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Answer :
The tension in the rope is approximately 74.21 N when the person is one-third up the ladder.
To find the tension in the rope when the person is one-third of the way up the ladder, we can break down the forces acting on the ladder at that point.
Let's denote:
- [tex]\( T \)[/tex] = tension in the rope
- [tex]\( F_{\text{person}} \)[/tex] = force exerted by the person on the ladder
- [tex]\( F_{\text{gravity}} \)[/tex] = force due to gravity acting on the person
- [tex]\( F_{\text{ladder}} \)[/tex] = force exerted by the ladder on the person
The ladder is in equilibrium, so the net force and torque acting on it must be zero.
First, let's find the force exerted by the person on the ladder, [tex]\( F_{\text{person}} \)[/tex]. This force is vertical and directed upwards and can be calculated using Newton's second law:
[tex]\[ F_{\text{person}} = m \cdot g = (18.6 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 182.28 \, \text{N} \][/tex]
Next, let's find the force due to gravity acting on the person, [tex]\( F_{\text{gravity}} \)[/tex]. This force is vertical and directed downwards:
[tex]\[ F_{\text{gravity}} = m \cdot g = (18.6 \, \text{kg}) \cdot (9.8 \, \text{m/s}^2) = 182.28 \, \text{N} \][/tex]
Now, let's resolve these forces into their components. Since the ladder is inclined at an angle of [tex]\( 66^\circ \)[/tex] with the horizontal, the force exerted by the person on the ladder and the force due to gravity can be resolved into horizontal and vertical components.
[tex]\[ F_{\text{person\_vertical}} = F_{\text{person}} \cdot \sin(\theta) \][/tex]
[tex]\[ F_{\text{person\_horizontal}} = F_{\text{person}} \cdot \cos(\theta) \][/tex]
[tex]\[ F_{\text{gravity\_vertical}} = F_{\text{gravity}} \cdot \cos(\theta) \][/tex]
[tex]\[ F_{\text{gravity\_horizontal}} = F_{\text{gravity}} \cdot \sin(\theta) \][/tex]
Where [tex]\( \theta = 66^\circ \)[/tex].
Now, the ladder is in equilibrium, so the sum of forces in both the horizontal and vertical directions must be zero:
[tex]\[ F_{\text{person\_horizontal}} + F_{\text{gravity\_horizontal}} + T = 0 \][/tex]
[tex]\[ F_{\text{person\_vertical}} - F_{\text{gravity\_vertical}} = 0 \][/tex]
We can use these equations to solve for [tex]\( T \)[/tex]:
[tex]\[ F_{\text{person\_horizontal}} + F_{\text{gravity\_horizontal}} + T = 0 \][/tex]
[tex]\[ F_{\text{person}} \cdot \cos(\theta) + F_{\text{gravity}} \cdot \sin(\theta) + T = 0 \][/tex]
[tex]\[ T = - F_{\text{person}} \cdot \cos(\theta) - F_{\text{gravity}} \cdot \sin(\theta) \][/tex]
Substitute the values:
[tex]\[ T = - (182.28 \, \text{N}) \cdot \cos(66^\circ) - (182.28 \, \text{N}) \cdot \sin(66^\circ) \][/tex]
[tex]Now, \( F_{\text{person}} = 182.28 \, \text{N} \) and \( F_{\text{gravity}} = 182.28 \, \text{N} \).[/tex]
Now, let's calculate [tex]\( T \)[/tex]:
[tex]\[ T = - (182.28 \, \text{N}) \cdot \cos(66^\circ) - (182.28 \, \text{N}) \cdot \sin(66^\circ) \][/tex]
[tex]\[ T = - (182.28 \, \text{N}) \cdot 0.4067 - (182.28 \, \text{N}) \cdot 0.9135 \][/tex]
[tex]\[ T \approx -74.21 \, \text{N} \][/tex]
The negative sign indicates that the tension in the rope acts in the opposite direction to what was assumed. We usually take the positive direction upwards, so we disregard the negative sign and consider the magnitude of the tension.
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