# Desperate for a help with tension in rope

1. Oct 6, 2004

### pinky2468

Desperate for a help with tension in rope!!

Two blocks are connected by a cable via a pulley which are massless and frictionless.
Given the following: m1 = 300.0 kg, m2 = 100.0 kg, theta=40degrees, uk=.350 ,what is the tension in the cable?

So have figured out the acceleration to be .304m/s the acceleration is going down the plane, but I don't know which equation to use for T:

T-m2g=m2a OR m1gsin40+T-Fk=m1a

I think it is the 1st one, but that doesn't take into account the friction?

Anyhelp? This is due in 2 hours!!!
Thanks!

2. Oct 6, 2004

### arildno

What's "theta"?
Is it the inclination of the plane, so that one of the blocks slides down it, whereas the other block is hangs from the rope and descends vertically?
(I take it uk is the kinematic coefficient of friction?)

3. Oct 6, 2004

### pinky2468

Yes the 300kg is on the plane and the 100kg is hanging down. "theta" is the angle, I don't know how else to write that and uk is the coefficient of friction.

4. Oct 6, 2004

### arildno

1) You cannot POSSIBLY have solved for the acceleration of the system correctly
2) You know that the frictional force is proportional to the NORMAL force, right?
Since the block on the plane doesn't fall through the plane, you should be able to figure out that the normal force from the plane on the block is $$m_{1}g\cos\theta$$
where $$m_{1}$$ is the mass of the block on the plane.
3) You must use BOTH equations to solve for a and T;
$$m_{2}g-T=m_{2}a$$

5. Oct 6, 2004

### pinky2468

We started out with two unknowns
m1gsin40-Fk-T=m1a and
T-m2g=m2a a
And worked thru them to get m1g(sin40-ukcos40)-m2g/m1+m2=a I have already worked thru the first part and I am pretty sure it is right, but I just can't figure out how to plug it back into T

6. Oct 6, 2004

### arildno

Your signs on the forces are wrong!
1) $$m_{1}g\sin(40)-F_{k}-T=m_{1}a$$
The tension T works in the same direction as gravity does:
$$m_{1}g\sin(40)-F_{k}+T=m_{1}a$$
This is correct; I thought this was the one you used.

Please note that the tension in this case is in the same direction as the acceleration.
2)You use:
$$T-m_{2}g=m_{2}a$$
This is wrong; for the falling block, the tension is in the OPPOSITE direction of the acceleration.
Hence, you must use EITHER:
$$T-m_{2}g=-m_{2}a$$
OR, equivalently:
$$-T+m_{2}g=m_{2}a$$
3) To solve for the tension once you've solved for the acceleration; use either one of your equation

7. Oct 6, 2004

### pinky2468

Ok, I see where you are with that, but the M1 is 300kg on the plane and is accelerating down the plane and M2 is 100kg so its direction is positive right? It is being pulled up? If what you are saying is right is my acceleration wrong too?

8. Oct 6, 2004

### arildno

What I'm saying is this:

We know that the system is dragged "downwards".
For the block on the plane, that means its pulled towards the pulley.
We choose to to signify that direction as "positive"
This means that the direction of tension and gravity components must be labeled positive, whereas "friction" is negative.
Similarly, we want "a" to be a positive number (how fast the block accelerates towards the pulley)
Let's now consider the falling block.
"a" is a positive number, but the block is falling downwards!!
This is the same direction as gravity, and opposite of tension.
Hence, the importance of the sign.

Here's a simple argument which shows why my sign convention is a correct one:
We may rewrite my equation as:
$$T=m_{2}(g-a)$$
If the rope is cut, a=g, which is CONSISTENT with the fact that T=0 in that case.
Your sign would say $$T=2m_{2}g$$ when the rope is cut, which is nonsense.

9. Oct 6, 2004

### pinky2468

I see what you are saying. Do I need to start over or do I use the acceleration of .304m/s that I found before in T-m2g=-m2a. This is due in about 45 min!!

10. Oct 6, 2004

### arildno

Since, you've been very good at posting your own thoughts, I'll give you bonus help ()
$$T+m_{1}g\sin\theta-\mu_{k}m_{1}g\cos\theta=m_{1}a$$
$$m_{2}g-T=m_{2}a$$
$$g(m_{2}+m_{1}(\sin\theta-\mu_{k}\cos\theta))=(m_{1}+m_{2})a$$
Or:
$$a=\frac{g(m_{2}+m_{1}(\sin\theta-\mu_{k}\cos\theta))}{m_{1}+m_{2}}$$
Which yields:
$$T=m_{2}g(1-\frac{(m_{2}+m_{1}(\sin\theta-\mu_{k}\cos\theta))}{m_{1}+m_{2}})$$

11. Oct 6, 2004

### pinky2468

Thank you so much for your help, one last question... is that assuming it is going up the incline or down. That is the question we have to answer before we even write the equation. Thanks again!

12. Oct 6, 2004

### arildno

Assuming it goes down the incline.

IMPORTANT:
I see now that you may have posted a bit too little information of the actual situation.
I have thought that the pulley has been placed at the downside of the block on the incline.
I've just realized that you might have had a problem in which the pulley was placed on the upside of the block on the plane. This would make the problem different; your original signs might have been correct.

(I apologize if this was your problem, but I cannot answer correctly if I've been given insufficient information)