two forces P and Q pass through a point A which is 4 ft to the right of and 3 ft above a moment center O. force P is 200 lb dire
1 answer:
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Answer:
a) τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Explanation:
a) The torque can be expressed as
τ = r x F
To tackle this equation, using the determinant approach is the most straightforward method
The resulting expression is
τ = i ^ (y - z
) + j ^ (z Fₓ - x
) + k ^ (x
- y Fₓ)
b) Now let's compute
τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)
τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)
τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m
c) To find angular acceleration, we use
τ = I α
α = τ / I
The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.
α = (-0.189i^ -5.6 j^ + 33.978k^) / 241
α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²
Answer:
A = 4.76 x 10⁻⁴ m²
Explanation:
Given data:
Person's weight = 625 N
Bike's weight = 98 N
Pressure per tire = 7.60 x 10⁵ Pa
Find: Contact area per tire
Total system weight = 625 + 98 = 723 N
Let F represent the force supported by each tire
2F = 723 N
Therefore, F = 361.5 N
Using the formula F = P × A
Contact area, A = 4.76 x 10⁻⁴ m²