Question

A 15 kg block rests on the surface of a plane inclined at an of 30 degrees to the horizontal. A light inextensible string
passing over a small, smooth pulley at the top of the plane connects the block to another 13 kg block hanging freely. The
coefficient of kinetic friction between the 15 kg block and the plane is 0.25. Find the acceleration of the blocks

Answer 1

Answer on Question #51064, Physics, Mechanics Kinematics Dynamics

A $15\mathrm{kg}$ block rests on the surface of a plane inclined at an of 30 degrees to the horizontal. A light inextensible string passing over a small, smooth pulley at the top of the plane connects the block to another $13\mathrm{kg}$ block hanging freely. The coefficient of kinetic friction between the $15\mathrm{kg}$ block and the plane is 0.25. Find the acceleration of the blocks.

Solution:

Fig.1

According to Newton's second law for the first solid

$$\vec {a} m _ {1} = \vec {T} + \vec {N} + \vec {m} _ {1} g + \vec {F} _ {f}$$

where $m_{1} = 15kg$ ; $g = 9.8m / s^{2}$ is the gravitational acceleration; $T$ is the tensile force; $F_{f} = N\mu$ friction force.

Than

$$\left\{ \begin{array}{l} O X: a m _ {1} = T - F _ {f} - m _ {1} g \sin \alpha \\ O Y: N = m _ {1} g \cos \alpha \end{array} \right.$$

From Eq. (2)

$$a m _ {1} = T - m _ {1} g (\mu \cos \alpha + \sin \alpha)$$

According to Newton's second law for the second solid

$$\vec {a} m _ {2} = \vec {m} _ {2} g + \vec {T}$$

where $m_{2} = 13kg$

Than

$$a m _ {2} = m _ {2} g - T$$

From Eq.(1) and Eq.(2)

$$a = g \cdot \frac {m _ {2} - m _ {1} (\mu \cos \alpha + \sin \alpha)}{(m _ {1} + m _ {2})}$$

So,

$$a = 9.8 \cdot \frac {13 - 15 \left(0.25 \cos 30 ^ {0} + \sin 30 ^ {0}\right)}{(13 + 15)} = 0.79 m / s ^ {2}$$

Answer: $a = g \cdot \frac{m_2 - m_1(\mu\cos\alpha + \sin\alpha)}{(m_1 + m_2)} = 0.79m / s^2$

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