Questions: Po hrapavem klancu strmine 30° drsi večja klada mase 100 kg. Koeficient trenja med podlago in klancem je 0.2. Vrvica preko škripca povezuje večje telo z manjšim. (Upoštevaj, da sta dva sistema. Zato zapiši sistem enačb za obe telesi) a) Pravilno nariši sile na posamezen sistem (Ločiti moraš zunanje sile in komponente!). b) Kolikšno maso mora imeti viseče telo, da se obe telesi gibljeta s pospeškom 2 m / s^2?

Po hrapavem klancu strmine 30° drsi večja klada mase 100 kg. Koeficient trenja med podlago in klancem je 0.2. Vrvica preko škripca povezuje večje telo z manjšim. (Upoštevaj, da sta dva sistema. Zato zapiši sistem enačb za obe telesi)
a) Pravilno nariši sile na posamezen sistem (Ločiti moraš zunanje sile in komponente!).
b) Kolikšno maso mora imeti viseče telo, da se obe telesi gibljeta s pospeškom 2 m / s^2?

Transcript text: Po hrapavem klancu strmine $30^{\circ}$ drsi večja klada mase 100 kg . Koeficient trenja med podlago in klancem je 0.2. Vrvica preko škripca povezuje večje telo z manjšim . (Upoštevaj, da sta dva sistema. Zato zapiši sistem enačb za obe telesi) a) Pravilno nariši sile na posamezen sistem (Ločiti moraš zunanje sile in komponente!). (1T) b) Kolikšno maso mora imeti viseče telo, da se obe telesi gibljeta s pospeškom $2 \mathrm{~m} / \mathrm{s} 2 ?(1 \mathrm{~T})

Solution

Solution Steps

Step 1: Identify the forces acting on the system

The larger block (mass $ m_1 = 100 $ kg) is on an inclined plane with an angle of $ 30^\circ $.
The coefficient of friction between the block and the plane is $ \mu = 0.2 $.
The smaller block (mass $ m_2 $) is hanging and connected to the larger block via a pulley.

Step 2: Draw the free-body diagrams

For the larger block on the incline:
- Gravitational force: $ m_1 g $
- Normal force: $ N $
- Frictional force: $ f = \mu N $
- Tension in the rope: $ T $
- Component of gravitational force parallel to the incline: $ m_1 g \sin(30^\circ) $
- Component of gravitational force perpendicular to the incline: $ m_1 g \cos(30^\circ) $
For the smaller block:
- Gravitational force: $ m_2 g $
- Tension in the rope: $ T $

Step 3: Write the equations of motion

For the larger block on the incline: \[ m_1 a = m_1 g \sin(30^\circ) - T - f \] where $ f = \mu m_1 g \cos(30^\circ) $.
For the smaller block: \[ m_2 a = T - m_2 g \]

Step 4: Solve for the unknowns

Given acceleration $ a = 2 $ m/s$^2$, solve for $ T $ and $ m_2 $.

Calculate the frictional force: \[ f = \mu m_1 g \cos(30^\circ) = 0.2 \times 100 \times 9.8 \times \cos(30^\circ) = 169.7 \text{ N} \]
Substitute into the equation for the larger block: \[ 100 \times 2 = 100 \times 9.8 \times \sin(30^\circ) - T - 169.7 \] \[ 200 = 490 - T - 169.7 \] \[ T = 490 - 169.7 - 200 = 120.3 \text{ N} \]
Substitute $ T $ into the equation for the smaller block: \[ m_2 \times 2 = 120.3 - m_2 \times 9.8 \] \[ 2m_2 + 9.8m_2 = 120.3 \] \[ 11.8m_2 = 120.3 \] \[ m_2 = \frac{120.3}{11.8} \approx 10.2 \text{ kg} \]