Block of mass M attached to rope of mass m

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SUMMARY

The discussion centers on the tension in a rope of mass m attached to a block of mass M, pulled by an external force F. The tension at the end of the rope, T(L), equals the applied force F, as it is responsible for accelerating both the block and the rope. The tension at the start of the rope, T(0), is derived from the equation T(0) = F(m(x/L) + M)/(m + M), confirming that T(0) is equal to Ma. The participants clarify the relationship between tension and mass distribution along the rope, emphasizing that T(0) and T(L) cannot be equal due to the rope's mass.

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AspiringPhysicist12
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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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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##?
 
AspiringPhysicist12 said:
I believe we have m(L/L)a = ma = F - T(L)
Why?
 
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.
 
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.
 
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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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.
 
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.
 
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.
 

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