Two blocks tied to a string with mass and accelerating

[RadiantL]

Homework Statement

The figure shows two 2.5 kg blocks connected by a rope that has a mass of 450g. The entire assembly is accelerated upwards at 7.4 m/s^

What is the tension at the bottom end of the rope, near block B

What is the tension at the at the top end of the rope, near block A

Homework Equations

I believe the relevant equation needed is just

Fnet = m * a

and Fnet = Sum of all forces

The Attempt at a Solution

So I am sure I got the first part correct, I made

Fnet = Tension - Mb * g
Mb * a = Tension - Mb * g
(Mb*a) + (Mb*g) = T
(2.5*7.4)+(2.5*9.8) = 43
Tension = 43

(correct me if I'm wrong for the above)

Now the second part I actually have no clue how to get started, I can't think up an equation that makes sense and could isolate for T

 

[PeterO]

Homework Statement

The figure shows two 2.5 kg blocks connected by a rope that has a mass of 450g. The entire assembly is accelerated upwards at 7.4 m/s^

What is the tension at the bottom end of the rope, near block B

What is the tension at the at the top end of the rope, near block A

Homework Equations

I believe the relevant equation needed is just

Fnet = m * a

and Fnet = Sum of all forces

The Attempt at a Solution

So I am sure I got the first part correct, I made

Fnet = Tension - Mb * g
Mb * a = Tension - Mb * g
(Mb*a) + (Mb*g) = T
(2.5*7.4)+(2.5*9.8) = 43
Tension = 43

(correct me if I'm wrong for the above)

Now the second part I actually have no clue how to get started, I can't think up an equation that makes sense and could isolate for T

I don't see any diagram attached?? but assuming it is two masses above each other, with the rope in between..

You could replace the situation you are presented with with a three mass system and mass-less strings.

The third mass is one of 450g in between the two 2.5 kg blocks.

The tension in the lower string would then be the same as the tension near block B that you have calculated [presumably correctly; I have not checked your working/answer].
The tension in the top string would be same as the tension near block A that you are seeking.
As I said, I have not checked your work so far, just offering a conceptual way to proceed with your calculations.

 

[RadiantL]

Oh wow, that's a really cool approach to this question... I got the correct answer through your way :D

But I'm kinda curious to why you could change the question like that and it would work... Could you explain please? THANKS!

 

[PeterO]

Oh wow, that's a really cool approach to this question... I got the correct answer through your way :D

But I'm kinda curious to why you could change the question like that and it would work... Could you explain please? THANKS!

The centre of the rope is only important if you are working out what is happening in the centre of the rope. Since you were calculating what happens at each end of the rope, you can include the rope as "just another 450g body" rather than a long object of some mass.

I have seen questions like this where they want to know "if the rope is 1m long, what is the tension 30 cm from body A?" or similar.

Even so you could then consider that 135g of the rope is above that point and 315g below; break in into a 4 body system and find the tension in the middle string, with the masses [from the bottom] 25. kg, 315g, 135g, and 2.5 kg.

note: 30cm is 30% of the 1m rope; 30% of 450g = 135g etc.

 

[asteriatic]

ension.


I would approach this problem by first identifying all the known variables and defining them with appropriate units. In this case, we know the masses of the blocks and the acceleration of the system, so we can write:

m1 = 2.5 kg (mass of block A)
m2 = 2.5 kg (mass of block B)
mr = 0.450 kg (mass of the rope)
a = 7.4 m/s^2 (acceleration of the system)

Next, we need to consider the forces acting on each block and the rope. The forces acting on block A are its weight (mg) and the tension force from the rope (T1). Similarly, the forces acting on block B are its weight (mg) and the tension force from the rope (T2).

Using Newton's second law, we can write equations for the net force on each block:

Fnet,A = ma = T1 - m1g
Fnet,B = ma = T2 - m2g

We can also consider the forces acting on the rope itself. The rope is experiencing a tension force at both ends, so we can write:

Fnet,rope = T1 + T2 - mr * g = ma

Since the rope is not accelerating vertically (it is being pulled up by the blocks), we can set the net force on the rope to zero and solve for T1 and T2:

T1 + T2 = mr * g = 0.450 kg * 9.8 m/s^2 = 4.41 N

Now, we can substitute this value for T2 in the equation for Fnet,A and solve for T1:

Fnet,A = ma = T1 - m1g
T1 = m1a + m1g = (2.5 kg * 7.4 m/s^2) + (2.5 kg * 9.8 m/s^2) = 43.5 N

So the tension at the top end of the rope, near block A, is 43.5 N. And the tension at the bottom end of the rope, near block B, is T2 = 4.41 N.

Overall, the important thing to remember is that the tension force in a rope is equal at both ends, but the net force on each block will be different depending on its mass and acceleration.