# Simple problem for anyone who knows what he is doing

#### Suicidal

I’m stuck on a problem involving a cord that has mass. I know how to do problems where the cord's mass is 0 but ones the factor of the cords mass is introduced I’m lost.

I’m stuck on the following problem:

A block (mass m1) on a smooth horizontal surface, connected by a cord that passes over a pulley to a second block (mass m2), which hangs vertically.

Determine a formula for the acceleration of the system if the cord has non-negligible mass mc specify in terms of L1 and L2 the lengths of cord from the respective masses to the pulley (total cord length L =L1+L2)

How do I approach this problem?

The answer (if it helps) is a=(m2+mc(L2/L))g/(m1+m2+mc)

Any help will be highly appreciated.
Thanks

Related Introductory Physics Homework Help News on Phys.org

#### HallsofIvy

Homework Helper
I think what you need to do is note that the cord has total length L= L1+ L2 and so has "linear density" mc/L= mc/(L1+ L2). That means the mass of the cord hanging vertically is
mc(L2/(L1+ L2))= mc(L2/L). That gives precisely the answer given.

From that you can calculate the total mass hanging down and total of all mass.

#### Suicidal

I already understood that the mass of the cord hanging down in L2/L.

How do you deal with the fact that the horizontal part of the cord is being pulled down and what about the tension in the cord?

I don’t understand how all the forces involved effect the system and therefore I don’t know how to set up an expression for the acceleration.

For the hanging down part of the system Net force = (m2+mc*L2/L)*g- Ft
Where Ft is the force of tension.

So
a=((m2+mc*L2/L)*g- Ft)/( m2+mc*L2/L)

I think I’m correct on that, but how do I find Ft.

#### HallsofIvy

Homework Helper
I thought you said you could do it if the cord was massless- once you know how to calculate the mass of each part of the cord, include that into the masses m1 and m2 and the rest of the problem is exactly as if the cord was massless.

The total mass "hanging down" is m2+ mc(L2/L). The downward pull is it's weight: (m2+ mc(L2/L)g. But that force (weight) has to accelerate the entire mass: m1+ m2+ mc. Using F= ma we have

(m2+ mc(L2/L)g= (m1+ m2+ mc)a.

Solve for a.

#### Suicidal

Thanks, the explanation helped a lot .
I think I understand now, but there is still one thing, the force of tension.

Last edited:

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The text passage contains a physics problem involving a system of two blocks connected by a cord with non-negligible mass. The problem asks for the formula for the acceleration of the system, and the answer is provided in the text.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Simple problem for anyone who knows what he is doing

#### Suicidal

I’m stuck on a problem involving a cord that has mass. I know how to do problems where the cord's mass is 0 but ones the factor of the cords mass is introduced I’m lost.

I’m stuck on the following problem:

A block (mass m1) on a smooth horizontal surface, connected by a cord that passes over a pulley to a second block (mass m2), which hangs vertically.

Determine a formula for the acceleration of the system if the cord has non-negligible mass mc specify in terms of L1 and L2 the lengths of cord from the respective masses to the pulley (total cord length L =L1+L2)

How do I approach this problem?

The answer (if it helps) is a=(m2+mc(L2/L))g/(m1+m2+mc)

Any help will be highly appreciated.
Thanks

Related Introductory Physics Homework Help News on Phys.org

#### HallsofIvy

Homework Helper
I think what you need to do is note that the cord has total length L= L1+ L2 and so has "linear density" mc/L= mc/(L1+ L2). That means the mass of the cord hanging vertically is
mc(L2/(L1+ L2))= mc(L2/L). That gives precisely the answer given.

From that you can calculate the total mass hanging down and total of all mass.

#### Suicidal

I already understood that the mass of the cord hanging down in L2/L.

How do you deal with the fact that the horizontal part of the cord is being pulled down and what about the tension in the cord?

I don’t understand how all the forces involved effect the system and therefore I don’t know how to set up an expression for the acceleration.

For the hanging down part of the system Net force = (m2+mc*L2/L)*g- Ft
Where Ft is the force of tension.

So
a=((m2+mc*L2/L)*g- Ft)/( m2+mc*L2/L)

I think I’m correct on that, but how do I find Ft.

#### HallsofIvy

Homework Helper
I thought you said you could do it if the cord was massless- once you know how to calculate the mass of each part of the cord, include that into the masses m1 and m2 and the rest of the problem is exactly as if the cord was massless.

The total mass "hanging down" is m2+ mc(L2/L). The downward pull is it's weight: (m2+ mc(L2/L)g. But that force (weight) has to accelerate the entire mass: m1+ m2+ mc. Using F= ma we have

(m2+ mc(L2/L)g= (m1+ m2+ mc)a.

Solve for a.

#### Suicidal

Thanks, the explanation helped a lot .
I think I understand now, but there is still one thing, the force of tension.

Last edited:

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
user
The text passage contains a physics problem involving a system of two blocks connected by a cord with non-negligible mass. The problem asks for the formula for the acceleration of the system, and the answer is provided in the text.

TYPE: SINGLE_QUESTION_WITH_ANSWER