Force Problem with tension, 2 blocks, and a rope that has mass.

[09jml90]

Homework Statement

Two blocks m₁= 9.4 kg and m₂ = 4.9 kg are connected by a homogeneous rope that has a mass of m_r = 1.42 kg. A constant vertical force, F = 184.9 N, is applied to the upper block.

What is the magnitude of the acceleration of the system?

What is the magnitude of the tension in the rope at the bottom end of the rope?

What is the magnitude of the tension in the rope at the top end of the rope?

Homework Equations

Newton's Second Law = \sum F = ma

The Attempt at a Solution

For the first part, I took the objects to be part of one system, used F=ma with 184.9=F, combined the 3 masses, and solved for a. I then subtracted 9.8 m/s^2 to get 1.96 m/s^2.

Tension Parts: I wasn't really sure how to do this from the get go, so I started trying some things. I'm also still not 100% sure about the relationships going on here. For the mag. of the tension of the rope at the bottom end of the rope, I took (4.9kg)(11.76 m/s^2) + (4.9kg)(9.8 m/s^2). I also tried it adding the mass of the rope on to the second block, so (5.62kg)(9.8m/s^2) + (9.4kg)(11.76 m/s^2)- I then wasn't sure if I needed to take into account block 1.

Because I couldn't figure out the bottom tension, I didn't attempt to do the tension of the top of the rope.

Any help would be appreciated. Thank you

 

[Doc Al]

For the first part, I took the objects to be part of one system, used F=ma with 184.9=F, combined the 3 masses, and solved for a. I then subtracted 9.8 m/s^2 to get 1.96 m/s^2.

You managed to get the right answer, but a better approach would be:
ΣF = 184.9 - Mg
Set ΣF = Ma and solve for a. (M is the total mass.)

Tension Parts: I wasn't really sure how to do this from the get go, so I started trying some things. I'm also still not 100% sure about the relationships going on here. For the mag. of the tension of the rope at the bottom end of the rope, I took (4.9kg)(11.76 m/s^2) + (4.9kg)(9.8 m/s^2). I also tried it adding the mass of the rope on to the second block, so (5.62kg)(9.8m/s^2) + (9.4kg)(11.76 m/s^2)- I then wasn't sure if I needed to take into account block 1.

Hint: To find the tension at the bottom of the rope, analyze the forces on the bottom block. The tension is one of the forces acting on that block. Apply ΣF = ma.

 

[09jml90]

Would the second force be weight?, w=mg (w=(4.9kg)(9.8m/s^2)
and then possibly

T-48.02=(mass of M1 + Mrope)(11.76 m/s^2) ?

 

[Doc Al]

Would the second force be weight?, w=mg (w=(4.9kg)(9.8m/s^2)

Yes. There are two forces acting on the bottom block: the rope tension pulling up and the weight pulling down.

and then possibly

T-48.02=(mass of M1 + Mrope)(11.76 m/s^2) ?

The left side of the equation is fine, but the right side has two problems:
(1) Since you're analyzing forces on the bottom block, the mass needed is just the mass of the bottom block.
(2) The acceleration of the system is not 11.76 m/s^2. You found the acceleration in post #1. Use it.

 

[09jml90]

I found the tension of the bottom of the rope to be 57.6 N, and I'm working on the top part.

 

[Doc Al]

I found the tension of the bottom of the rope to be 57.6 N, and I'm working on the top part.

Good.

 

[09jml90]

Do I assume the mass of the rope to be included with the mass of the first block? to give the force going up as (9.4kg + 1.42kg)(1.95 m/s^2) = 21.1 N

 

[09jml90]

What kind of martial arts do you train in/interested in?

 

[Doc Al]

Do I assume the mass of the rope to be included with the mass of the first block?

Not if you want to find the tension at the top of the rope. That end of the rope must be at the edge of your system. You can either: analyze forces on the top block or analyze forces on the 'rope + bottom block'.

 

[09jml90]

So, block 1 has a force going up which is 184.9 N. The force going down is weight. Is weight here just the weight of block 1 or the rest of the system of the rope and block 2.

If so, the weight of the rope and block 2 has a force of 61.936 N. And the block going up has a force of 184.9 N. So added together the magnitude of the force is 246.84 N?

 

[09jml90]

Never-mind that is wrong too.

 

[09jml90]

If you are allowed to offer me any more hints, here is what I've come to until I'm stuck.

In analyzing the top block, the force going upwards is 18.33 N (9.4kg)(1.95m/s^2). There is a force going downwards, the weight, which I'm not sure about. Is the weight needed for only block 1 or the weight of the rope and block 2 added together. In either case, if I'm doing the calculations correctly for either block 1 or block 2 + rope, I am getting the wrong answer.

 

[Doc Al]

In analyzing the top block, the force going upwards is 18.33 N (9.4kg)(1.95m/s^2).

Don't call that a force, that's just mass X acceleration, which Newton's 2nd law tells us equals the net force on the object.

There is a force going downwards, the weight, which I'm not sure about. Is the weight needed for only block 1 or the weight of the rope and block 2 added together.

The weight acting on the top block is just the weight of the top block. Note: There are 3 forces acting on the top block. What are they? Set their sum equal to 'ma'.

 

[09jml90]

Sorry, the computer says I finally got it right. I don't know why I kept getting caught up in thinking about the other objects instead of just block 1. You told me that I had calculated the net force when I multiplied block 1 (9.4kg)(1.95m/s^2) = 18.33 N, which is = ma. The weight of block 1 was (9.4kg)(9.8) = 92.12 N.

T + 92.12 N = 18.33 N
T = -73.79 N
Magnitude = 73.79 N

Thank you for you help.

 

[Doc Al]

Sorry, the computer says I finally got it right. I don't know why I kept getting caught up in thinking about the other objects instead of just block 1. You told me that I had calculated the net force when I multiplied block 1 (9.4kg)(1.95m/s^2) = 18.33 N, which is = ma. The weight of block 1 was (9.4kg)(9.8) = 92.12 N.

T + 92.12 N = 18.33 N
T = -73.79 N
Magnitude = 73.79 N

Thank you for you help.

Sorry, but this solution is not correct. (Even though your answer is close.)

Try this:

ΣF = ma
F - T - mg = ma (where F here is the applied force).

 

[kkhz]

sorry i have a similar problem to this; on the last part, are both of the "m"s the summation of all of the masses? so F - T - (summation masses x g) = (summation masses x a)?

 

[kkhz]

nevermind I got it, thanks for the help