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(Note: The title would better describe the subject if written like this:

"2 blocks on a horizontal frictionless surface connected by a massless string vs. pulley system")Hello, I have a problem regarding pulling two blocks of different masses connected by a string horizontally with opposing forces and evaluating the tension. Then, I would like to find out if this problem is set in such a way that it is analogous to two blocks on a pulley.

Note: Let mass of block 1 be A and mass of block 2 be B. So m1 = A and m2 = B.

Question 1: In what direction would the system accelerate?

Question 2: What is the net force of the system?

Question 3: What is the acceleration of the system?

Question 4: What would be the tension in the massless cable connected to:

1. the internal portion of A (T1 on to m1)

2. the internal portion of B (T1 on to m2)

Question 5:

If you were to take the internal cable of this system and drape it over a pull, with Block A hanging down on the left side of the pulley and Block B hanging down on the right side of the pulley, would your values for tension change?

F=ma

Answer 1

Then the block B would require more force:

FA = [A(a) = 10kg(9.8m/s^2) = 98N] < FB = [B(a) = 100kg(9.8m/s^2) = 980N]

Answer 2

So the net force on the system would be Fnet = FA + FB = B(a) - A(a) = 980N - 98N = 882N

Answer 3

The acceleration would be F/(A+B) = 882N/110kg = 8.0181m/s^2

Answer 4

For Tensions,

First establish net force of each block:

Fnet_A = A(a) = 10kg(8.018m/s^2) = 80.18N

Fnet_B = B(a) = 100kg(8.018m/s^2) = 801.81N,

Then Tension in strings are:

Tension of left most string on Block A

FA = 10kg(9.8m/s^2) = 98N = T1/A

Tension of internal string on Block A

FA = T2/A - T1/A ==> T2/A = F1 + T1/A = A(a2) + A(a1) = A(a2+a1) = 10kg(9.8+8.018) = 178.18N

Tension of internal string on Block B

FB = T3/B - T2/B ==> T2/B = T3/B - F2 = B(a1) - B(a2) = B(a1 - a2) = 100kg(9.8 - 8.018) = 178.2N

Tension of right most string on Block B

FB = 100kg(9.8m/s^2) = 980N = T3/B

Question 5

I think that if you were to drape the internal string over a pulley, the tensions in the string would remain exactly the same because gravity, at 9.8m/s^2, would be pulling on each box just as the blocks are being pulled from opposite directions horizontally.

Thank you

"2 blocks on a horizontal frictionless surface connected by a massless string vs. pulley system")Hello, I have a problem regarding pulling two blocks of different masses connected by a string horizontally with opposing forces and evaluating the tension. Then, I would like to find out if this problem is set in such a way that it is analogous to two blocks on a pulley.

**If you pull to the left with a force such that if Block A were an isolated mass, it would accelerate 9.8m/s^2 ..and you pull to the right of Block B with such a force that if Block B were an isolated mass, it would accelerate 9.8m/s^2.**

1. Homework Statement1. Homework Statement

Note: Let mass of block 1 be A and mass of block 2 be B. So m1 = A and m2 = B.

Question 1: In what direction would the system accelerate?

Question 2: What is the net force of the system?

Question 3: What is the acceleration of the system?

Question 4: What would be the tension in the massless cable connected to:

1. the internal portion of A (T1 on to m1)

2. the internal portion of B (T1 on to m2)

Question 5:

If you were to take the internal cable of this system and drape it over a pull, with Block A hanging down on the left side of the pulley and Block B hanging down on the right side of the pulley, would your values for tension change?

## Homework Equations

F=ma

## The Attempt at a Solution

[/B]Answer 1

Then the block B would require more force:

FA = [A(a) = 10kg(9.8m/s^2) = 98N] < FB = [B(a) = 100kg(9.8m/s^2) = 980N]

Answer 2

So the net force on the system would be Fnet = FA + FB = B(a) - A(a) = 980N - 98N = 882N

Answer 3

The acceleration would be F/(A+B) = 882N/110kg = 8.0181m/s^2

Answer 4

For Tensions,

First establish net force of each block:

Fnet_A = A(a) = 10kg(8.018m/s^2) = 80.18N

Fnet_B = B(a) = 100kg(8.018m/s^2) = 801.81N,

Then Tension in strings are:

Tension of left most string on Block A

FA = 10kg(9.8m/s^2) = 98N = T1/A

Tension of internal string on Block A

FA = T2/A - T1/A ==> T2/A = F1 + T1/A = A(a2) + A(a1) = A(a2+a1) = 10kg(9.8+8.018) = 178.18N

Tension of internal string on Block B

FB = T3/B - T2/B ==> T2/B = T3/B - F2 = B(a1) - B(a2) = B(a1 - a2) = 100kg(9.8 - 8.018) = 178.2N

Tension of right most string on Block B

FB = 100kg(9.8m/s^2) = 980N = T3/B

Question 5

I think that if you were to drape the internal string over a pulley, the tensions in the string would remain exactly the same because gravity, at 9.8m/s^2, would be pulling on each box just as the blocks are being pulled from opposite directions horizontally.

Thank you

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