Two blocks connected by string on inclined plane — find acceleration and tension

Question

Two blocks of masses $m_1 = 5$ kg and $m_2 = 3$ kg are connected by a light string over a frictionless pulley at the top of a frictionless inclined plane (angle 30°). $m_1$ hangs vertically and $m_2$ is on the inclined surface. Find the acceleration of the system and tension in the string. (g = 10 m/s²)

Solution — Step by Step

The system consists of:

$m_1 = 5$ kg: Hanging freely, weight $m_1 g = 50$ N downward
$m_2 = 3$ kg: On incline (30°), weight $m_2 g = 30$ N vertically downward

Assume $m_1$ moves down (and $m_2$ moves up the incline) — take this as positive direction.

Forces on $m_1$ : weight $m_1g$ downward, tension $T$ upward.

m_1g - T = m_1 a

50 - T = 5a \quad ...(1)

Forces on $m_2$ along the incline:

Component of weight down the incline: $m_2 g\sin 30° = 30 \times 0.5 = 15$ N (opposing upward motion)
Tension $T$ up the incline (pulling it up)

Net force on $m_2$ up the incline = $T - m_2g\sin\theta$

T - m_2g\sin 30° = m_2 a

T - 15 = 3a \quad ...(2)

Add equations (1) and (2):

(50 - T) + (T - 15) = 5a + 3a

35 = 8a

a = \frac{35}{8} = 4.375 \text{ m/s}^2 \approx 4.4 \text{ m/s}^2

Substitute back into (2):

T = 15 + 3 \times 4.375 = 15 + 13.125 = 28.125 \text{ N} \approx 28.1 \text{ N}

From (1): $50 - T = 5a = 5 \times 4.375 = 21.875$

$T = 50 - 21.875 = 28.125$ N ✓

Answer: Acceleration = 4.4 m/s², Tension = 28.1 N

Why This Works

The two blocks are connected by an inextensible string, so they have the same magnitude of acceleration. By applying Newton’s 2nd law to each block separately and adding the equations, the unknown tension cancels, giving us acceleration directly.

The inclined plane problem reduces to a simpler form because the surface is frictionless. Only the component of gravity along the incline ( $m_2 g\sin\theta$ ) acts as resistance; the normal force ( $m_2 g\cos\theta$ ) is balanced by the surface’s normal reaction and plays no role in acceleration.

General formula for this type of problem: when mass $m_1$ hangs vertically and mass $m_2$ is on an incline at angle $\theta$ (frictionless):

a = \frac{m_1 g - m_2 g\sin\theta}{m_1 + m_2} = \frac{(m_1 - m_2\sin\theta)g}{m_1 + m_2}

Substituting: $a = \frac{(5 - 3 \times 0.5) \times 10}{8} = \frac{3.5 \times 10}{8} = 4.375$ m/s². Same answer, faster route once you derive the formula.

Common Mistake

Students frequently use the full weight $m_2 g$ for the inclined block instead of $m_2 g\sin\theta$ . The full weight acts downward; only its component ALONG the incline opposes the string tension. The component perpendicular to the incline ( $m_2 g\cos\theta$ ) is balanced by the normal force — it plays NO role in the motion along the incline. This is the most common error in Atwood-on-incline problems, and it gives a completely wrong answer.

Question

Solution — Step by Step

Why This Works

Common Mistake

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