Answer :
An 18.6kg person climbs up a uniform 143n ladder, the upper and lower ends of the ladder rest on frictionless surfaces, the tension in the rope when the person one-third of the way up the ladder is is approximately [tex]\( 310.65 \, \text{N} \).[/tex]
To solve this problem, we can apply Newton's second law of motion.
First, let's find the forces acting on the ladder-person system when the person is one-third of the way up the ladder.
Let:
- m be the mass of the person, ( m = 18.6 ) kg
- g be the acceleration due to gravity, g = 9.8 m/s²
- T be the tension in the rope
- N be the normal force exerted by the ground on the ladder
- [tex]\( F_{\text{gravity}} \)[/tex] be the gravitational force acting on the ladder-person system
- [tex]\( F_{\text{person}} \)[/tex] be the force exerted by the person on the ladder
- [tex]\( F_{\text{ladder}} \)[/tex] be the force exerted by the ladder on the person
For equilibrium in the vertical direction, the sum of vertical forces equals zero:
[tex]\[ N + F_{\text{person}} - F_{\text{gravity}} = 0 \][/tex]
For equilibrium in the horizontal direction, the tension in the rope balances the horizontal component of the gravitational force:
[tex]\[ T = F_{\text{gravity}} \cdot \sin(66^\circ) \][/tex]
We need to find [tex]\( F_{\text{gravity}} \)[/tex] first.
The gravitational force acting on the ladder-person system is:
[tex]\[ F_{\text{gravity}} = (m + m_{\text{ladder}}) \cdot g \][/tex]
Given that the ladder has a weight of 143 N, and since weight (W) equals mass (m) times gravitational acceleration (g), we can find the mass of the ladder as follows:
[tex]\[ m_{\text{ladder}} = \frac{143 \, \text{N}}{g} \][/tex]
Now we can find the gravitational force:
[tex]\[ F_{\text{gravity}} = (m + m_{\text{ladder}}) \cdot g \][/tex]
[tex]\[ F_{\text{gravity}} = \left(18.6 \, \text{kg} + \frac{143 \, \text{N}}{g}\right) \cdot g \][/tex]
Now we can calculate T using the horizontal equilibrium equation:
[tex]\[ T = F_{\text{gravity}} \cdot \sin(66^\circ) \][/tex]
[tex]\[ T = \left(18.6 \, \text{kg} + \frac{143 \, \text{N}}{g}\right) \cdot g \cdot \sin(66^\circ) \][/tex]
[tex]\[ T \approx \left(18.6 \, \text{kg} + \frac{143 \, \text{N}}{9.8 \, \text{m/s}^2}\right) \times 9.8 \, \text{m/s}^2 \times \sin(66^\circ) \][/tex]
[tex]\[ T \approx \left(18.6 \, \text{kg} + \frac{143 \, \text{N}}{9.8 \, \text{m/s}^2}\right) \times 9.8 \, \text{m/s}^2 \times 0.9135 \][/tex]
[tex]\[ T \approx \left(18.6 \, \text{kg} + 14.59 \, \text{kg}\right) \times 9.8 \, \text{m/s}^2 \times 0.9135 \][/tex]
[tex]\[ T \approx 33.19 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.9135 \][/tex]
[tex]\[ T \approx 310.65 \, \text{N} \][/tex]
So, the tension in the rope when the person is one-third of the way up the ladder is approximately [tex]\( 310.65 \, \text{N} \).[/tex]
