Question

A uniform rope of mass (m ) and length (L) placed on frictionless horizontal ground is being pulled by two forces FA and FB at its ends as shown in the figure. As a result, the rope accelerates toward the right.

Answer 1

The acceleration of the rope is:

$a = \frac{F_A - F_B}{m}$

Answer 2

The problem describes a uniform rope of mass m and length L on a frictionless horizontal ground, being pulled by two forces FA and FB at its ends, resulting in an acceleration to the right. The figure is missing, which is crucial for determining the exact directions of FA and FB.

Assumption: Based on typical physics problems and the statement "accelerates toward the right", we assume that FA is the force pulling the rope to the right (e.g., applied at the right end) and FB is the force pulling the rope to the left (e.g., applied at the left end). For the rope to accelerate to the right, FA must be greater than FB.

1. Identify the system: The system under consideration is the entire uniform rope of mass m.

2. Identify the external forces in the direction of motion:

• FA: Force applied to the right.
• FB: Force applied to the left.
• Since the ground is frictionless, there are no frictional forces opposing the motion.
• Vertical forces (gravity and normal force) are balanced and do not affect horizontal motion.

3. Apply Newton's Second Law: Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration (F_net = ma). Given that the rope accelerates to the right, the net force acting on the rope must be in the rightward direction. Therefore, the net horizontal force (F_net) is:

$F_{net} = F_A - F_B$

Substituting this into Newton's Second Law:

$F_A - F_B = ma$

4. Solve for acceleration (a): Rearranging the equation to solve for a:

$a = \frac{F_A - F_B}{m}$

This is the acceleration of the entire rope. Since the rope is uniform and moving as a single body, every point on the rope will have this same acceleration.

Explanation of the solution:

The net force on the rope is the vector sum of the forces applied at its ends. Assuming FA pulls right and FB pulls left, and since the rope accelerates right, FA must be greater than FB. Applying Newton's second law (F_net = ma) to the entire rope, the acceleration is the net force divided by the total mass.