Question

The blocks a and b are connected by a piece of string. block b rests on an inclined plane of 40 and block a hangs vertically. The coefficient of friction between block b and the inclined plane is 0.29. Calculate the acceleration of the system if the mass of block a is 0.15 ks and that of block b is 7.5 kg

Answer 1

Answer on Question #63936-Physics-Classical Mechanics

The blocks a and b are connected by a piece of string. Block b rests on an inclined plane of 40 and block a hangs vertically. The coefficient of friction between block b and the inclined plane is 0.29. Calculate the acceleration of the system if the mass of block a is 0.15 kg and that of block b is 7.5 kg

Solution

$$m _ {b} a = m _ {b} g - T$$ $$m _ {a} a = T - m _ {a} g \sin 4 0 {}^ {\circ} - F _ {f r}$$ $$F _ {f r} = \mu N = \mu m _ {a} g \cos 4 0 {}^ {\circ}$$

So,

$$m _ {b} a = m _ {b} g - T$$ $$m _ {a} a = T - m _ {a} g \sin 4 0 {}^ {\circ} - \mu m _ {a} g \cos 4 0 {}^ {\circ}$$

Add these tow equations:

$$(m _ {a} + m _ {b}) a = m _ {b} g + m _ {a} g \sin 4 0 {}^ {\circ} - \mu m _ {a} g \cos 4 0 {}^ {\circ}$$ $$a = \frac {m _ {b} - m _ {a} \sin 4 0 {}^ {\circ} - \mu m _ {a} \cos 4 0 {}^ {\circ}}{(m _ {a} + m _ {b})} g = \frac {7 . 5 - 0 . 1 5 \sin 4 0 {}^ {\circ} - 0 . 2 9 \cdot 0 . 1 5 \cos 4 0 {}^ {\circ}}{(0 . 1 5 + 7 . 5)} 9. 8 = 9. 4 \frac {m}{s ^ {2}}.$$

Answer: 9.4 $\frac{m}{s^2}$.

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