Two Blocks on a Ramp with a Massive Pulley (Friction & Rotation)

We have a classic coupled system:

• Block 1: mass $m_1 = 2.45\,\text{kg}$ on a rough ramp at $\theta = 30^\circ$.
• Block 2: mass $m_2 = 5.80\,\text{kg}$ hanging vertically.
• Pulley: solid disk, radius $R = 0.25\,\text{m}$, mass $M = 10.0\,\text{kg}$.
• Coefficient of kinetic friction: $\mu_k = 0.360$ for both surfaces.
• String is massless but the pulley has rotational inertia, so tensions differ on each side.

We want:

1. The magnitude of the common acceleration $a$ of the blocks.
2. The tension on the left side of the pulley (ramp side), $T_1$.
3. The tension on the right side of the pulley (hanging side), $T_2$.

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1. Force analysis and directions

We assume (and later confirm) that:

• Block $m_2$ moves downward.
• Block $m_1$ moves up the ramp.

Choose positive directions:

• For $m_1$: along the ramp, up the incline.
• For $m_2$: downward.
• For rotation of the pulley: direction corresponding to $m_2$ going down is positive (so side with $T_2$ tends to rotate it).

Forces on $m_1$ (on the ramp)

Along the incline:

• Tension $T_1$ up the ramp.
• Gravity component $m_1 g \sin\theta$ down the ramp.
• Kinetic friction $f_{k1}$ down the ramp (because motion is up the ramp).

Normal force: $N_1 = m_1 g \cos\theta$.

Friction:

$$f_{k1} = \mu_k N_1 = \mu_k m_1 g \cos\theta$$

Newton's 2nd law along the ramp (positive up-slope):

$$T_1 - m_1 g \sin\theta - f_{k1} = m_1 a$$

Substitute $f_{k1}$:

$$T_1 - m_1 g \sin\theta - \mu_k m_1 g \cos\theta = m_1 a \tag{1}$$

Forces on $m_2$ (hanging)

Take downward as positive:

• Gravity $m_2 g$ (downward, +)
• Tension $T_2$ upward (negative).

Newton's 2nd law:

$$m_2 g - T_2 = m_2 a \tag{2}$$

Rotation of the pulley

A solid disk has moment of inertia

$$I = \frac{1}{2} M R^2$$

Torques about the center (positive in direction of $T_2$ creating the motion):

• Torque from $T_2$: $+T_2 R$.
• Torque from $T_1$: $-T_1 R$.

Net torque:

$$(T_2 - T_1) R = I \alpha$$

No slipping of the string, so

$$a = \alpha R \implies \alpha = \frac{a}{R}$$

Thus

$$(T_2 - T_1) R = I \frac{a}{R}$$ $$T_2 - T_1 = \frac{I}{R^2} a = \frac{\tfrac{1}{2} M R^2}{R^2} a = \frac{1}{2} M a \tag{3}$$

Now we have 3 unknowns $a, T_1, T_2$ and 3 equations (1), (2), (3).

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2. Solve symbolically

From (2):

$$T_2 = m_2 g - m_2 a \tag{2'}$$

From (1):

$$T_1 = m_1 a + m_1 g \sin\theta + \mu_k m_1 g \cos\theta \tag{1'}$$

Plug (1') and (2') into (3):

$$(m_2 g - m_2 a) - (m_1 a + m_1 g \sin\theta + \mu_k m_1 g \cos\theta) = \frac{1}{2} M a$$

Group terms in $a$ and constant terms:

$$m_2 g - m_1 g \sin\theta - \mu_k m_1 g \cos\theta - m_2 a - m_1 a = \frac{1}{2} M a$$

Move all $a$ terms to one side:

$$m_2 g - m_1 g \sin\theta - \mu_k m_1 g \cos\theta = m_2 a + m_1 a + \frac{1}{2} M a$$

Factor $a$ on the right:

$$m_2 g - m_1 g \sin\theta - \mu_k m_1 g \cos\theta = \left(m_1 + m_2 + \tfrac{1}{2} M \right) a$$

So

$$a = \dfrac{m_2 g - m_1 g \sin\theta - \mu_k m_1 g \cos\theta}{m_1 + m_2 + \tfrac{1}{2} M} \tag{4}$$

Then use (1') and (2') to get $T_1, T_2$.

