Question

A 3-kg on an inclined plane of angle 30 degrees is connected by a cord over a frictionless pulley to a second load which has a mass of 2 kg. What is the acceleration of each load? What is the tension on the cord?

Answer 1

$$m _ {1} = 3 k g$$ $$m _ {2} = 2 k g$$ $$m _ {1} \ddot {a} = \ddot {T} + m _ {1} \ddot {g}$$ $$m _ {2} \ddot {a} = \ddot {T} + m _ {2} \ddot {g}$$

where:

$T$ - tension on the cord

$a$ - acceleration

In projection on $X$ :

$-m_{1}a = T + (-m_{1}g\sin (\alpha))$

$-m_{2}a = -T + (-m_{2}g\sin (\alpha))$

Add the first and th second one string:

$-m_{1}a - m_{2}a = g\sin (\alpha)((-m_{1} - m_{2})$

$a = g\sin (\alpha)$

$$a = 1 0 \cdot 0. 5 = 5 \frac {m}{s ^ {2}}$$ $$T = - m _ {1} a + m _ {1} g \sin (\alpha)$$ $$T = 0 N$$

ANSWER:

$$T = 0 N; a = 5 \frac {m}{s ^ {2}}$$