# Tension between two objects

1. Oct 17, 2007

### ~christina~

1. The problem statement, all variables and given/known data
2 blocks are connected by a rope of negligible mass are being dragged by a horizontal force. Suppose F= 68.0N, m1= 12.0kg, m2= 18.0kg and the coefficient of kinetic friction between each block annd the surface is 0.100.

a.) draw free body diagram for each block
b.) determine the tension T and the magnitude of the acceleration of the system.

2. Relevant equations
Ffk= uk * Fn (Ffk=force of kinetic friction) (uk= coefficient of friction) (Fn= normal force)

F-T= m2*a

T-f= m1a or is it T= m1*a (I changed the equation that I saw for a system with objects being pulled but they said it was T= m1*a however I'm thinking that this doesn't include friction into it since the one with the first box connected directly to the rope being pulled does take into account the tension force and it subtracts it from the force that pulls the box ...so I was thinking maybe they didn't include the friction since it would also be a opposing force to the tension..am I correct?)

a= F/m1 + m2

3. The attempt at a solution

a)for the free body diagram all I did was draw it as

b) this is where I have trouble...

since they give the coeffiecent if friction doesn't that equal to the applied force while the object is moving? ..I was wondering why they gave that but since i have the mass wouldn't it be equal to in the equation..

Ffk= uk* Fn

uk= 0.1
m1= 12.0kg
m2= 18.0kg

For m1
Fn=mg Fn= 12.0kg(9.8m/s^2)= 117.6N

plugging in..

Ffk= 0.1 (117.6N) = 11.76 N

Now I don't know what to do with that number...

well going along and doing the same for m2 gives me..

For m2
Fn= 18.0kg(9.8m/s^2)= 176.4N
plugging in

Ffk= 0.1(176.4N)= 17.64N

Once again I'm not sure however I"m thinking that this is the force that is applied but ...Not sure sinc they say it's 68N...well I was thinking that I plug into ..

F-T= m2*a for F and find T However I find that I need the acceleration and I get that from

a= F/m1 + m2

a= 68N/ (12.0kg + 18.0kg)= 2.27m/s^2

and I'm officially stuck...
_______________________________________________________________________

MY first attempt before I started to think about the coefficent of friction and that equation...was...based on that other example but that didn't include the coefficient of friction..

F-T= m2*a

T=m1*a (I still think it's supposed to include friction with T-f= m1*a so PLEASE TELL ME WHICH IS CORRECT)

a= F/(m1 + m2) ====> would it be the same if there was friction?? I would guess so...(68N total force applied)

m1= 12.0kg
m2= 18.0kg

F= 68.0N

u= 0.1

a= 68.0N/(12.0kg + 18.0kg)= 2.27m/s^2

T-f= m1*a =============> well I didn't have f so....using the other one which I stillt think doesn't include friction... but is the frictional force found if I use the coefficent of friction and Fn of the m1 box ?

T= m1*a

T= (12.0kg)*(2.27m/s^2)
T= 27.24N

same thing with box 2

F-T= m2*a
T= 27.14 N ------> why is this less than the other..?

Well...I would like you to answer the question as to which is correct and which equation is fine for tension..since I can't figure out that...'

Please..I really need to get this clarified...

Thank you

2. Oct 17, 2007

### l46kok

You forgot the frictional force on your free body diagram. Go ahead and change that first.

3. Oct 17, 2007

### ~christina~

Actually I did add it....for which box are you refering to...
m1 has friction on the other side... but are you saying that there would be friction for th first box as well plus the tension???

I need to figure this out by this morning unfortunately...and I notice your not online..

4. Oct 17, 2007

### Staff: Mentor

Yes, friction acts on both boxes.

Once you fix your free body diagram, just apply Newton's 2nd law to each mass.

Last edited: Oct 17, 2007
5. Oct 17, 2007

### ~christina~

Well I fixed it..but I still don't know about what do I do after I find the kinetic frictional force..does that only apply to the m1?? as the opposing force?

confusion stems from:
T is not same as frictional force (kinetic) right?

and Fnet that is pulling on the boxes causing them to move is not I know equal to the kinetic frictional force...

so is it the only place I place kinetic frictional force that I find in my equation that I created?

T-f= m1*a or is my equation incorrect?

( I thought about the frictional force opposing the tensional force pulling the boxes in the other direction while the tensional force of m2 first box opposes the Fnet)

Please tell me if I'm correct and if my equation that I kind of created is correct...

6. Oct 17, 2007

### Staff: Mentor

Since both boxes move across the floor, they each experience a friction force acting to the left (since they are moving to the right). The friction that they experience is $\mu m_1 g$ for mass 1 and $\mu m_2 g$ for mass 2. (I recommend that you don't plug in numbers until the last step, whenever possible.)

That's right. T is the tension force from the connecting rope.

For m1, this equation is correct, where f is the friction force. I'd write it this way:
$$\Sigma F_x = m_1 a$$
$$T - \mu m_1 g = m_1 a$$

Now write a similar equation for m2. The forces are friction, tension, and the applied force (call it F for now).

7. Oct 17, 2007

### ~christina~

For m2 ...hm...since it includes friction as well plus the tensin in string...wouldn't it be added together?

for m2
$$\sum Fx= m2*ax$$ ==> not sure if this is correct...

so then... since it's if I'm correct.. Tension + Frictional kinetic force..

$$\ F- (T + \mu m2g) = m2*a$$

well...is this ok?

8. Oct 17, 2007

### Staff: Mentor

Perfect! It might be easier to see how to combine those two equations if you write it like this:
$$F -T - \mu m_2 g = m_2 a$$

Now combine this equation with the one for m1 and you can solve for the two unknowns: T & a.

9. Oct 17, 2007

### ~christina~

yay..Okay..so

$$\sum F_x = m_1 a$$
$$T - \mu m_1 g = m_1 a$$

$$\sum F_x= m_2*ax$$
$$\ F- T - \mu m_2g) = m_2a$$

combining them I'd get..

Well I'm confused so I'll check with you first...
I wrote it out and got this..

$$F- \mu m_1g - \mu m_2g = m_1a + m_2a$$

10. Oct 17, 2007

### Staff: Mentor

Excellent! You added the two equations, which eliminated T. Now you can solve for "a". (Then plug it back in to one of the original equations to solve for T.)

11. Oct 17, 2007

### ~christina~

Alright...

$$F- \mu m_1g - \mu m_2g = m_1a + m_2a$$

so..

$$a= (F- \mu m_1g - \mu m_2g)/ m_1 + m_2$$

a= (68.0N)-(11.76)-(17.64) / 30.0kg

a= 1.287 m/s^2

so plugging into one of the other original equations
$$\Sigma F_x = m_1 a$$
$$T - \mu m_1 g = m_1 a$$

$$T= m_1a + \mu m_1g$$

T= (12.0kg)(1.287m/s^2) + 11.76 m/s^2
T= 27.204 N

And I think that's it..is it okay?
(just checking)

12. Oct 17, 2007

### Staff: Mentor

Magnifico!

13. Oct 17, 2007

### ~christina~

YAAAY!!

THANKS VERY MUCH Doc Al !!