Question:
One end of a long chain is lifted vertically from flat ground to a height H with constant speed v by a force of magnitude F. Assume that the length of the chain is greater than H and that it has a uniform mass per unit length \(\rho\). The magnitude of the force F at height H is
(g is the acceleration due to gravity)
One end of a long chain is lifted vertically from flat ground to a height H with constant speed v by a force of magnitude F. Assume that the length of the chain is greater than H and that it has a uniform mass per unit length \(\rho\). The magnitude of the force F at height H is
(g is the acceleration due to gravity)
(g is the acceleration due to gravity)
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In problems involving lifting or accumulating mass at a constant velocity (variable mass systems), the total required force is always the sum of two parts: the weight of the mass already in the system (\(mg\)) and a dynamic "thrust" term (\(v \frac{dm}{dt}\)) that accounts for the momentum change of the incoming mass.
Updated On: Sep 8, 2025
- \(\rho(gH + v^2)\)
- \(\rho(gH + 2v^2)\)
- \(\rho(2gH + v^2)\)
- \(\frac{\rho}{2}(gH + v^2)\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
This is a problem involving a system with variable mass. The applied force \(F\) has to serve two purposes: first, to support the weight of the part of the chain that is already suspended in the air, and second, to continuously provide an upward impulse to the new segments of the chain being lifted off the ground, accelerating them from rest to a constant speed \(v\).
Step 2: Key Formula or Approach:
We can analyze the forces acting on the suspended length of the chain. The total force \(F\) can be written as the sum of the force required to support the weight (\(F_w\)) and the force required to change the momentum of the mass being lifted (\(F_p\)). \[ F = F_w + F_p \] The rate of change of momentum for a variable mass system is given by Newton's second law in the form \(F_{ext} = \frac{dP}{dt}\).
Step 3: Detailed Explanation:
1. Force to support the weight (\(F_w\)): At the instant when a length \(H\) of the chain is off the ground, the mass of this suspended part is \(m = \rho H\). The gravitational force (weight) on this part is \(F_w = mg = \rho g H\). 2. Force to change momentum (\(F_p\)): The chain is being lifted at a constant speed \(v\). This means that in a small time interval \(dt\), a small length \(dL = v \, dt\) of the chain is lifted off the ground. The mass of this small segment is \(dm = \rho \, dL = \rho v \, dt\). This mass \(dm\) is accelerated from a velocity of 0 to a velocity of \(v\). The change in its momentum is \(dp = (dm)v = (\rho v \, dt)v\). The force required to produce this change in momentum is the rate of change of momentum: \[ F_p = \frac{dp}{dt} = \frac{(\rho v^2 \, dt)}{dt} = \rho v^2 \] This is often called a thrust force. 3. Total Force (F): The total applied force \(F\) must be equal to the sum of the force supporting the weight and the force required to change the momentum. \[ F = F_w + F_p = \rho g H + \rho v^2 \] Factoring out \(\rho\), we get: \[ F = \rho(gH + v^2) \] Step 4: Final Answer:
The magnitude of the force F at height H is \(\rho(gH + v^2)\). This corresponds to option (A).
This is a problem involving a system with variable mass. The applied force \(F\) has to serve two purposes: first, to support the weight of the part of the chain that is already suspended in the air, and second, to continuously provide an upward impulse to the new segments of the chain being lifted off the ground, accelerating them from rest to a constant speed \(v\).
Step 2: Key Formula or Approach:
We can analyze the forces acting on the suspended length of the chain. The total force \(F\) can be written as the sum of the force required to support the weight (\(F_w\)) and the force required to change the momentum of the mass being lifted (\(F_p\)). \[ F = F_w + F_p \] The rate of change of momentum for a variable mass system is given by Newton's second law in the form \(F_{ext} = \frac{dP}{dt}\).
Step 3: Detailed Explanation:
1. Force to support the weight (\(F_w\)): At the instant when a length \(H\) of the chain is off the ground, the mass of this suspended part is \(m = \rho H\). The gravitational force (weight) on this part is \(F_w = mg = \rho g H\). 2. Force to change momentum (\(F_p\)): The chain is being lifted at a constant speed \(v\). This means that in a small time interval \(dt\), a small length \(dL = v \, dt\) of the chain is lifted off the ground. The mass of this small segment is \(dm = \rho \, dL = \rho v \, dt\). This mass \(dm\) is accelerated from a velocity of 0 to a velocity of \(v\). The change in its momentum is \(dp = (dm)v = (\rho v \, dt)v\). The force required to produce this change in momentum is the rate of change of momentum: \[ F_p = \frac{dp}{dt} = \frac{(\rho v^2 \, dt)}{dt} = \rho v^2 \] This is often called a thrust force. 3. Total Force (F): The total applied force \(F\) must be equal to the sum of the force supporting the weight and the force required to change the momentum. \[ F = F_w + F_p = \rho g H + \rho v^2 \] Factoring out \(\rho\), we get: \[ F = \rho(gH + v^2) \] Step 4: Final Answer:
The magnitude of the force F at height H is \(\rho(gH + v^2)\). This corresponds to option (A).
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