Concept Card: Pulley Systems & Atwood Machines
1. The Ideal Pulley-String Model:
- Ideal Pulley: Massless ($m_p = 0$), moment of inertia zero ($I = 0$), and frictionless ($\mu = 0$).
Consequence: Tension in the cord passing over an ideal pulley is completely uniform on both sides ($T_1 = T_2 = T$). Net force on a massless pulley must be zero ($\sum \vec{F}_p = \vec{0}$). - Ideal Cord: Massless and inextensible ($\frac{dL}{dt} = 0, \frac{d^2L}{dt^2} = 0$).
Consequence: Tension is uniform along its length, and connected bodies obey strict geometric acceleration constraints.
2. Simple Atwood Machine Dynamics:
Two masses $m_1$ and $m_2$ ($m_1 > m_2$) are suspended across a fixed ideal pulley:
- Acceleration:
$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right) g$ - Cord Tension:
$T = \frac{2 m_1 m_2}{m_1 + m_2} g$ - Thrust on Pulley Support:
$F_{\text{clamp}} = 2T = \frac{4 m_1 m_2}{m_1 + m_2} g < (m_1 + m_2)g$
High-Yield Note: A scale supporting an accelerating Atwood machine reads strictly less than the total static weight $(m_1 + m_2)g$. - Acceleration of Center of Mass ($\vec{a}_{\text{cm}}$):
$\vec{a}_{\text{cm}} = -\left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g\hat{j}$
Crucial Fact: The center of mass of an Atwood machine always accelerates vertically downwards!
3. Modified Atwood Systems:
- Horizontal Table with Hanging Mass: Block $m_1$ on a smooth table connected to hanging mass $m_2$:
$a = \frac{m_2 g}{m_1 + m_2}, \quad T = \frac{m_1 m_2 g}{m_1 + m_2}, \quad F_{\text{pulley}} = T\sqrt{2} = \frac{\sqrt{2} m_1 m_2 g}{m_1 + m_2}$ - Single Incline with Hanging Mass: Block $m_1$ on a smooth incline $\theta$ connected to hanging mass $m_2$:
$a = \frac{(m_2 – m_1\sin\theta)g}{m_1 + m_2}, \quad T = \frac{m_1 m_2 g (1 + \sin\theta)}{m_1 + m_2}$ - Double Incline System: Masses $m_1$ on slope $\alpha$ and $m_2$ on slope $\beta$:
$a = \frac{(m_1\sin\alpha – m_2\sin\beta)g}{m_1 + m_2}, \quad T = \frac{m_1 m_2 g(\sin\alpha + \sin\beta)}{m_1 + m_2}$
4. Movable Pulleys & Constraint Relations:
- Movable Pulley Kinematic Rule:
$\vec{a}_p = \frac{\vec{a}_1 + \vec{a}_2}{2}$ - Virtual Work Shortcut (Tension Power Method):
Since an ideal cord performs zero net work on a closed system:
$\sum_{i} \vec{T}_i \cdot \vec{a}_i = 0$
Rule: If tension on block $A$ is $T$ and on block $B$ is $2T$, then $T a_A – 2T a_B = 0 \implies a_A = 2 a_B$.
Solved Examples
Example 1 (Standard Atwood Machine & Center of Mass Dynamics):
Two masses $m_1 = 6.0\text{ kg}$ and $m_2 = 4.0\text{ kg}$ are connected by a light inextensible cord passing over a smooth fixed pulley. Taking $g = 10\text{ m/s}^2$:
(a) Find the acceleration $a$ of the masses and the tension $T$ in the cord.
(b) Find the reaction force exerted by the pulley axle on the ceiling clamp.
(c) Calculate the magnitude and direction of the acceleration of the center of mass of the system.
Solution:
(a) $a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right) g = \left(\frac{6.0 – 4.0}{6.0 + 4.0}\right)(10) = 2.0\text{ m/s}^2$.
$T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2(6.0)(4.0)(10)}{10.0} = 48.0\text{ N}$.