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3. Numerical evaluation

Use:

• $m_1 = 2.45\,\text{kg}$
• $m_2 = 5.80\,\text{kg}$
• $M = 10.0\,\text{kg}$
• $\theta = 30^\circ$: $\sin 30^\circ = 0.5$, $\cos 30^\circ = \sqrt{3}/2 \approx 0.866$
• $\mu_k = 0.360$
• $g \approx 9.80\,\text{m/s}^2$ (you can also use 9.81; the result changes slightly)

Compute the denominator

$$m_1 + m_2 + \tfrac{1}{2} M = 2.45 + 5.80 + 0.5 \cdot 10.0 = 2.45 + 5.80 + 5.00 = 13.25\,\text{kg}$$

Compute the numerator

First, compute each term:

1. $m_2 g \approx 5.80 \cdot 9.80 \approx 56.84\,\text{N}$
2. $m_1 g \sin\theta \approx 2.45 \cdot 9.80 \cdot 0.5 \approx 2.45 \cdot 4.9 \approx 12.0\,\text{N}$
3. $\mu_k m_1 g \cos\theta \approx 0.360 \cdot 2.45 \cdot 9.80 \cdot 0.866$.
   • $2.45 \cdot 9.80 \approx 24.01$
   • $24.01 \cdot 0.866 \approx 20.8$
   • $0.360 \cdot 20.8 \approx 7.5\,\text{N}$ (approx).

So the numerator is

$$m_2 g - m_1 g \sin\theta - \mu_k m_1 g \cos\theta \approx 56.84 - 12.0 - 7.5 \approx 37.3\,\text{N}$$

Acceleration

$$a \approx \frac{37.3}{13.25} \approx 2.8\,\text{m/s}^2$$

So the magnitude of the acceleration is about

$a \approx 2.82\,\text{m/s}^2$ (depending on rounding).

Direction:

• $m_2$ moves downward.
• $m_1$ moves up the ramp.

Tensions

Using $a \approx 2.82\,\text{m/s}^2$.

1. From (2$'$):

   $$T_2 = m_2 g - m_2 a \approx 5.80(9.80 - 2.82) \approx 5.80 \cdot 6.98 \approx 40.5\,\text{N}$$

   So $T_2 \approx 40\text{–}41\,\text{N}$.

2. From (1$'$):

   $$T_1 = m_1 a + m_1 g \sin\theta + \mu_k m_1 g \cos\theta$$
   • $m_1 a \approx 2.45 \cdot 2.82 \approx 6.9\,\text{N}$
   • $m_1 g \sin\theta \approx 12.0\,\text{N}$ (from above)
   • $\mu_k m_1 g \cos\theta \approx 7.5\,\text{N}$ (from above)

   Then

   $$T_1 \approx 6.9 + 12.0 + 7.5 \approx 26.4\,\text{N}$$

So numerically (rounded):

• Acceleration: $a \approx 2.8\,\text{m/s}^2$ (magnitude)
• Tension (left / ramp side): $T_1 \approx 26\,\text{N}$
• Tension (right / hanging side): $T_2 \approx 41\,\text{N}$

The difference $T_2 - T_1$ is what produces the angular acceleration of the massive pulley.

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4. What the visualization shows

Use the controls to:

• Change $m_1, m_2, M, \mu_k, \theta$.
• See blocks move in real time.
• Watch friction and gravity components on $m_1$ along the ramp.
• Watch both tensions and the net torque on the pulley.

Conceptually:

• Increasing $m_2$ makes the system accelerate more.
• Increasing $m_1$, $\mu_k$, or $\theta$ (to increase $m_1 g\sin\theta$) can slow or even reverse the motion.
• Increasing $M$ (the pulley's inertia) does not change the net driving force, but it makes more mass "participate" in acceleration, so $a$ becomes smaller for the same forces.

You can also use the "Show Field Lines" toggle to overlay a simplified diagram of force components along the ramp for $m_1$, helping connect the algebra to the geometry of the situation.