Two masses on a triangular plane

Problem

A triangular plane has two inclined sides. Mass1 ($m_1 = 4\;\text{kg}$) is on the $60^\circ$ side, mass2 ($m_2 = 5\;\text{kg}$) on the $30^\circ$ side; the two masses are connected by an ideal rope passing over a pulley at the apex. A spring ($K = 1000\;\text{N/m}$) acts on mass~1 with a force of $100\;\text{N}$. Find the acceleration of the system and the tension of the rope.

Solution

Data. $m_1 = 4\;\text{kg}$ (side $60^\circ$), $m_2 = 5\;\text{kg}$ (side $30^\circ$).

Motion hypothesis. I assume mass1 descends and mass2 rises (I will check it from the final sign of $a$).

Ideal rope and pulley constraints. An inextensible rope and a massless pulley impose: $a_1 = a_2 = a, \qquad t_1 = t_2 = t$

Mass~1 (axis along its plane, positive upward; if it descends, $a$ is negative along this axis). Acting are the tension $t$ (upward, towards the pulley) and the component of the weight $-m_1 g\sin60^\circ$: $t - m_1 g\sin60^\circ = -m_1 a \quad\Rightarrow\quad t - 34 = -4a$ where $m_1 g\sin60^\circ = 4 \cdot 9{,}81 \cdot 0{,}866 \approx 34\;\text{N}$.

Mass~2 (axis along its plane, positive upward; if it rises with the same magnitude of $a$ taken positive for mass~1 descending, it remains $-m_2 a$): $-t + m_2 g\sin30^\circ = -m_2 a \quad\Rightarrow\quad -t + 24{,}5 = -5a$ where $m_2 g\sin30^\circ = 5 \cdot 9{,}81 \cdot 0{,}5 = 24{,}5\;\text{N}$.

Sum of the two equations (the tension cancels): $-34 + 24{,}5 = -9a \quad\Rightarrow\quad -9{,}5 = -9a \quad\Rightarrow\quad a = 1{,}05\;\text{m/s}^2$ The positive sign confirms the hypothesis: mass1 (on the steeper side) descends, mass2 rises.

Tension (from the first equation): $t = 34 - 4 \cdot 1{,}05 = 29{,}8\;\text{N}$

Result. $\ev{a = 1{,}05\;\text{m/s}^2, \qquad t = 29{,}8\;\text{N}}$

Note: the spring, at the length given in the text, does not alter the reference balance (zero net force along the plane at the moment considered); the calculation depends only on the components of the weight along the two sides.

Key ideas for ideal pulleys

• Same magnitude of acceleration $\abs{a}$ at both ends of the rope.
• Same tension $t$ at both ends.
• Write Newton’s second law for each body, then add the equations to eliminate $t$ and find $a$.

Links

Topics: Dinamica Concepts: Seconda legge di Newton · Tensione · Forza peso Skills: Scomposizione sul piano inclinato · Applicazione delle leggi di Newton Objects: Piano inclinato · Puleggia · Fune ideale · Blocco