# M1 pulleys :(

#1
I have a way of working them out which makes sense to me (finding the net downward force then dividing by the total mass of the system to find the acceleration of the system) but they don't like it! they ask me questions about the tension of the string, which confuses me a lot.

could someone explain what's going on with tension in these questions?
0
8 years ago
#2
Who are 'they'?
0
8 years ago
#3
Yes, Tension is simply the force in the rope. Here is an example:

Consider the image above. I will explain how to solve using Tension. The general process is to consider the forces acting on each block, and then combine the equations.

1. Consider forces acting on the left block:

F_net = ma = UpForce - DownForce = Tension - mg (Here the rope is providing the upward force, so this is the tension force).

Therefore,

a = (T - mg)/m [where T = Tension]

2. Consider forces acting on the right block:

F_net = Ma = UpForce - DownForce = Tension - Mg (Notice Tension is the same as in the previous example, this is a key feature of Tension. Tension is always constant for the same rope. (if it wasn't the same, the rope would pull itself apart) Because this rope is the same as the rope attached to the left block, the tension in this rope is the same, so I cause the same variable T.)

Therefore,

a = (T - Mg)/M

3. Combine equations. Because as one side moves down, the other moves up at the same rate, the accelerations will be equal and opposite to each other.

Therefore,

(T - Mg)/M = (mg - T)/m

So,

m(T - Mg) = M(mg - T)

So,

mT - Mmg = Mmg - TM

So,

(m+M)T = 2Mmg

Therefore,

T = 2Mmg/(m+M)

Now that we know tension, we can calculate acceleration by simply substituting our answer here into one of the equations from either step 1 or step 2.

This process may seem tedious, but it is essential to know how to do this for more complicated problems that can be created.
1
8 years ago
#4
Yes in most problems at this level the tension will be constant for the whole rope/string.

But it is actually only constant when

- the rope is massless (often referred to as "light" in questions.
- it is inextensible. (If it extends then you have tension due to extension to add in.)
- there is no friction between the string and the pulley

So in more complicated problems you need to be on the lookout!
0
#5
(Original post by Stonebridge)
Yes in most problems at this level the tension will be constant for the whole rope/string.

But it is actually only constant when

- the rope is massless (often referred to as "light" in questions.
- it is inextensible. (If it extends then you have tension due to extension to add in.)
- there is no friction between the string and the pulley

So in more complicated problems you need to be on the lookout!
I'm only doing m1&2 (Edexcel). Surely Hooke's Law/whatever other physics concepts involved won't up..?
0
#6
(Original post by Doctor_Einstein)
Yes, Tension is simply the force in the rope. Here is an example:

Consider the image above. I will explain how to solve using Tension. The general process is to consider the forces acting on each block, and then combine the equations.

1. Consider forces acting on the left block:

F_net = ma = UpForce - DownForce = Tension - mg (Here the rope is providing the upward force, so this is the tension force).

Therefore,

a = (T - mg)/m [where T = Tension]

2. Consider forces acting on the right block:

F_net = Ma = UpForce - DownForce = Tension - Mg (Notice Tension is the same as in the previous example, this is a key feature of Tension. Tension is always constant for the same rope. (if it wasn't the same, the rope would pull itself apart) Because this rope is the same as the rope attached to the left block, the tension in this rope is the same, so I cause the same variable T.)

Therefore,

a = (T - Mg)/M

3. Combine equations. Because as one side moves down, the other moves up at the same rate, the accelerations will be equal and opposite to each other.

Therefore,

(T - Mg)/M = (mg - T)/m

So,

m(T - Mg) = M(mg - T)

So,

mT - Mmg = Mmg - TM

So,

(m+M)T = 2Mmg

Therefore,

T = 2Mmg/(m+M)

Now that we know tension, we can calculate acceleration by simply substituting our answer here into one of the equations from either step 1 or step 2.

This process may seem tedious, but it is essential to know how to do this for more complicated problems that can be created.
Thank you! I'm not sure if I have the ~intuition~ for the constant tension thing (thinking about Newton's third generally confuses me a lot), but that doesn't really matter.
0
8 years ago
#7
(Original post by genuinelydense)
I'm only doing m1&2 (Edexcel). Surely Hooke's Law/whatever other physics concepts involved won't up..?
Not in a rope-on-a-pulley type question like this.

Comment was aimed mainly at the other poster.
0
1 year ago
#8
mafs g
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Have you done work experience at school/college?

Yes (155)
41.67%
Not yet, but I will soon (69)
18.55%
No (148)
39.78%

View All
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.