# Sliding blocks on an inclined plane

1. Jan 19, 2010

### Fanta

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

Two blocks of mass m1 = 1kg and m2 = 2kg, connected through a rope, slide across an inclined plane with inclination of

$$\theta = \frac{\pi}{6}$$

The friction coefficient equals 0.5

3.1 Calculate the total work done by the friction force that acts on the first block after it has moved 1m along the inclined plane.

3.2 Determine the accelaration of the system and the rope tension.

2. Relevant equations

$$\theta = \frac{\pi}{6}$$

$$\mu = 0.5$$

$$F_{f} = \m \cdot u \cdot N$$

$$F = m \cdot a$$

g = 9.81

3. The attempt at a solution

3.1 -

To calculate the friction force in the first block, i think we dont have to consider the second block, so:

$$F_{f} = \mu \cdot N$$

$$N = m_{1} \cdot g \cdot cos \theta$$

$$F_{f} = \mu \cdot m_{1} \cdot g \cdot cos \theta$$

then we multiply that for the displacement (1m), and we get the work.

Is this correct?

3.2 Determine the acceleration of the system and the rope tension.

First, to determine the acceleration, I considered the whole system as a whole. So, consideing only external forces, I sum the masses 1 and 2, giving me M -> mass of the whole system.

$$Fr = F_{gx} - F{f}$$

$$M \times a = (M g sin \theta) - (M g cos \theta)$$

$$a = \frac{(M g sin \theta) - (M g cos \theta)}{M}$$

now, for the tension:

$$F{r1} = (m{1} g sin \theta) - (m{1} g cos \theta)$$

$$F{r2} = (m{2} g sin \theta) - (m{2} g cos \theta)$$

$$Tension = F{r2} - F{r1}$$

Is everything right?

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