A plane is connected to a 4 kg mass by a light inextensible string passing over a smooth pulley at the top of the plane (Figure 2.12).

Kwa masi $m_1$ (kwenye ndege iliyoinama):
Nguvu inayovuta $m_1$ chini ya ndege ni sehemu ya uzito wake: $F_{1, \text{parallel}} = m_1 g \sin\theta$.
Nguvu ya mvutano kwenye kamba ni $T$.

Kwa masi $m_2$ (inayoning'inia):
Nguvu inayovuta $m_2$ chini ni uzito wake: $F_{2, \text{weight}} = m_2 g$.
Nguvu ya mvutano kwenye kamba ni $T$.

Linganisha nguvu zinazowezekana za kusababisha mwendo:

• Nguvu inayovuta $m_1$ chini ya ndege: $m_1 g \sin\theta = 10 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(30^\circ) = 10 \times 9.8 \times 0.5 = 49 \text{ N}$.
• Nguvu inayovuta $m_2$ chini: $m_2 g = 4 \text{ kg} \times 9.8 \text{ m/s}^2 = 39.2 \text{ N}$. Kwa kuwa $m_1 g \sin\theta > m_2 g$, mfumo utaenda kwa njia ambayo $m_1$ inashuka chini ya ndege na $m_2$ inapanda juu.

Mlinganyo wa mwendo kwa $m_1$ (kuchukua chini ya ndege kama chanya):

$$m_1 g \sin\theta - T = m_1 a \quad (1)$$

Mlinganyo wa mwendo kwa $m_2$ (kuchukua juu kama chanya):

$$T - m_2 g = m_2 a \quad (2)$$

Jumlisha Mlinganyo (1) na Mlinganyo (2) ili kuondoa $T$:

$$(m_1 g \sin\theta - T) + (T - m_2 g) = m_1 a + m_2 a$$ $$m_1 g \sin\theta - m_2 g = (m_1 + m_2) a$$

Toa $g$ nje na utatue kwa $a$:

$$a = \frac{g (m_1 \sin\theta - m_2)}{m_1 + m_2}$$

Badilisha thamani zilizotolewa:

$$a = \frac{9.8 \text{ m/s}^2 (10 \text{ kg} \times \sin(30^\circ) - 4 \text{ kg})}{10 \text{ kg} + 4 \text{ kg}}$$ $$a = \frac{9.8 (10 \times 0.5 - 4)}{14}$$ $$a = \frac{9.8 (5 - 4)}{14}$$ $$a = \frac{9.8 \times 1}{14}$$ $$a = 0.7 \text{ m/s}^2$$

Kuongeza kasi ya mfumo ni: $\boxed{\mathbf{0.7 \text{ m/s}^2}}$

Tumia Mlinganyo (2):

$$T - m_2 g = m_2 a$$ $$T = m_2 a + m_2 g$$ $$T = m_2 (a + g)$$

Badilisha thamani za $m_2$, $a$, na $g$:

$$T = 4 \text{ kg} (0.7 \text{ m/s}^2 + 9.8 \text{ m/s}^2)$$ $$T = 4 \text{ kg} (10.5 \text{ m/s}^2)$$ $$T = 42 \text{ N}$$

Mvutano kwenye kamba ni: $\boxed{\mathbf{42 \text{ N}}}$