Question
A 2200-kg
airplane pulls two gliders, the first of mass 350 kg
and the second of mass 250 kg
, down the runway with an acceleration of 1.3 m/s2
(Figure 1). Neglect the mass of the two ropes and any frictional forces. Determine the magnitude of the tension force in the second rope.
airplane pulls two gliders, the first of mass 350 kg
and the second of mass 250 kg
, down the runway with an acceleration of 1.3 m/s2
(Figure 1). Neglect the mass of the two ropes and any frictional forces. Determine the magnitude of the tension force in the second rope.
Answers
GPT-4o mini
To determine the tension force in the second rope, we can use Newton's second law of motion, \(F = ma\).
### Step 1: Calculate the total mass being pulled.
The total mass consists of the airplane and both gliders:
\[
m_{\text{total}} = m_{\text{airplane}} + m_{\text{glider 1}} + m_{\text{glider 2}} = 2200\, \text{kg} + 350\, \text{kg} + 250\, \text{kg} = 2800\, \text{kg}
\]
### Step 2: Calculate the total force acting on the system.
Using Newton's second law, we can find the total force required to accelerate this mass:
\[
F_{\text{total}} = m_{\text{total}} \cdot a = 2800\, \text{kg} \cdot 1.3\, \text{m/s}^2 = 3640\, \text{N}
\]
### Step 3: Consider the forces on the second glider.
Now, we examine the forces acting on the second glider (the glider with mass \(250\, \text{kg}\)). The total force acting on it can be expressed as:
\[
F_{\text{tension}} - F_{\text{glider}} = m_{\text{glider}} \cdot a
\]
Where:
- \(F_{\text{tension}}\) is the tension in the second rope (which we need to find),
- \(F_{\text{glider}} = m_{\text{glider}} \cdot g\) only applies if we are considering weight in a vertical direction, and since we neglect friction and are in horizontal linear motion, we skip that consideration here.
### Step 4: Apply Newton's second law to the second glider.
Substituting our values for the second glider \(F_{\text{tension}} - 0 = m_{\text{glider}} \cdot a\):
\[
F_{\text{tension}} = m_{\text{glider}} \cdot a
\]
Where:
- \( m_{\text{glider}} = 250\, \text{kg} \)
- \( a = 1.3\, \text{m/s}^2 \)
### Step 5: Calculate the tension force.
\[
F_{\text{tension}} = 250\, \text{kg} \cdot 1.3\, \text{m/s}^2 = 325\, \text{N}
\]
### Conclusion
The magnitude of the tension force in the second rope is:
\[
\boxed{325\, \text{N}}
\]
### Step 1: Calculate the total mass being pulled.
The total mass consists of the airplane and both gliders:
\[
m_{\text{total}} = m_{\text{airplane}} + m_{\text{glider 1}} + m_{\text{glider 2}} = 2200\, \text{kg} + 350\, \text{kg} + 250\, \text{kg} = 2800\, \text{kg}
\]
### Step 2: Calculate the total force acting on the system.
Using Newton's second law, we can find the total force required to accelerate this mass:
\[
F_{\text{total}} = m_{\text{total}} \cdot a = 2800\, \text{kg} \cdot 1.3\, \text{m/s}^2 = 3640\, \text{N}
\]
### Step 3: Consider the forces on the second glider.
Now, we examine the forces acting on the second glider (the glider with mass \(250\, \text{kg}\)). The total force acting on it can be expressed as:
\[
F_{\text{tension}} - F_{\text{glider}} = m_{\text{glider}} \cdot a
\]
Where:
- \(F_{\text{tension}}\) is the tension in the second rope (which we need to find),
- \(F_{\text{glider}} = m_{\text{glider}} \cdot g\) only applies if we are considering weight in a vertical direction, and since we neglect friction and are in horizontal linear motion, we skip that consideration here.
### Step 4: Apply Newton's second law to the second glider.
Substituting our values for the second glider \(F_{\text{tension}} - 0 = m_{\text{glider}} \cdot a\):
\[
F_{\text{tension}} = m_{\text{glider}} \cdot a
\]
Where:
- \( m_{\text{glider}} = 250\, \text{kg} \)
- \( a = 1.3\, \text{m/s}^2 \)
### Step 5: Calculate the tension force.
\[
F_{\text{tension}} = 250\, \text{kg} \cdot 1.3\, \text{m/s}^2 = 325\, \text{N}
\]
### Conclusion
The magnitude of the tension force in the second rope is:
\[
\boxed{325\, \text{N}}
\]