Question

In the diagram shown (Fig. (1)), $m_1 = 1$ kg, $m_2 = 1$ kg and coefficient of friction, both static and dynamic, between $m_1$ and plane is $\mu = 0.6$. The two masses are connected by a light inextensible string passing over a light frictionless pulley. Take $g = 10$ m$\cdot$s$^{-2}$.

(a) Find the acceleration of the system.

(b) Find the force of friction and the magnitude of the tension in the string.

Answer 1

(a) The acceleration of the system is 0 m/s$^2$. (b) The force of friction is 5 N (acting down the incline), and the magnitude of the tension in the string is 10 N.

Answer 2

1. Calculate the component of gravity down the incline for $m_1$ ($W_{1\parallel} = 5$ N) and the weight of $m_2$ ($W_2 = 10$ N).
2. Calculate the normal force on $m_1$ ($N = 5\sqrt{3}$ N) and the maximum static friction ($f_{s,max} = 3\sqrt{3}$ N $\approx 5.196$ N).
3. Assume static equilibrium. For $m_2$, tension $T = m_2 g = 10$ N.
4. For $m_1$ to be in equilibrium, $T = W_{1\parallel} + f$ (if friction is down) or $W_{1\parallel} + f = T$ (if friction is up).
5. If $T = 10$ N and $W_{1\parallel} = 5$ N, then for equilibrium, the required friction is $f = |T - W_{1\parallel}| = |10 - 5| = 5$ N.
6. Since the required friction (5 N) is less than the maximum static friction ($f_{s,max} \approx 5.196$ N), the system remains in static equilibrium.
7. Therefore, acceleration $a = 0$ m/s$^2$.
8. Tension $T = 10$ N.
9. The force of friction is 5 N. Its direction is down the incline, as $T > W_{1\parallel}$, indicating a tendency for $m_1$ to move up, which friction opposes.