What is the tension equation for masses connected by a cord?

[Caitlin Galla]

Homework Statement

There is a 1.0 kg block on the left connected by a cord to a 2.5kg block on the right.
If this is being pulled from the right by a 50n force. What is the tension on the cord connecting the two masses.
There is also a small 2.0kg block in front of the 1.0 kg block that is not attached to the cord.
But if the blocks are pulled the 2.0 kg block will have to go with it.

Homework Equations

I found the acceleration of this system to be 9.1 m/s2.

I just need to find the tension equation to use for this problem.
Any ideas? Thanks

The Attempt at a Solution

 

[dark adonis]

Homework Statement

There is a 1.0 kg block on the left connected by a cord to a 2.5kg block on the right.
If this is being pulled from the right by a 50n force. What is the tension on the cord connecting the two masses.
There is also a small 2.0kg block in front of the 1.0 kg block that is not attached to the cord.
But if the blocks are pulled the 2.0 kg block will have to go with it.

Homework Equations

I found the acceleration of this system to be 9.1 m/s2.

I just need to find the tension equation to use for this problem.
Any ideas? Thanks

The Attempt at a Solution

Try the following approach:
1) What is the net force on the block on the 2.5 kg block?
2) What forces are directly acting on the 2.5 kg block?

Answer these questions and I think it should be clear how to procede

 

[Caitlin Galla]

this is a picture. it still seems unclear to me. i thought the net force was 50 N . Is that not the net force?

[img=http://img442.imageshack.us/img442/9350/blocksun4.gif]

 

[dark adonis]

this is a picture. it still seems unclear to me. i thought the net force was 50 N . Is that not the net force?

[img=http://img442.imageshack.us/img442/9350/blocksun4.gif]

50 N is the net force acting on the blocks system, but it isn't the net force acting on the single block. To get that you will need to look at the acceleration of the block. Essentially you will need F = m a.

 

[Caitlin Galla]

so if i use the mass of the block as 2.5 and times that by calculated acceleration it would be 2.5 x 9.1= 22.75. would that be the force acting on the single block? And it still isn't clear to me where to go from here.

 

[dark adonis]

so if i use the mass of the block as 2.5 and times that by calculated acceleration it would be 2.5 x 9.1= 22.75. would that be the force acting on the single block? And it still isn't clear to me where to go from here.

Alright we have the net force, this is equivalent to the sum of all of the forces acting on the block. Now we need only list the forces acting on the block. Set up an algebraic expression like: $$22.75 N = F_{Tension}+F_{Other Force}$$ where the other forces are known and solve for $$F_{Tension}$$

 

[Caitlin Galla]

which other forces are included here? is that just the known 50 N of force on the whole system?

 

[dark adonis]

which other forces are included here? is that just the known 50 N of force on the whole system?

Bingo The only other force that could be acting is the 50 N pull

 

[w3390]

I have two questions for you:
(1) Is there any friction between the blocks and whatever they are resting on?
(2) If the 50N force pulling the right block is constant, then there shouldn't be
any acceleration.

I would approach this problem by first labeling all of the forces. If I understand correctly, all the blocks are resting on a frictionless table. Therefore, the sum of all forces in the y direction is zero. You then set a direction of motion as positive and the opposite direction as negative. In this case, it makes sense to set motion to the right as the +x direction. Since the force that the 1kg box exerts on the 2kg box is equal to the force that the 2kg box exerts on the 1kg box, there is no net force you have to worry about regarding the two boxes to the left. Therefore, your equation for the sum of all forces in the x direction will be
F(net)=ma
F(net)= F(pull)-F(tension)
F(pull)-F(tension)=ma=0
F(pull)= F(tension)= 50N

 

[Caitlin Galla]

thanks so much. i am just struggling with physics a lot more than any other science class i have taken. i appreciate the help!

 

[dark adonis]

I have two questions for you:
(1) Is there any friction between the blocks and whatever they are resting on?
(2) If the 50N force pulling the right block is constant, then there shouldn't be
any acceleration.

I would approach this problem by first labeling all of the forces. If I understand correctly, all the blocks are resting on a frictionless table. Therefore, the sum of all forces in the y direction is zero. You then set a direction of motion as positive and the opposite direction as negative. In this case, it makes sense to set motion to the right as the +x direction. Since the force that the 1kg box exerts on the 2kg box is equal to the force that the 2kg box exerts on the 1kg box, there is no net force you have to worry about regarding the two boxes to the left. Therefore, your equation for the sum of all forces in the x direction will be
F(net)=ma
F(net)= F(pull)-F(tension)
F(pull)-F(tension)=ma=0
F(pull)= F(tension)= 50N

Err... the block definitely accelerates, a constant force definitely means a constant acceleration...