# Acceleration for Masses on a slope/plane connected by a string

1. Oct 7, 2012

### AJKing

I think I've solved this, I'm looking for an answer check.

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

Given the picture above, find the acceleration of the block and the tension in the string. Assume the string preforms ideally and the pulley is frictionless.

2. Relevant equations

Fnet = ƩFx + ƩFy = ma

3. The attempt at a solution

Looking at M2, the horizontal forces are the tension in the string and the force of friction.
Ff = mg*0.47 = 10.6N

Looking at M1, the horizontal forces are the tension in the string in the opposite direction and the horizontal portion of the force pulling M1 down the hill.
F = -mg*sinθ = -18.7N
Fx = 18.7N*cosθ = - 16.66N

An ideal string transfers the forces equally, so we may ignore the string and consider the masses to be one system.
ƩFx = -16.66N+10.6N = -6.06N
ƩFy = -18.7N*sinθ = -8.49N
Fnet = √((-6.06)2+(-8.49)2) = -10.4N 27° below horizontal

-10.4N = ma = (m1+m2)a
a = -10.4N/(4.2+2.3)kg = -1.60m/s2 27° below horizontal

----

There are 2 forces on m1, the tension in the string and the force of gravity down the slope.
the sum of those forces must equal -10.4N
(-10.4 + 18.7)N = 8.3N = Ts

Is this all correct?

2. Oct 7, 2012

### howie8594

Sadly, I don't think you did this exactly right.

This equation is never true. You cannot add x and y components to get a net force.

Secondly, no y-components of force are necessary in this problem. No components at all. Just the forces by themselves.

The only reason you'd need components here is to find the force pulling the 4.2kg block down the slope. You did trig for that and came up with mgsinθ, which is correct. Once you found that though, you went further and found the x-component of that force, which you don't do.

It might sound confusing because the 4.2kg object is being pulled in a different direction than the 2.3kg object, but remember that a pulley simply changes the direction of forces. Therefore, you could think of this as both objects being pulled in the same direction. So no components are necessary.

Now you need to figure out the acceleration of the system before you do much else, so simply add the force of gravity parallel to the slope that you found with mgsinθ and the force of friction of the 2.3kg object. Once you do that, it's easy to come up with the acceleration of the two blocks, but don't forget to add the two masses together.

After that, then find the tension in the rope. You only need to consider the 2.3kg block when doing this. How much force does it take to move a 2.3kg object accelerating at x m/s/s? And that's your tension.

Once again, no components are involved here.

Hope this helps

3. Oct 8, 2012

### AJKing

Ah, that makes sense. I'll find more questions like this to check my understanding.
Thanks for the response

You most certainly can!
But I did write it lazily: this is what I meant and I'm certain it's true.
If I'm wrong I'd really appreciate the correction.

$\vec{F}_{net} = \Sigma\vec{F}_{x} + \Sigma\vec{F}_{y} = m\vec{a}$

4. Oct 8, 2012

### howie8594

I actually don't know what that means, but if it means you get an x component of force and a y component of force, then do pythagorean's theorum to get a resultant force, then that would be correct.

I was saying that if for instance, you add x components together and get 10 N, and then add y components and get 13 N, the resultant net force is not 23 N.

I'm sure that's what you meant.

5. Oct 8, 2012

### AJKing

Ok, we're both correct on that front then, I was a little shocked :P.
The arrows indicate vector addition which requires the pythagorean theorum in this case