A car is moving at a constant speed along a straight line. Which statement is true about the forces acting on the car?
2 answers:
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(a) The equilibrant C for force of vector A and B is 3.43 N.
(b) The equilibrant C for fx of vector A and B is 2.1 N.
(c) The equilibrant C, for fy of vector A and B is 2.12 N.
<h3>What is equilibrant force?</h3>An equilibrant force is a single force that will bring other bodies into equilibrium.
<h3>From configuration 1:</h3>Vector A: mass = 0.2 kg, θ = 20⁰
Vector B: mass = 0.15 kg, θ = 80⁰
Fx = mg cosθ
Fy = mg sinθ
where;
- m is mass
- g is acceleration due to gravity
Force of A due to its weight
F(A) = mg
F(A) = 0.2 x 9.8 = 1.96 N
Fx = (0.2 x 9.8) cos(20) = 1.84 N
Fy = (0.2 x 9.8) sin(20) = 0.67 N
<h3>Resultant force</h3>R = √(0.67² + 1.84²)
R = 1.96 N
<h3>Vector B</h3>Force of B due to its weight
F(B) = mg
F(B) = 0.15 x 9.8
F(B) = 1.47 N
Fx = (0.15 x 9.8) cos(80) = 0.26 N
Fy = (0.15 x 9.8) sin(80) = 1.45 N
<h3>Resultant force </h3>R = √(0.26² + 1.45²)
R= 1.47 N
<h3>Equilibrant C of vector A and B</h3>Equilibrant force:
Force, C = 1.96 N + 1.47 N
Force, C = 3.43 N
Equilibrant FX:
Fx, C = Fx(A) + Fx(B)
Fx, C = 1.84 N + 0.26 N = 2.1 N
Equilibrant FY:
Fy, C = Fy(A) + Fy(B)
Fy, C =0.67 N + 1.45 N = 2.12 N
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Answer:
a = 1.41 m/s²
Explanation:
Given that
mass ,m= 41 kg
F₁ = 65 N , θ = 59°
F₂ = 35 N ,θ = 32°
The component of Force F₁
F₁x= F₁cos59° i
F₁x= 65 x cos59° i = 33.47 i
F₁y= - F₁ sin 59° j
F₁y= - 65 x sin 59° j = - 55.71 j
The component of Force F₂
F₂x= F₂ sin 32° i
F₂x= 35 x sin 32° i = 18.54 i
F₂y= F₂ cos 32° j
F₂y= 35 x cos 32° j = 29.68 j
The total force F
F= 33.47 i + 18.54 i - 55.71 j + 29.68 j
F= 52.01 i - 26.03 j
The magnitude of the force F
F=58.16 N
We know that
F= m a
a= Acceleration
m=mass
58.16 = 41 x a
a = 1.41 m/s²
