A rope of length L has its mass per unit lenght lambda varies according to function lambda (x) = e raised to x/L. the rope is pulled by constant force of 1N on a smooth horizontal surface. Find the tension in rope at x = L/2.

To determine the tension in the rope at the point \( x = \frac{L}{2} \), we first need to understand how the mass per unit length, denoted as \( \lambda(x) \), affects the distribution of tension along the rope. Given that \( \lambda(x) = e^{x/L} \), we can calculate the total mass of the rope segment from the start up to \( x \) and then use that to find the tension at the specified point.

Step 1: Calculate the Mass of the Rope Segment

The mass of a small segment of the rope, \( dm \), can be expressed as:

dm = \lambda(x) \, dx = e^{x/L} \, dx

To find the total mass of the rope from \( 0 \) to \( x \), we integrate \( dm \):

\[ m(x) = \int_0^x e^{t/L} \, dt \]

To solve this integral, we can use the substitution \( u = \frac{t}{L} \), which gives \( dt = L \, du \). The limits change accordingly from \( 0 \) to \( \frac{x}{L} \):

\[ m(x) = L \int_0^{x/L} e^u \, du = L \left[ e^u \right]_0^{x/L} = L \left( e^{x/L} - 1 \right) \]

Step 2: Find the Mass at \( x = \frac{L}{2} \)

Now, substituting \( x = \frac{L}{2} \) into the mass equation:

\[ m\left(\frac{L}{2}\right) = L \left( e^{1/2} - 1 \right) \]

Step 3: Calculate the Tension in the Rope

The tension \( T \) at a point in the rope is determined by the force needed to accelerate the mass of the rope segment to the right. Since the rope is being pulled by a constant force of \( 1 \, N \), the net force acting on the mass of the rope segment from \( 0 \) to \( \frac{L}{2} \) is given by:

\[ T = F - m\left(\frac{L}{2}\right) \cdot a \]

Here, \( F \) is the pulling force (1 N), and \( a \) is the acceleration of the rope. The acceleration can be found using Newton's second law:

\[ F = m \cdot a \implies a = \frac{F}{m} = \frac{1}{m\left(\frac{L}{2}\right)} = \frac{1}{L \left( e^{1/2} - 1 \right)} \]

Substituting \( a \) back into the tension equation:

\[ T = 1 - m\left(\frac{L}{2}\right) \cdot \frac{1}{L \left( e^{1/2} - 1 \right)} \]

Now substituting \( m\left(\frac{L}{2}\right) \) into the equation:

\[ T = 1 - L \left( e^{1/2} - 1 \right) \cdot \frac{1}{L \left( e^{1/2} - 1 \right)} = 1 - (e^{1/2} - 1) = 2 - e^{1/2} \]

Final Result

Thus, the tension in the rope at the point \( x = \frac{L}{2} \) can be expressed as:

\[ T = 2 - e^{1/2} \]

This result indicates how the varying mass distribution along the rope influences the tension at different points, particularly under a constant pulling force. The exponential function for mass per unit length leads to a non-uniform tension distribution, which is a fascinating aspect of dynamics in continuous systems.