Tension in a rope at the midpoint

[James2911]

Homework Statement

Calculate Tension (Tp) and (TQ) (P is the mid point)

Homework Equations

T - mg = ma

The Attempt at a Solution

For Q (massless rope)
Tq - 1.9g - 0.2g - 2.9g = 5a
Tq = 5(g+a)
Tq = 5 * 10 = 50N

Given answer = 50N

(Calculation of Tension for Q is clear to me, but I am getting 2 answers for tension of P)

Answer 1 - For P (rope with mass, hence tension can't be considered equal on all points)
Tp - .1g - 2.9g = 3a
Tp = 3(g+a) = 30N

Answer 2 - For P
Tp - 1.9g - 0.1g = 2a
Tp = 2(g+a) = 20N

The given answer is Tp = 30N
Now here's my doubt.
In answer 1 - we are calculating Tension at point P. We considered the weights 0.1kg and 2.9kg and ignored that the 1.9kg weight. But I think the weight 1.9kg would apply a force 1.9g which should be considered.(Here we come up from down and include one weight (1.9kg and .1kg)

In answer 2 - we considered the weights 1.9kg and 0.1kg but didn't include 2.9kg. So, here we ignored 2.9kg like it didn't exist. But wouldn't the 2.9g exert a force? (Here we go up from down and include 2.9kg and 0.1kg(half of mass of rope)

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Out of answer 1 and 2, I consider Answer 2 i.e., 20N to be more right than the second one because, we are asked to find Tension at the midpoint. So we should just only consider the forces that are acting on it.

Please help. I've tried to find more solutions of this question online. Tq always comes out to be 50N.
But the value of Tp seems to be either 20N or 30N on different forums. Which is the correct value?

(reiteration: Tp = 30N is the given answer)

 

[Orodruin]

What free body diagrams are you considering in the two different cases? Remember to include all forces acting on the system, this includes all tensile forces.

Edit: Until you feel more secure: Always draw free body diagrams.

 

[Chestermiller]

Your answer 1 is correct. Your force balance for answer 2 is incorrect.

 

[PeroK]

Homework Statement

Calculate Tension (Tp) and (TQ) (P is the mid point)
View attachment 238043

Homework Equations

T - mg = ma

The Attempt at a Solution

For Q (massless rope)
Tq - 1.9g - 0.2g - 2.9g = 5a
Tq = 5(g+a)
Tq = 5 * 10 = 50N

Given answer = 50N

(Calculation of Tension for Q is clear to me, but I am getting 2 answers for tension of P)

Answer 1 - For P (rope with mass, hence tension can't be considered equal on all points)
Tp - .1g - 2.9g = 3a
Tp = 3(g+a) = 30N

Answer 2 - For P
Tp - 1.9g - 0.1g = 2a
Tp = 2(g+a) = 20N

The given answer is Tp = 30N
Now here's my doubt.
In answer 1 - we are calculating Tension at point P. We considered the weights 0.1kg and 2.9kg and ignored that the 1.9kg weight. But I think the weight 1.9kg would apply a force 1.9g which should be considered.(Here we come up from down and include one weight (1.9kg and .1kg)

In answer 2 - we considered the weights 1.9kg and 0.1kg but didn't include 2.9kg. So, here we ignored 2.9kg like it didn't exist. But wouldn't the 2.9g exert a force? (Here we go up from down and include 2.9kg and 0.1kg(half of mass of rope)

--------------------------------------------------------------------------------------------------------------------
Out of answer 1 and 2, I consider Answer 2 i.e., 20N to be more right than the second one because, we are asked to find Tension at the midpoint. So we should just only consider the forces that are acting on it.

Please help. I've tried to find more solutions of this question online. Tq always comes out to be 50N.
But the value of Tp seems to be either 20N or 30N on different forums. Which is the correct value?

(reiteration: Tp = 30N is the given answer)

In general, does the tension in a hanging system depend on the mass above or mass below?

It seems to me that if the tension at P is 20N, then shouldn't the tension at Q be 0N?

 

[haruspex]

Answer 2 - For P
Tp - 1.9g - 0.1g = 2a
Tp = 2(g+a) = 20N

You appear to be analysing the system between P and Q to arrive at this. But that system also has the tension at Q acting, so that should feature in the equation. Also, some signs are wrong.

 

[Chestermiller]

Guys, the OP was last seen at 2:28 pm viewing this thread, so he has seen all posts except haruspex's post. No response on his part.

 

[James2911]

You appear to be analysing the system between P and Q to arrive at this. But that system also has the tension at Q acting, so that should feature in the equation. Also, some signs are wrong.

Thank you! It worked!

if we consider the system between Q and P;
Tq - Tp - 0.1g - 1.9g = 2a
50 - 2(9.8+0.2) = Tp
Tp = 50 - 20
Tp = 30

This did bring out the right answer. But instead of my traditional method of drawing FBD of a body, I considered a system between P and Q and took the forces acting on it (Like Tq acting upwards, and the rest 3 acting downwards).

But, how do I relate the Tension Q to FBD of P? There is a block (1.9kg) in between.

Guys, the OP was last seen at 2:28 pm viewing this thread, so he has seen all posts except haruspex's post. No response on his part.

I apologize.

 

[Orodruin]

But, how do I relate the Tension Q to FBD of P?

There is not such a thing as a FBD of a point. You need to consider a system that you draw a contour around. Your reaction here seems to suggest that you need to revisit FBDs and how they are used.

The FBD is a FBD including everything between P and Q.

 

[James2911]

There is not such a thing as a FBD of a point. You need to consider a system that you draw a contour around. Your reaction here seems to suggest that you need to revisit FBDs and how they are used.

The FBD is a FBD including everything between P and Q.

Thank you sir!