# Tension homework question

1. Oct 10, 2004

### Alem2000

I am having trouble figuring this out. " Two blocks are connected by a uniform

rope with a mass of $$4.00kg$$. An upward force of $$200N$$ is

applied. ...a)what is the $$\vec{a}$$ of the system and what is the

tension at the top of the rope.

The $$\vec{a}$$ of the system would have to be

$$\frac{\sum{\vec{F}}}{m}=a$$

$$\sum{\vec{F}}=200N-(m_1+m_2+m_3)g$$

which is $$200N-147N=53N$$

and then $$\frac{53N}{15.0g}=\vec{a}$$

so now for the tension of the top of the rope. What are the forces acting on the top of the rope...there is the $$200N$$...and weight of the box hanging from the rope?

Last edited: Oct 10, 2004
2. Oct 10, 2004

### Gokul43201

Staff Emeritus
Since you haven't provided all the information from the question, I'll have to make some assumptions...but basically you can model this system as a single block of mass (m1+m2) being pulled upwards by a single rope of mass m3. Now the upward force on this system is T, while the total downward force is (m1+m2+m3)g. Use the Second Law to find T.

3. Oct 10, 2004

### Staff: Mentor

Apparently you know the masses of the two blocks and the rope. (Since you were able to find the acceleration.)

To find the tension in the top of the rope, consider the top block as your object and apply Newton's 2nd law to it.

4. Oct 10, 2004

### Alem2000

Yes I see. But then it asks me to find the tension at the midpoint of the rope. I

start to solve for$$T_t$$ which is tension at the top

$$F-m_1g-T_t=ma$$ where the mass is that of the top block and

acceleration is that of the system. The book says the correct set up is

$$T_t=F-m(g+a)$$ which is the same as mine except the accelaration is

gravity plus the systems...why? I can understand why you would do that.

5. Oct 10, 2004

### Staff: Mentor

Those two equations are equivalent.

6. Oct 10, 2004

### Alem2000

so to find the $$T_m$$ tension in the middle of the rope would you

calculate the forces as that of the pull, weight of 1st box, and weight of 1st half

of rope? Like this

$$\sum\vec{F}=200N-W_1-\frac{1}{2}W_2$$ and

$$ma=(m_1+\frac{1}{2}m_2)a$$

7. Oct 11, 2004

### Staff: Mentor

I would apply Newton's 2nd law to the top box plus half the rope.
Almost. But you forgot the tension itself:
$$\sum\vec{F}=200N-W_1-\frac{1}{2}W_2 - T_m$$

Yes.