A stunt man is being pulled, at a constant velocity, along a rough road by a cable attached to a truck. The cable is parallel to
the ground. The mass of the stunt man is 109 kg, and the coefficient of kinetic friction between the road and the stunt man is 0.870. Find the tension in the cable.
2 answers:
Answer:
the tension in the cable= 929.33N
Explanation:
the tension in the cable is equivalent to the frictional force experienced between the road and the stunt man
coefficient of friction= μ=
μ=0.870
N= Normal force =m × g
m= 109kg
g= acceleration due to gravity =9.8m/s²
= μ × N
= μ × m × g
= 0.870 × 109 × 9.8
=929.33N
The tension in the cable is 929.33N
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Answer:
The horizontal component of the vector ≈ -16.06
The vertical component of the vector ≈ 19.15
Explanation:
The magnitude of the vector, = 25 units
The direction of the vector, θ = 130°
Therefore, we have;
The horizontal component of the vector, Rₓ = × cos(θ)
∴ Rₓ = 25 × cos(130°) ≈ -16.06
<em>The horizontal component of the vector, Rₓ ≈ -16.06</em>
The vertical component of the vector, R =
× sin(θ)
∴ R = 25 × sin(130°) ≈ 19.15
<em>The vertical component of the vector, R</em><em> ≈ 19.15</em>
(The vector, R = Rₓ + R
= Rₓ·i + R
·j
∴ ≈ -16.07·i + 19.15j)
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Explanation:
Detailed solution of the question is given in the attached files.
