Based on the calculation of the resultant of vector forces:
- the resultant force due to the quadriceps is 1795 N
- the resultant force due to the quadriceps is 1975 N. Training and strengthening the vastus medialis results in a greater force of muscle contraction.
<h3>What is the resultant force due to the quadriceps?</h3>
The resultant of more than two vector forces is given by:
where:
- Fₓ is the sum of the horizontal components of the forces
- Fₙ is the sum of the vertical components of the forces
- Fx = F₁cosθ + F₂cosθ + F₃cosθ + F₄cosθ
- Fₙ = F₁sinθ + F₂sinθ + F₃sinθ + F₄sinθ
- F₁ = 680N, θ = 90 = 30 = 120°
- F₂ = 220 N, θ = 90 + 16 = 106°
- F₃ = 600 N, θ = 90 + 15 = 105°
- F₄ = 480 N, θ = 90 - 35 = 55°
then:
Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 480 * cos 55
Fx = -280.6 N
Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 480 * sin 55
Fₙ = 1773.1 N
then:
F = √(-280.6)² + ( 1773.1)²
F = 1795.16 N
F ≈ 1795 N
Therefore, the resultant force due to the quadriceps is 1795 N
<h3>What would happen if the vastus medialis was trained and strengthened to contract with 720N of force?</h3>
From the new information provided:
- F₁ = 680N, θ = 90 = 30 = 120°
- F₂ = 220 N, θ = 90 + 16 = 106°
- F₃ = 600 N, θ = 90 + 15 = 105°
- F₄ = 720 N, θ = 90 - 35 = 55°
then:
Fx = 680 * cos 120 + 220 * cos 106 + 600 * cos 105 + 720 * cos 55
Fx = -142.95 N
Fₙ = 680 * sin 120 + 220 * sin 106 + 600 * sin 105 + 720 * sin 55
Fₙ = 1969.72 N
then:
F = √(-142.95)² + ( 1969.72)²
F = 1974.9 N
F ≈ 1975 N
Therefore, the resultant force due to the quadriceps is 1975 N.
Training and strengthening the vastus medialis results in a greater force of muscle contraction.
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The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
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According to the net force, the acceleration of the book is 16.47 m/s².
We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as
∑F = m . a
where F is force, m is mass and a is acceleration
From the question above, we know that
m = 3 kg
g = 9.8 m/s²
F1 = 20 N
Find the net force
∑F = F1 + W
∑F = 20 + m . g
∑F = 20 + 3 . 9.8
∑F = 20 + 29.4
∑F = 49.4 N
Find the acceleration
∑F = m . a
49.4 = 3 . a
a = 16.47 m/s²
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