Block of mass M attached to rope of mass m

In summary: I'm not sure what you're asking.Can you please clarify?I think at the "end" of the rope the tension must be ## F ## because it is accelerating the mass ## M ## and all of the mass of the rope ## m_r ## at ##a##. Near the block, the tension force is a reaction to accelerating the mass ## M ## at ##a##, and thus will be less than ##F##
  • #1
AspiringPhysicist12
16
6
Homework Statement
A uniform rope of mass m and length l is attached to a block of
mass M. The rope is pulled with force F. Find the tension at distance
x from the end of the rope. Neglect gravity.
Relevant Equations
T(0) = T(L) = Ma
We have the string's mass constraint m(x) = m(x/L). The block is accelerated right by force T(x=0) = Ma, where a = F / m + M.

But at the point where the force is applied (i.e. x = L), I believe we have m(L/L)a = ma = F - T(L). But this would imply T(L) = F - ma = F - m(F / m + M) = F(M / m + M) = Ma. I know it doesn't make sense to have T(0) = T(L) since the rope has mass, but what am I doing wrong here?
 

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  • #2
AspiringPhysicist12 said:
Homework Statement:: A uniform rope of mass m and length l is attached to a block of
mass M. The rope is pulled with force F. Find the tension at distance
x from the end of the rope. Neglect gravity.
Relevant Equations:: T(0) = T(L) = Ma

We have the string's mass constraint m(x) = m(x/L). The block is accelerated right by force T(x=0) = Ma, where a = F / m + M.

But at the point where the force is applied (i.e. x = L), I believe we have m(L/L)a = ma = F - T(L). But this would imply T(L) = F - ma = F - m(F / m + M) = F(M / m + M) = Ma. I know it doesn't make sense to have T(0) = T(L) since the rope has mass, but what am I doing wrong here?
I can't follow what you are doing. Why not start with the tension at ##x = 0##?
 
  • #3
AspiringPhysicist12 said:
I believe we have m(L/L)a = ma = F - T(L)
Why?
 
  • #4
Sorry let me reword it a bit. If I use the constraint T(0) = Ma, then that gives me a final answer of T(x) = F(m(x/L) + M)/(m + M). This would mean T(0) = Ma as expected, but also that T(L) = F. I don't understand why T(L) = F.
 
  • #5
AspiringPhysicist12 said:
Homework Statement:: A uniform rope of mass m and length l is attached to a block of
mass M. The rope is pulled with force F. Find the tension at distance
x from the end of the rope. Neglect gravity.
Relevant Equations:: T(0) = T(L) = Ma

We have the string's mass constraint m(x) = m(x/L). The block is accelerated right by force T(x=0) = Ma, where a = F / m + M.

But at the point where the force is applied (i.e. x = L), I believe we have m(L/L)a = ma = F - T(L). But this would imply T(L) = F - ma = F - m(F / m + M) = F(M / m + M) = Ma. I know it doesn't make sense to have T(0) = T(L) since the rope has mass, but what am I doing wrong here?
I believe you want to start by presenting FBD's of the block and an arbitrary length of rope ## x ##, and one of the remaining length of rope.
 
  • #6
AspiringPhysicist12 said:
Sorry let me reword it a bit. If I use the constraint T(0) = Ma, then that gives me a final answer of T(x) = F(m(x/L) + M)/(m + M). This would mean T(0) = Ma as expected, but also that T(L) = F. I don't understand why T(L) = F.
What do you think it should be?

Hint: consider that the end of the rope has a small handle of small mass ##m_0##. Find the tension in the rope attached to the handle. Now take ##m_0 = 0##.
 
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  • #7
Right, it makes sense in that case but that would mean the mass at the end of the rope is 0? Why do we take the rope's mass as zero there and as the entire mass m where the rope and block meet.
 
  • #8
AspiringPhysicist12 said:
Right, it makes sense in that case but that would mean the mass at the end of the rope is 0? Why do we take the rope's mass as zero there and as the entire mass m where the rope and block meet.
It was just to give you an idea on why the tension must be ##F## at the end of the rope where there is no more mass.
 
  • #9
AspiringPhysicist12 said:
Right, it makes sense in that case but that would mean the mass at the end of the rope is 0? Why do we take the rope's mass as zero there and as the entire mass m where the rope and block meet.
I think at the "end" of the rope the tension must be ## F ## because it is accelerating the mass ## M ## and all of the mass of the rope ## m_r ## at ##a##. Near the block, the tension force is a reaction to accelerating the mass ## M ## at ##a##, and thus will be less than ##F##

Does that explain it, or have I not understood it properly myself?
 
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  • #10
AspiringPhysicist12 said:
I don't understand why T(L) = F.
Umm.. because that is the external force that is being applied there, perhaps?
 
  • #11
Ok I got it now thanks.
 

1. What is the purpose of studying a block of mass M attached to a rope of mass m?

The purpose of studying this system is to understand the dynamics and behavior of objects connected by a rope or string. This system is commonly used in physics experiments and real-life scenarios, such as lifting objects with a pulley system.

2. How does the mass of the block and rope affect the system?

The mass of the block and rope affects the system by changing the overall weight and inertia of the system. This can impact the acceleration and velocity of the system, as well as the tension in the rope.

3. What is the significance of the rope's length in this system?

The length of the rope can affect the motion of the block and the tension in the rope. A longer rope may result in a slower acceleration and lower tension, while a shorter rope may result in a faster acceleration and higher tension.

4. How does the angle of the rope affect the system?

The angle of the rope can impact the direction of the tension force and the motion of the block. A steeper angle may result in a greater tension force and a more vertical motion, while a shallower angle may result in a smaller tension force and a more horizontal motion.

5. What are some real-life applications of this system?

This system can be seen in various real-life applications, such as elevators, cranes, and rock climbing. Understanding the dynamics of this system can help engineers design more efficient and safe mechanisms for lifting and moving objects.

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