More physics help

[Nate]

I am COMPLETELY baffled by this.

Two blocks made of different materials connected together by a thin cord, slide down a plane ramp inclined at an angle (theta) to the horizontal. Block 2 is above block 1. The masses of the blocks are equal (5.0 kg). The coefficients of friction are (myu)1 = 0.20 and (myu)2 = 0.30. Determine the acceleration of the blocks.

Now, I defined my coordinates so that the x direction is parallell to the surface. Since it isn't moving in the y direction, the net force there is zero, meaning normal force is equal to the mass of the block times gravity times the cosine of (theta).

What I know is that the acceleration of the two blocks will be equal. The masses are equal. The tension for the cord is the same. But without knowing the value of theta, how can I solve for acceleration? Without theta I can't get the normal force, and therefore I can't get the force of friction.

What am I supposed to do?

[Warrior4Christ]

Perhaps solve not for a numerical value, but solve the acceleration in terms of theta. Eg. a = a(theta) (acceleration is a function of theta)

[Nate]

Well, the problem is, the answer to this one (being an odd numbered problem) is in the back of the book, and the answer is 2.8 m/s^2

So apparently I can get a numerical value for acceleration, I just don't see how without knowing the value of theta.

[Nate]

Never mind. I'm an idiot. I failed to see the thing after part b that said, "for an angle theta = 30 degrees.

Though they should have put that in the BEGINNING of the problem, rather than after the second part, WHICH I HADN'T STARTED YET.

*stabs stupid textbook writers*

[Slater]

lol, well this problem is pretty straight forward now, simple algebra.

are you studying from Giancoli by any chance?

[Warrior4Christ]

Also, myu is spelt 'mu'.

[Nate]

are you studying from Giancoli by any chance?

Yes, yes I am. Physics For Scientists and Engineers (the college level course). I actually did this problem about ten times last night even after knowing the value of theta, and couldn't get the answer in the back (which was 2.8 m/s^2). I kept getting 2.1

I'm pretty sure there's something I'm missing, but I'm not sure what...so if someone could show me how to work out this problem, I'd appreciate it.

[Slater]

remember...
Force of Friction is the Normal Force multiplied by the coefficient of friction.

Also remember that forces are vector quantities.

[Nate]

Right, I'm aware of that.

Bleh, I wish there was a way to type out equations the way I write them, but I'll try and make do...all numbers given in the equations are supposed to be subscripts.

For block one, the force is mass times acceleration. This is equal to the forces acting on block one which is tension of the cord, and friction. Both tension and friction are going to oppose the movement of block one. Thus, the equation would be:

m1a = T + F(Fr1)

Right?

Since I defined my coordinate system, F(Fr), like you said, is going to be normal force times the friction coefficient. The normal force on block one would be, as I said before, mg * cos(theta)

This gives me a value of 8.49 N for the force of friction on block one.

Now, for block 2, it's the same. Tension is going to aid the movement of block 2, but friction will still oppose.

m2a = F(Fr2) - T

Right?

Normal force equation will be the same for block 2, with a different coefficient of friction.

Force of friction on block 2 is 12.73 N

So anyway, to find the acceleration...take the equation for block one, rearrange block 2's equation to find the value of tension, and you get:

m1a = F(Fr2) - m2a + F(Fr1)

Right?

So then once you get all your algebra done rearranging for a, the equation should be:

a = (F[Fr2] + F[Fr1])/(m1+m2)

Right?

But this gives me 2.1, and the answer in the back of the book says 2.8

Where is my error?

[SP1]

Doesn't gravity act on both blocks with acceleration = g * sin(theta) ? I don't see that you've accounted for that.

g = 9.8 m/s/s

In your solution for a, what you have actually done is solve for the friction force, not the acceleration vector. That is, you have an equivalent of 2.1 m/s/s of acceleration (up the ramp). Since sin(30) = 0.5, g*sin(30) = 4.9.

4.9 - 2.1 = 2.8 m/s/s. Or at least that's how I thing it should work. You can also visualize this problem as a single mass with two different friction coefficients. Since the upper mass has a higher coefficient, they will remain apart. This is like a rigid system, not two masses attached by a thread. If you look at your equation for a, its basically a weighted average of friction coefficients.