A roller coaster has 4 cars, each with the same number of seats. There are 21 passengers seated in the roller coaster, but 3 sea
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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:
# m∠A = 65.3°
# m∠A = 25.8°
# m∠A = 22.7°
Step-by-step explanation:
* Lets revise the trigonometry function to solve the problem
- In any right angle triangle:
# The side opposite to the right angle is called the hypotenuse
# The other two sides are called the legs of the right angle
* If the name of the triangle is ABC, where B is the right angle
∴ The hypotenuse is AC
∴ AB and BC are the legs of the right angle
- ∠A and ∠C are two acute angles
- For angle A
# sin(A) = opposite/hypotenuse
∵ The opposite to ∠A is BC
∵ The hypotenuse is AC
∴ sin(A) = BC/AC
# cos(A) = adjacent/hypotenuse
∵ The adjacent to ∠A is AB
∵ The hypotenuse is AC
∴ cos(A) = AB/AC
# tan(A) = opposite/adjacent
∵ The opposite to ∠A is BC
∵ The adjacent to ∠A is AB
∴ tan(A) = BC/AB
* Lets solve the problems
# In Δ ABC
∵ m∠B = 90°
∵ AB = 2.3 ⇒ adjacent to angle A
∵ BC = 5 ⇒ apposite to angle A
- To find m∠A use the tangent function because we have opposite
and adjacent sides
∴ tan A = BC/AB
∴ tan A = 5/2.3 ⇒ use tan^-1 to find m∠A
∴ m∠A =
* m∠A = 65.3°
# In Δ ABD
∵ m∠B = 90°
∵ AB = 5.4 ⇒ adjacent to angle A
∵ DA = 6 ⇒ the hypotenuse
- To find m∠A use the cosine function because we have adjacent
and hypotenuse sides
∴ cos A = AB/DA
∴ cos A = 5.4/6 ⇒ use cos^-1 to find m∠A
∴ m∠A =
* m∠A = 25.8°
# In Δ ABE
∵ m∠B = 90°
∵ EB = 2.4 ⇒ opposite to angle A
∵ EA = 6.8 ⇒ the hypotenuse
- To find m∠A use the sine function because we have opposite
and hypotenuse sides
∴ sin A = EB/EA
∴ sin A = 2.4/6.8 ⇒ use sin^-1 to find m∠A
∴ m∠A =
* m∠A = 22.7°