewise, for the third term:
Let's solve for <em>d</em> for each of the equations.
Subtracting in the first equation yields:
And for the second equation:
To avoid fractions, let's multiply the first equation by 2. Hence:
Therefore:
Simplifying yields:
Solve for <em>p</em>. We can factor:
Factor:
Grouping:
Zero Product Property:
Then, we can use the second equation to solve for <em>d</em>. So:
Substituting:
So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.
Likewise, for the second case:
So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.
By using the values, we can determine our series.
For Case 1, we will have:
17, 15, 13.
For Case 2, we will have:
-11/2, -15/4, -2.