# Two blocks on a frictionless horizontal surface connected by a massless string

• physics120
In summary, two blocks connected by a massless string are pulled towards each other with a horizontal force. The tensions in the two strings are calculated using Newton's second law.
physics120

## Homework Statement

Two blocks on a frictionless horizontal surface are connected together by a massless string. A second massless string is attached to the block of mass m2, as shown. Suppose you pull on this second string with a horizontal force directed to the right.

Determine the ratio of the tensions for the two strings.

NOTE: THE PICTURE SHOWS ONE BLOCK LARGER THAN THE OTHER. HOWEVER, IN THIS QUESTION WE ACTUALLY DO NOT KNOW WHICH BLOCK IS LARGER.

Let T= tension

use ratio: T1/T2

## The Attempt at a Solution

So, would it just simply be (T(m2onm1))/(T(Fonm2) - T(m1onm2))

where m2onm1 is the subscript of the 1st T, Fonm2 is the subscript of the 2nd T, and m1onm2 is the subscript on the 3rd T.
(These are the action reaction forces I obtained by completing a free body diagram- Tm2onm1 with Tm1onm2. The T(Fonm2) is the force the second string applies on block m2).

Or is the question asking more than that? Are they asking, what does T1/T2 equal ? I am not sure if I understood the question correctly..

Thank-you!

#### Attachments

• blocks_dragged.gif
1.7 KB · Views: 1,501
They want the exact ratio T1/T2, not just generalities. Your answer should be in terms of m1 & m2.

Apply Newton's 2nd law! (Twice.)

Hi,

well I tried what you suggested, and I got: T1/T2= (0.5)(m1/m2)

where m1 is the mass of the first block,
m2 is the mass of the 2nd block,
T1 is the tension from the first massless string,
T2 is the tension from the second massless string on block m2

Is that correct? Or am I still not even understanding the question correctly?

physics120 said:
well I tried what you suggested, and I got: T1/T2= (0.5)(m1/m2)
Please describe how you got that result.

(It's not quite right, but if you show your work perhaps we can see where you went wrong.)

OK, here it goes:

n= normal force
w= gravitational force

After drawing the blocks' free body diagrams, I obtained:

For m1:

Fy: n - w =0
n=mg
Fx: T(m2onm1)=m1a1

For m2:

Fy: n - w=0
n=mg
Fx: T(Fonm2) - T(m1onm2)=m2a2

a1=a2=a

T(m2onm1)=T(m1onm2)=T

a= T/m1

a= (T(Fonm2) - T)/m2

putting them together i get:

(T/m1) = ((T(Fonm2) - T)/m2)

multiply both sides by m2:

((m2/m1)T) = (T(Fonm2) - T)

divide both sides by T:

(m2/m1) = ((T(Fonm2)/T) - T/T)

(m2/m1) = ((T(Fonm2)/T) - 1)

((T(Fonm2)/T) = ((m2/m1) + 1)

For some reason, now I am getting ((m2/m1) + 1). Although I really don't think that is correct either. Especially since this time, for the ratio, I got the second tension (2nd massless rope " T(Fonm2)") over the 1st tension (from 1st massless rope "T"). Shouldn't it be the other way around? Or does that not matter?

Thank-you for helping, btw!

physics120 said:
For some reason, now I am getting ((m2/m1) + 1).
Excellent.
Although I really don't think that is correct either. Especially since this time, for the ratio, I got the second tension (2nd massless rope " T(Fonm2)") over the 1st tension (from 1st massless rope "T"). Shouldn't it be the other way around? Or does that not matter?
You found T2/T1. If you want T1/T2, just invert it.

Here's how I'd do this one.

First I'd consider both masses together:
T2 = (m1 + m2)a

Then m1 by itself:
T1 = (m1)a

Then take any ratio you like:
T2/T1 = (m1 + m2)/m1
T1/T2 = m1/(m1 + m2)

OK, thanks a lot!

## 1. What is the purpose of connecting two blocks on a frictionless horizontal surface with a massless string?

The purpose of connecting two blocks on a frictionless horizontal surface with a massless string is to study the relationship between the acceleration of the blocks and the force applied on one of the blocks.

## 2. What is the role of friction in this scenario?

In this scenario, there is no friction present on the horizontal surface. This is to eliminate any external forces that may affect the motion of the blocks and to simplify the experiment.

## 3. How does the mass of the blocks affect their acceleration in this setup?

The mass of the blocks does not affect their acceleration in this setup. This is because the string connecting the blocks is massless, so the force applied on one block is directly transmitted to the other block without any change in magnitude.

## 4. Can the blocks move in opposite directions in this setup?

Yes, the blocks can move in opposite directions in this setup. This can happen if the force applied on one block is greater than the force applied on the other block, causing the blocks to accelerate in opposite directions.

## 5. How does changing the length of the string affect the acceleration of the blocks?

The length of the string does not affect the acceleration of the blocks in this setup. This is because the tension in the string, which is the force that causes the blocks to accelerate, is independent of the length of the string as long as the string remains taut.

• Introductory Physics Homework Help
Replies
8
Views
1K
• Introductory Physics Homework Help
Replies
29
Views
4K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
3K