Question

An ice tube is kept on an inclined plane of angle 30degree . Coefficient of kinetic friction between block and inclined plane is 1 upon underroot 3 . What is the acceleration of block?

Answer 1

Answer on Question #72118, Physics / Mechanics | Relativity

Question:

An ice tube is kept on an inclined plane of angle 30 degree. Coefficient of kinetic friction between block and inclined plane is 1 upon under root 3. What is the acceleration of block?

Solution:

According to $2^{\text{nd}}$ Newton's law: $m \vec{a} = m \vec{g} + \overrightarrow{F_{friction}} + \vec{N}$

In projections:

$$ma = mg * \sin(30{}^\circ) - F_{friction}$$ $$0 = N - mg \cos(30{}^\circ)$$

Also we use that: $F_{friction} = \mu N$ ($\mu = \frac{1}{\sqrt{3}}$)

So, $a = g \sin(30{}^\circ) - \mu g \cos(30{}^\circ) = 9.8 \left(0.5 - \frac{\sqrt{3}}{2} \frac{1}{\sqrt{3}}\right) = 0$

So, the acceleration of the ice tube is zero

Answer: 0

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