Determine whether the container will be delivered safely when the system is released from rest.

Kwa uzito wa $m_1 = 300 \, \text{kg}$ (sanduku kwenye usawa):
Nguvu ya kawaida $N_1 = m_1 g$.
Nguvu ya msuguano $f_1 = \mu_1 N_1 = \mu_1 m_1 g$.
Mlinganyo wa mwendo (kwa mwelekeo wa usawa):

$$T - f_1 = m_1 a$$ $$T - \mu_1 m_1 g = m_1 a \quad (1)$$

Kwa uzito wa $m_2 = 500 \, \text{kg}$ (kontena kwenye mteremko):
Nguvu ya kawaida $N_2 = m_2 g \cos\theta$.
Nguvu ya msuguano $f_2 = \mu_2 N_2 = \mu_2 m_2 g \cos\theta$.
Mlinganyo wa mwendo (kwa mwelekeo wa mteremko, ukichukulia inashuka):

$$m_2 g \sin\theta - T - f_2 = m_2 a$$ $$m_2 g \sin\theta - T - \mu_2 m_2 g \cos\theta = m_2 a \quad (2)$$

Ongeza mlinganyo (1) na (2) ili kuondoa $T$:

$$(T - \mu_1 m_1 g) + (m_2 g \sin\theta - T - \mu_2 m_2 g \cos\theta) = m_1 a + m_2 a$$ $$m_2 g \sin\theta - \mu_1 m_1 g - \mu_2 m_2 g \cos\theta = (m_1 + m_2)a$$ $$a = \frac{g(m_2 \sin\theta - \mu_1 m_1 - \mu_2 m_2 \cos\theta)}{m_1 + m_2}$$

Weka namba:
$\sin 30^\circ = 0.5$
$\cos 30^\circ \approx 0.8660$

$$a = \frac{9.81 \, \text{m/s}^2 (500 \, \text{kg} \times 0.5 - 0.3 \times 300 \, \text{kg} - 0.1 \times 500 \, \text{kg} \times 0.8660)}{300 \, \text{kg} + 500 \, \text{kg}}$$ $$a = \frac{9.81 (250 - 90 - 43.30)}{800}$$ $$a = \frac{9.81 \times 116.7}{800}$$ $$a = \frac{1144.707}{800}$$ $$a \approx 1.4309 \, \text{m/s}^2$$

Kasi ya kontena baada ya sekunde 3 (kutoka mapumziko, $u=0$):

$$v = u + at$$ $$v = 0 + (1.4309 \, \text{m/s}^2) \times (3 \, \text{s})$$ $$v = 4.2927 \, \text{m/s}$$

Kasi iliyohesabiwa ($4.2927 \, \text{m/s}$) ni chini ya kasi ya juu inayoruhusiwa ($5 \, \text{m/s}$). Kwa hivyo, sharti hili limetimizwa.

Tumia mlinganyo (1):

$$T = m_1 a + \mu_1 m_1 g$$ $$T = (300 \, \text{kg} \times 1.4309 \, \text{m/s}^2) + (0.3 \times 300 \, \text{kg} \times 9.81 \, \text{m/s}^2)$$ $$T = 429.27 \, \text{N} + 882.9 \, \text{N}$$ $$T = 1312.17 \, \text{N}$$

Nguvu ya mmenyuko kwenye puli ($R$) ni matokeo ya nguvu mbili za mvutano $T$ zinazofanya kazi kwenye puli. Nguvu moja inafanya kazi kwa usawa kushoto, na nyingine inafanya kazi chini ya mteremko kwa pembe ya $30^\circ$ chini ya usawa. Pembe kati ya nguvu hizi mbili ni $180^\circ - 30^\circ = 150^\circ$.

$$R = 2T \cos\left(\frac{\phi}{2}\right)$$

Ambapo $\phi$ ni pembe kati ya kamba mbili zinazovuta puli. Hapa $\phi = 150^\circ$.

$$R = 2 \times 1312.17 \, \text{N} \times \cos\left(\frac{150^\circ}{2}\right)$$ $$R = 2 \times 1312.17 \, \text{N} \times \cos(75^\circ)$$

$\cos(75^\circ) \approx 0.2588$

$$R = 2 \times 1312.17 \times 0.2588$$ $$R \approx 679.50 \, \text{N}$$

Nguvu ya mmenyuko iliyohesabiwa ($679.50 \, \text{N}$) ni chini ya nguvu ya juu inayoruhusiwa ($700 \, \text{N}$). Kwa hivyo, sharti hili limetimizwa.

Hitimisho:
Kwa kuwa masharti yote mawili ya usalama yametimizwa (kasi ya kontena baada ya sekunde 3 ni $4.2927 \, \text{m/s}$ ambayo ni chini ya $5 \, \text{m/s}$, na nguvu ya mmenyuko kwenye puli ni $679.50 \, \text{N}$ ambayo ni chini ya $700 \, \text{N}$), kontena litasafirishwa kwa usalama.

The container will be delivered safely.