Question

a transport plane takes off from a level landing field with two gliders in tow, one behind the other. the mass of each glider is 700 kg, and the total resistance (air drag plus friction with the runway) on each may be assumed constant and equal to 1900 n. the tension in the towrope between the transport plane and the first glider is not to exceed 12000 n. part b what is the tension in the towrope between the two gliders while they are accelerating for the takeoff? express your answer in newtons.

Explanation:

Step1: Analyze the total system (plane + two gliders)

Let the mass of each glider be \( m = 700\space kg \), so total mass of two gliders is \( 2m = 1400\space kg \). The tension between plane and first glider is \( T_1 = 12000\space N \), and resistance on each glider is \( f = 1900\space N \), so total resistance on two gliders is \( 2f = 3800\space N \). Using Newton's second law \( F_{net}=ma \), for the total system: \( T_1 - 2f=(M + 2m)a \), but wait, actually the plane's mass? Wait, no, the problem is about the two gliders? Wait, no, the first part (maybe part A) was about the tension between plane and first glider, now part B is between two gliders. Let's correct: Let's consider the second glider. Let the tension between two gliders be \( T_2 \). The mass of the second glider is \( m = 700\space kg \), resistance on it is \( f = 1900\space N \). First, find acceleration from the total system (plane, first glider, second glider). Wait, the tension between plane and first glider is \( T_1 = 12000\space N \), resistance on first glider is \( f = 1900\space N \), resistance on second glider is \( f = 1900\space N \), mass of first glider \( m_1 = 700\space kg \), mass of second glider \( m_2 = 700\space kg \). Wait, maybe the plane is pulling the first glider, which is pulling the second glider. So total force on the two gliders: \( T_1 - f - f=(m_1 + m_2)a \). So \( 12000 - 1900 - 1900=(700 + 700)a \). Calculate left side: \( 12000 - 3800 = 8200\space N \). Right side: \( 1400a \). So \( a=\frac{8200}{1400}=\frac{41}{7}\approx5.857\space m/s^2 \). Now, for the second glider: the net force is \( T_2 - f = m_2a \). So \( T_2 = f + m_2a \). Substitute \( f = 1900\space N \), \( m_2 = 700\space kg \), \( a=\frac{41}{7}\space m/s^2 \). So \( T_2 = 1900 + 700\times\frac{41}{7} \). Simplify: \( 700\times\frac{41}{7}=100\times41 = 4100 \). So \( T_2 = 1900 + 4100 = 6000\space N \). Wait, alternatively, maybe the first glider has mass \( m \), second glider mass \( m \), tension between plane and first is \( T_1 \), between first and second is \( T_2 \). For the second glider: \( T_2 - f = ma \). For the first glider: \( T_1 - T_2 - f = ma \). Add both equations: \( T_1 - 2f = 2ma \), which is the same as before. Then from the second equation: \( T_2 = f + ma \). And from \( T_1 - 2f = 2ma \), we get \( ma=\frac{T_1 - 2f}{2} \). So \( T_2 = f + \frac{T_1 - 2f}{2}=\frac{2f + T_1 - 2f}{2}=\frac{T_1}{2} \). Wait, that's a simpler way! Because \( T_1 - 2f = 2ma \), so \( ma=\frac{T_1 - 2f}{2} \), then \( T_2 = f + ma = f + \frac{T_1 - 2f}{2}=\frac{2f + T_1 - 2f}{2}=\frac{T_1}{2} \). Oh! Because the two gliders have the same mass and same resistance, so the tension between them should be half of the tension between plane and first glider? Wait, let's check with numbers. \( T_1 = 12000\space N \), so \( T_2 = 6000\space N \). Let's verify with acceleration. \( a=\frac{T_1 - 2f}{2m}=\frac{12000 - 3800}{1400}=\frac{8200}{1400}\approx5.857\space m/s^2 \). Then for second glider: \( T_2 - f = ma \), \( T_2 = 1900 + 700\times5.857\approx1900 + 4100 = 6000\space N \). Yes, that works. So the key insight is that since both gliders have the same mass and same resistance, the tension between them is half of the tension between the plane and the first glider.

Step1: Recognize symmetric system (optional)

Since both gliders have mass \( m = 700\space kg \) and resistance \( f = 1900\space N \), the tension between them (\( T_2 \)) will be half of the tension between the plane and the first glider (\( T_1 \)) because the net force on each glider segment (first glider + second glider, and second glider alone) must account for equal acceleration.

Step2: Calculate tension between gliders

Given \( T_1 = 12000\space N \), for two identical gliders with same resistance, the tension between them is \( T_2=\frac{T_1}{2} \) (derived from Newton's second law for each glider, showing that the force needed to accelerate the second glider plus its resistance is half of the total force from the plane minus total resistance, due to equal mass and resistance).
\[ T_2=\frac{12000}{2}=6000\space N \]

Answer:

\( 6000 \)