(b) Force on ceiling clamp: $F_{\text{clamp}} = 2T = 2(48.0) = 96.0\text{ N}$ (strictly less than total weight $100\text{ N}$).
(c) Acceleration of center of mass:
$\vec{a}_{\text{cm}} = \frac{m_1(-a\hat{j}) + m_2(+a\hat{j})}{m_1 + m_2} = \frac{6.0(-2.0\hat{j}) + 4.0(+2.0\hat{j})}{10.0} = \frac{-4.0\hat{j}}{10.0} = -0.40\hat{j}\text{ m/s}^2$.
Magnitude is $0.40\text{ m/s}^2$, directed vertically downwards.
Example 2 (Horizontal Table with Modified Atwood System):
A block $A$ of mass $m_1 = 3.0\text{ kg}$ lies on a smooth horizontal table. A light cord connected to $A$ runs horizontally over a smooth corner pulley and suspends block $B$ of mass $m_2 = 2.0\text{ kg}$. Taking $g = 10\text{ m/s}^2$:
(a) Find the acceleration of the blocks and the tension in the cord.
(b) Find the magnitude and direction of the net force exerted by the cord on the pulley.
(c) If the cord leaves block $A$ at an angle of $37^\circ$ above the horizontal to reach the pulley, calculate the new acceleration and verify that contact is maintained.
Solution:
(a) $a = \frac{m_2 g}{m_1 + m_2} = \frac{2.0(10)}{3.0 + 2.0} = \frac{20.0}{5.0} = 4.0\text{ m/s}^2$.
$T = m_1 a = 3.0(4.0) = 12.0\text{ N}$.
(b) $F_{\text{pulley}} = \sqrt{T^2 + T^2} = T\sqrt{2} = 12\sqrt{2}\text{ N} \approx 16.97\text{ N}$ directed at $45^\circ$ below the horizontal.
(c) Cord at $37^\circ$:
$T\cos(37^\circ) = m_1 a \implies 0.8 T = 3.0 a$.
$m_2 g – T = m_2 a \implies 20.0 – T = 2.0 a \implies a = 10.0 – 0.5 T$.
Substitute $a$: $0.8 T = 3.0(10.0 – 0.5 T) = 30.0 – 1.5 T \implies 2.3 T = 30.0 \implies T \approx 13.04\text{ N}$.
$a = \frac{0.8(13.04)}{3.0} \approx 3.48\text{ m/s}^2$.
Normal force: $N = m_1 g – T\sin(37^\circ) = 30.0 – (13.04)(0.6) = 30.0 – 7.82 = 22.18\text{ N} > 0$. Contact is maintained.
Example 3 (Movable Pulley System with 2:1 Constraint):
Block $A$ of mass $m_A = 2.0\text{ kg}$ rests on a smooth horizontal table and is attached to a cord that passes over a fixed pulley and down around a light movable pulley. Block $B$ of mass $m_B = 3.0\text{ kg}$ is suspended from the axle of the movable pulley. The other end of the cord is anchored to the ceiling. Taking $g = 10\text{ m/s}^2$:
(a) Establish the constraint between $a_A$ and $a_B$.
(b) Calculate the acceleration of each block.
(c) Find the tension in the cord.
Solution:
(a) Virtual work method: $T a_A – (2T) a_B = 0 \implies a_A = 2 a_B$.
(b) Equations of motion:
On $A$: $T = m_A a_A = 2.0(2 a_B) = 4.0 a_B$.
On $B$: $m_B g – 2T = m_B a_B \implies 30.0 – 2(4.0 a_B) = 3.0 a_B \implies 30.0 = 11.0 a_B$.
$a_B = \frac{30.0}{11.0} \approx 2.73\text{ m/s}^2$.
$a_A = 2 a_B = \frac{60.0}{11.0} \approx 5.45\text{ m/s}^2$.
(c) $T = 4.0 a_B = 4.0\left(\frac{30.0}{11.0}\right) = \frac{120.0}{11.0} \approx 10.91\text{ N}$.
