Block of mass M attached to rope of mass m

  • #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?
 

Attachments

  • 1653505282845.png
    1653505282845.png
    711 bytes · Views: 36

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,758
15,364
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##?
 
  • #4
AspiringPhysicist12
16
6
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
erobz
Gold Member
1,479
660
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,758
15,364
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##.
 
  • Like
Likes AspiringPhysicist12
  • #7
AspiringPhysicist12
16
6
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
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2022 Award
23,758
15,364
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
erobz
Gold Member
1,479
660
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?
 
  • Like
Likes AspiringPhysicist12
  • #11
AspiringPhysicist12
16
6
Ok I got it now thanks.
 

Suggested for: Block of mass M attached to rope of mass m

Replies
20
Views
606
Replies
31
Views
418
Replies
4
Views
651
Replies
4
Views
337
  • Last Post
Replies
2
Views
613
Replies
10
Views
508
Replies
5
Views
287
  • Last Post
Replies
15
Views
343
Replies
6
Views
518
Top