Answer:
The tension of the rope is T= 172.52 N
Explanation:
m= 21 kg
α= 21º
μ= 0.285
a= 2.09 m/s²
g= 9.81 m/s²
W= m*g
W=206.01 N
Fr= μ*W*cos(α)
Fr= 54.81 N
Fx= W * sin(α)
Fx= 73.82 N
T-Fr-Fx = m*a
T= m*a + Fr + Fx
T= 172.52 N
The magnitude of the force of motion of an object pulled by a rope up an inclined plane is the tension in the rope less the frictional forces as well as the component of the weight of the object acting along the plane
The tension in the rope is approximately 172.53 Newtons
Known:
The mass of the box, m = 21.0 kg
The angle of inclination of the ramp, θ = 21.0°
The coefficient of kinetic friction, μk= 0.285
The acceleration of the box up the ramp, a = 2.09 m/s²
The acceleration due to gravity, g = 9.81 m/s²
Required:
The tension in the rope, T
Solution;
The normal reaction of the box, N = m × g × cos(θ)
∴ N = 21.0 × 9.81 × cos(21.0°) ≈ 192.33
The normal reaction, N ≈ 192.33 N
The frictional force, [tex]F_f[/tex] = μ × N
∴ [tex]F_f[/tex] = 0.285 × 192.33 ≈ 54.81405
The frictional force, [tex]F_f[/tex] ≈ 54.81405 N
The force pulling the box, F = m×a = T - [tex]F_f[/tex] - The component of the weight acting along the plane
The component of the weight acting along the plane = m·g·sin(θ)
∴ m×a = T - [tex]F_f[/tex] - m·g·sin(θ)
T = m×a + [tex]F_f[/tex] + m·g·sin(θ)
Which gives;
T = 21.0 × 2.09 + 54.81405 + 21.0 × 9.81 × sin(21.0°) ≈ 172.53
The tension in the rope, T ≈ 172.53 N
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