Example 4 (Sub-Atwood System on an Incline):
A block $M = 10.0\text{ kg}$ rests on a frictionless plane inclined at $\theta = 30^\circ$. A cord attached to $M$ passes over a fixed apex pulley and suspends a movable pulley $P$ of negligible mass. From pulley $P$, two masses $m_1 = 2.0\text{ kg}$ and $m_2 = 3.0\text{ kg}$ hang in a sub-Atwood arrangement. Taking $g = 10\text{ m/s}^2$:
(a) Find the tension $T_{\text{sub}}$ in the lower string supporting $m_1$ and $m_2$.
(b) Determine the tension $T_0$ in the main cable attached to $M$.
(c) Find the acceleration of block $M$ along the incline.
Solution:
(a) Let pulley $P$ descend with acceleration $a_0 = a_M$ (where $a_M$ is upward acceleration of $M$).
In frame of pulley $P$, effective gravity is $g_{\text{eff}} = g – a_M$.
$T_{\text{sub}} = \frac{2 m_1 m_2}{m_1 + m_2} (g – a_M) = \frac{2(2.0)(3.0)}{5.0} (g – a_M) = 2.4(g – a_M)$.
(b) On massless pulley $P$: $T_0 = 2 T_{\text{sub}} = 4.8(g – a_M)$.
(c) On block $M$ up the incline: $T_0 – M g \sin(30^\circ) = M a_M$.
$4.8(10 – a_M) – 10.0(10)(0.5) = 10.0 a_M$
$48.0 – 4.8 a_M – 50.0 = 10.0 a_M \implies -2.0 = 14.8 a_M \implies a_M = -\frac{2.0}{14.8} \approx -0.135\text{ m/s}^2$.
Block $M$ accelerates DOWN the incline with magnitude $0.135\text{ m/s}^2$.
$T_0 = 4.8(10 – (-0.135)) \approx 48.65\text{ N}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
In a simple Atwood machine, two masses $m_1$ and $m_2$ ($m_1 > m_2$) are connected by a light string over a frictionless pulley. If the acceleration of the system is $\frac{g}{3}$, the ratio $\frac{m_1}{m_2}$ is:
(A) $2:1$
(B) $3:1$
(C) $3:2$
(D) $4:3
Problem 2 (JEE Main – Single Correct):
Two masses $m_1 = 5\text{ kg}$ and $m_2 = 3\text{ kg}$ hang over a fixed smooth pulley. The downward force exerted by the string on the pulley axle while the masses accelerate is:
(A) $80\text{ N}$
(B) $75\text{ N}$
(C) $37.5\text{ N}$
(D) Zero
Problem 3 (JEE Main – Single Correct):
A block of mass $m_1 = 4\text{ kg}$ on a smooth horizontal table is connected via a string over a frictionless corner pulley to a hanging block $m_2 = 1\text{ kg}$. The acceleration of the system is:
(A) $\frac{g}{4}$
(B) $\frac{g}{5}$
(C) $\frac{g}{3}$
(D) $g$
Problem 4 (JEE Main – Single Correct):
A body of mass $m$ is suspended from a movable pulley. The other end of the string is pulled upward with an acceleration $a_0$. The upward acceleration of the body is:
(A) $a_0$
(B) $\frac{a_0}{2}$
(C) $2a_0$
(D) $\frac{a_0}{4}$
Problem 5 (JEE Main – Single Correct):
Two equal masses $m_1 = m_2 = 5.0\text{ kg}$ hang in equilibrium on an Atwood machine. An additional rider of mass $\Delta m = 2.0\text{ kg}$ is placed on $m_1$. Taking $g = 10\text{ m/s}^2$, the tension in the string during motion is:
(A) $50\text{ N}$
(B) $58.3\text{ N}$
(C) $70\text{ N}$
(D) $60\text{ N}$
Problem 6 (JEE Advanced – One or More Correct):
For an ideal Atwood machine with masses $m_1 > m_2$ accelerating under gravity:
(A) The tension $T$ in the string satisfies $m_2 g < T < m_1 g$.
