# Two Masses and the Tension in a rod

1. Jun 19, 2012

### Rohaaan

1. The problem statement, all variables and given/known data

Two masses, m1 = 1.65kg and m2 = 3.30kg, attached by a massless rod parallel to the inclined plane on which they both slide, travel along the plane with m1 trailing m2. The angle of incline is 30°. The coefficient of kinetic friction between m1 and the incline is μ1 = 0.226; that between m2 and the incline is μ2 = 0.113.

What is the tension in the rod?

2. Relevant equations

Ff = μk × FN

FN = mgcosθ

3. The attempt at a solution

So, for m1:

Ff1 = μ1 × FN
Ff1 = 0.226 × mgcos30
Ff1 = 0.226 × 14.29
Ff1 = 3.23N

Force Down Incline = mgsin30
Force Down Incline = 8.25N

Therefore, Tension 1 = Force Down Incline - Ff1 = 5.02N

And for m2:

Ff1 = μ1 × FN
Ff1 = 0.113 × mgcos30
Ff1 = 0.226 × 28.58
Ff1 = 3.23N

Force Down Incline = mgsin30
Force Down Incline = 16.5N

Therefore, Tension 2 = Force Down Incline - Ff2 = 13.27N

Finally, Total Tension = Tension 2 - Tension 1 = 8.25N

There are no answers for this question, so I came on here hoping that someone may be able to correct me if I have done something wrong, or possibly verify my answer.

Any help greatly appreciated!

Thanks :)

2. Jun 19, 2012

### azizlwl

What is the tension in the rod?
Tension 1= Tensiion 2

3. Jun 19, 2012

### Rohaaan

What do you mean sorry?

4. Jun 19, 2012

### azizlwl

If you hang an object of mass m by a massless rod , what is the tension on the rod?

5. Jun 20, 2012

### Rohaaan

That is my question haha.

6. Jun 20, 2012

### azizlwl

The upper mass will pull up the lower mass.
The reaction of this the lower will pull the upper mass too.
So action and reaction are the same tension on the rod.
Use fbd for the lower mass and the upper mass to find the tension where tension pull are the same, only different direction.