# Blocks, incline, tension and acceleration

1. Dec 12, 2009

### noob314

1. The problem statement, all variables and given/known data
A large block of mass M = 6kg is on an incline which is at an angle of 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the incline is 0.3. The block is attached to a string which runs over a pulley and is connected to a smaller block of mass m = 2.4 kg.

If the block is sliding up the incline, find a) the acceleration (magnitude and direction) of the smaller block m and b) the tension in the string.

3. The attempt at a solution
I need someone to confirm if what I'm doing is correct.

a)
$$\Sigma \vec{F}_{M} = Ma = T - f_{k} - Mgsin\theta$$
$$T = M(gsin\theta + a + \mu_{k}gcos\theta)$$
$$\Sigma\vec{F}_{m} = ma = mg - T$$
$$= mg - M(gsin\theta + a + \mu_{k}gcos\theta)$$
$$= ma + Ma = mg - Mgsin\theta - \mu_{k}Mgcos\theta$$
$$= a(m+M) = mg - Mgsin\theta - \mu_{k}Mgcos\theta$$
$$= a = \frac{mg - Mgsin\theta - \mu_{k}Mgcos\theta}{m+M}$$
$$= a = -2.52\frac{m}{s^{2}}$$
I'm guessing this is the acceleration in the y-direction, and the acceleration in the x-direction would be 0, so the magnitude is 2.52
$$= 2.52\frac{m}{s^{2}}$$ down

b)
$$\Sigma\vec{F}_{m} = mg - T = ma$$
$$T = mg - ma$$
$$= (2.4kg)(9.80 \frac{m}{s^{2}}) - (2.4kg)(2.52\frac{m}{s^{2}})$$
$$= 17.47N$$

Which looks about right, since in order for m to be moving down, mg needs to be greater than T, which in this case it is.

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2. Dec 12, 2009

### PhanthomJay

I haven't checked your equations, but you should first check the wording of the problem. The block can't be sliding up the plane, the hanging mass is too small to haul it up. It can't be sliding down the plane, either. Did you copy the problem down correctly?

3. Dec 12, 2009

### noob314

Yes, the problem is copied down correctly.

4. Dec 12, 2009

### PhanthomJay

Then someone made a boo-boo. The problem makes no sense as written.

5. Dec 12, 2009

### noob314

Is it really impossible? I assumed some unknown force allowed it to overcome whatever static friction there may have been which would have allowed mass m to overcome the kinetic friction.

6. Dec 12, 2009

### PhanthomJay

What unknown force?

7. Dec 12, 2009

### noob314

I assume it would be a force that's barely able to overcome the static friction and then disappear after that.

8. Dec 12, 2009

### PhanthomJay

The more you try to make sense of this problem, the more it will confuse you. Just rip it up and try another problem.

9. Dec 12, 2009

### noob314

Yeah, you're right. I rechecked my calculations and the numbers didn't add up. I was just curious because it was on one of my old exams.