Incline with pulley and hanging mass

Problem

A block of mass $m_1 = 3{,}0\;\mathrm{kg}$ is on a frictionless incline of $30^\circ$ and is connected, via an ideal rope passing over an ideal pulley placed at the top, to a block $m_2 = 2{,}0\;\mathrm{kg}$ hanging vertically. Determine the direction and magnitude of the acceleration of the system and the tension in the rope.

Solution

Free-body diagram. On $m_1$ (on the frictionless incline) act: weight, normal reaction, and tension $T$ directed along the incline towards the pulley. The component of the weight along the incline is $m_1 g\sin\theta$. On $m_2$ (hanging) act: weight $m_2 g$ downwards and tension $T$ upwards.

Direction of motion. Compare the two “driving” forces: $m_2 g = 2{,}0\cdot 9{,}8 = 19{,}6\;\mathrm{N}\qquad m_1 g\sin\theta = 3{,}0\cdot 9{,}8\cdot 0{,}50 = 14{,}7\;\mathrm{N}$ Since $m_2 g > m_1 g\sin\theta$, the hanging block descends and $m_1$ rises up the incline.

Newton’s second law (positive axis in the direction of motion, same magnitude $a$ for both blocks since the rope is inextensible): $\begin{aligned} m_2 g - T &= m_2 a \\ T - m_1 g\sin\theta &= m_1 a \end{aligned}$

Adding the two equations eliminates $T$: $a = \frac{m_2 g - m_1 g\sin\theta}{m_1 + m_2} = \frac{19{,}6 - 14{,}7}{5{,}0} = \frac{4{,}9}{5{,}0}\;\mathrm{m/s^2}$

$\ev{a \approx 0{,}98\;\mathrm{m/s^2}\ \text{(m}_2\text{ descends, m}_1\text{ rises)}}$

Tension. From the first equation: $T = m_2(g - a) = 2{,}0\cdot(9{,}8 - 0{,}98) = 2{,}0\cdot 8{,}82\;\mathrm{N}$

$\ev{T \approx 17{,}6\;\mathrm{N}}$

Consistency check: $m_1 g\sin\theta + m_1 a = 14{,}7 + 3{,}0\cdot 0{,}98 = 17{,}6\;\mathrm{N}$, equal to $T$. The tension lies between $m_1 g\sin\theta$ and $m_2 g$, as expected for an accelerating system.

Links

Topics: Dynamics Concepts: Newton’s second law · Tension · Weight force Skills: Resolution on an incline · Applying Newton’s laws · Free-body diagram Methods: Free-body diagram method Objects: Inclined plane · Block · Pulley · Ideal rope