(B) The total downward force on the pulley support is strictly less than $(m_1 + m_2)g$.
(C) The center of mass of the system accelerates vertically downward.
(D) The total mechanical energy (kinetic + gravitational potential) of the two masses is conserved.
Problem 7 (JEE Advanced – One or More Correct):
In any system of ideal massless pulleys and light inextensible strings:
(A) The total virtual work done by string tensions on the system is zero ($\sum \vec{T}_i \cdot \vec{a}_i = 0$).
(B) A movable pulley doubles the effective mechanical advantage while halving the distance moved.
(C) The tension is identical throughout any single continuous light string passing over frictionless pulleys.
(D) The net force acting on any massless pulley is always zero.
Problem 8 (JEE Advanced – One or More Correct):
A block $m_1$ rests on a frictionless plane inclined at $\theta$ to the horizontal and is connected via a light string over an apex pulley to a hanging mass $m_2$. Which of the following is/are correct?
(A) If $m_2 > m_1\sin\theta$, block $m_1$ moves up the incline.
(B) If $m_2 = m_1\sin\theta$, the system is in static equilibrium with $T = m_2 g$.
(C) The tension in the string is $T = \frac{m_1 m_2 g (1 + \sin\theta)}{m_1 + m_2}$.
(D) If $\theta = 90^\circ$, the system reduces mathematically to the standard Atwood machine.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
In an Atwood machine, the two suspended masses are $m_1 = 7.0\text{ kg}$ and $m_2 = 3.0\text{ kg}$. Taking $g = 10\text{ m/s}^2$, determine the magnitude of acceleration of the masses in $\text{m/s}^2$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m_1 = 4.0\text{ kg}$ on a smooth horizontal table is attached to a light cord passing over a pulley to a hanging mass $m_2 = 6.0\text{ kg}$. Taking $g = 10\text{ m/s}^2$, find the tension in the cord in Newtons.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (B) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 4 | 10. 24
Solution 1:
$a = \left(\frac{m_1 – m_2}{m_1 + m_2}\right) g = \frac{g}{3} \implies 3m_1 – 3m_2 = m_1 + m_2 \implies 2m_1 = 4m_2 \implies \frac{m_1}{m_2} = 2$.
Correct Answer: (A)
Solution 2:
$T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2(5)(3)(10)}{8} = 37.5\text{ N}$.
Downward force on axle: $F_{\text{clamp}} = 2T = 2(37.5) = 75\text{ N}$.
Correct Answer: (B)
Solution 3:
$a = \frac{m_2 g}{m_1 + m_2} = \frac{1 \cdot g}{4 + 1} = \frac{g}{5}$.
Correct Answer: (B)
Solution 4:
$a_{\text{body}} = \frac{a_{\text{fixed}} + a_0}{2} = \frac{0 + a_0}{2} = \frac{a_0}{2}$.
Correct Answer: (B)
Solution 5:
$M_1 = 7.0\text{ kg}, M_2 = 5.0\text{ kg}$.
$T = \frac{2(7.0)(5.0)(10)}{12.0} = \frac{700}{12} = 58.33\text{ N}$.
Correct Answer: (B)
Solution 6:
All statements (A, B, C, D) are verified fundamental properties of the Atwood machine.
Correct Answer: (A, B, C, D)
Solution 7:
All statements (A, B, C, D) are established principles of ideal pulley constraints and virtual work.
Correct Answer: (A, B, C, D)
Solution 8:
All statements (A, B, C, D) are valid properties of the modified incline Atwood system.
Correct Answer: (A, B, C, D)
Solution 9:
$a = \left(\frac{7.0 – 3.0}{7.0 + 3.0}\right)(10) = 4.0\text{ m/s}^2$.
Correct Answer: 4
Solution 10:
$T = \frac{m_1 m_2 g}{m_1 + m_2} = \frac{(4.0)(6.0)(10)}{10.0} = 24.0\text{ N}$.
Correct Answer: 24