Concept Card: Free Body Diagrams (FBD) & Connected Body Mechanics
1. Free Body Diagram (FBD) – Principles & Construction Algorithm:
A Free Body Diagram is an isolated sketch of a chosen body (or subsystem) where all surrounding objects are removed, and every external force acting on the body is explicitly drawn as a vector.
- Isolate the System: Choose the particle, block, or composite subsystem.
- Field Forces: Draw the downward gravitational force $\vec{W} = m\vec{g}$ acting through the center of gravity.
- Contact Forces: At every physical boundary point, identify:
- Normal Contact Force ($\vec{N}$): Always acts perpendicular to the surface of contact, directed inward (pushing).
- Tension Force ($\vec{T}$): For light flexible strings, tension acts along the string, pulling away from the body.
- Frictional Force ($\vec{f}$): Acts parallel to the interface, opposing relative motion or impending slippage.
- Spring Restoring Force ($\vec{F}_s$): $\vec{F}_s = -k\vec{x}$.
- Coordinate Axis Selection: Align one axis along the direction of actual or anticipated acceleration $\vec{a}$ to minimize trigonometric resolution.
- Golden Rule: Never include forces exerted by the body on external entities. Only forces acting ON the isolated body belong on its FBD.
2. Mechanics of Connected Bodies:
- Bodies in Direct Contact on a Frictionless Floor:
For $n$ blocks in contact pushed by a horizontal force $F$:
Common acceleration: $a = \frac{F}{\sum_{i=1}^n m_i}$.
For two blocks ($m_1$ and $m_2$) pushed from $m_1$:
Contact normal force: $N_c = m_2 a = \frac{m_2}{m_1 + m_2} F$.
Directional Asymmetry: If pushed from $m_2$ towards $m_1$, $N_c’ = m_1 a = \frac{m_1}{m_1 + m_2} F \neq N_c$. - Blocks Connected by Light Strings:
For three blocks ($m_1, m_2, m_3$) connected by light strings pulled by force $F$ applied to $m_3$:
$a = \frac{F}{m_1 + m_2 + m_3}$
Tension between $m_1$ and $m_2$: $T_1 = m_1 a = \frac{m_1}{m_1 + m_2 + m_3} F$
Tension between $m_2$ and $m_3$: $T_2 = (m_1 + m_2) a = \frac{m_1 + m_2}{m_1 + m_2 + m_3} F$ - Connected Bodies on an Incline:
For blocks $m_1$ and $m_2$ pulled up a smooth incline of angle $\theta$ by force $F$ on $m_2$:
Acceleration: $a = \frac{F – (m_1 + m_2)g\sin\theta}{m_1 + m_2} = \frac{F}{m_1 + m_2} – g\sin\theta$
Connecting string tension: $T = m_1(a + g\sin\theta) = \frac{m_1}{m_1 + m_2} F$
Remarkable Property: The string tension is completely independent of $g$ and incline angle $\theta$! - Heavy (Massive) Ropes:
When a rope of mass $M_r$ and length $L$ accelerates, tension varies continuously along its length:
$T(x) = \left[m_{\text{trail}} + M_r\left(1 – \frac{x}{L}\right)\right] a$
where $x$ is the distance from the pulled end and $m_{\text{trail}}$ is the trailing mass.
Solved Examples
Example 1 (Three Blocks in Direct Contact):
Three blocks of masses $m_1 = 2.0\text{ kg}, m_2 = 3.0\text{ kg}, m_3 = 5.0\text{ kg}$ are kept in contact on a smooth horizontal floor. A horizontal force $F = 40.0\text{ N}$ is pushed against $m_1$ towards $m_2$.
(a) Find the common acceleration $a$ of the blocks.
(b) Draw the FBD for each individual block and find the contact normal forces $N_{12}$ and $N_{23}$.
(c) If the force $F = 40.0\text{ N}$ is applied from the right pushing $m_3$ towards $m_2$, find the new contact forces $N_{23}’$ and $N_{12}’$.
Solution:
(a) Total mass $M = 2.0 + 3.0 + 5.0 = 10.0\text{ kg}$.
Acceleration: $a = \frac{F}{M} = \frac{40.0\text{ N}}{10.0\text{ kg}} = 4.0\text{ m/s}^2$.
(b) FBD analysis:
– On block 3 ($m_3 = 5.0\text{ kg}$): $N_{23} = m_3 a = (5.0)(4.0) = 20.0\text{ N}$.
– On block 2 ($m_2 = 3.0\text{ kg}$): $N_{12} – N_{23} = m_2 a \implies N_{12} = 20.0 + (3.0)(4.0) = 32.0\text{ N}$.
– Check on block 1 ($m_1 = 2.0\text{ kg}$): $F – N_{12} = 40.0 – 32.0 = 8.0\text{ N} = (2.0)(4.0)$.
(c) Pushed from the right on $m_3$:
– On block 1: $N_{12}’ = m_1 a = (2.0)(4.0) = 8.0\text{ N}$.
– On block 2: $N_{23}’ – N_{12}’ = m_2 a \implies N_{23}’ = 8.0 + (3.0)(4.0) = 20.0\text{ N}$.
(Note: $N_{12} = 32\text{ N} \neq N_{12}’ = 8\text{ N}$, illustrating contact force dependence on application point).
Example 2 (Connected Blocks with Pull at an Angle):
Two blocks $m_1 = 4.0\text{ kg}$ and $m_2 = 6.0\text{ kg}$ are connected by a light string on a smooth horizontal floor. A pull $F = 50.0\text{ N}$ is applied to $m_2$ at an angle $\theta = 37^\circ$ above the horizontal ($\cos 37^\circ = 0.8, \sin 37^\circ = 0.6$). Taking $g = 10\text{ m/s}^2$:
(a) Verify whether block $m_2$ loses contact with the floor.
(b) Find the acceleration of the system and the tension $T$ in the string.
(c) Find the normal force on each block from the floor.
Solution:
(a) Upward component of pull on $m_2$: $F_y = F\sin(37^\circ) = 50.0(0.6) = 30.0\text{ N}$.
Weight of $m_2$: $m_2 g = 6.0 \times 10 = 60.0\text{ N}$.
Since $F_y < m_2 g$, $m_2$ remains firmly on the floor.
(b) Horizontal equation of motion: $F_x = F\cos(37^\circ) = 50.0(0.8) = 40.0\text{ N}$.
Total mass $M = 4.0 + 6.0 = 10.0\text{ kg} \implies a = \frac{40.0}{10.0} = 4.0\text{ m/s}^2$.
Tension in string (from FBD of $m_1$): $T = m_1 a = (4.0)(4.0) = 16.0\text{ N}$.
(c) Normal forces:
$N_1 = m_1 g = 40.0\text{ N}$.
$N_2 = m_2 g – F\sin(37^\circ) = 60.0 – 30.0 = 30.0\text{ N}$.
Example 3 (Double Incline with Connected String over Apex Pulley):
A fixed wedge has two smooth inclined faces at angles $\theta_1 = 30^\circ$ and $\theta_2 = 60^\circ$ to the horizontal. Blocks $m_1 = 4.0\text{ kg}$ and $m_2 = 8.0\text{ kg}$ lie on slopes $\theta_1$ and $\theta_2$ respectively, connected by a light string passing over a frictionless apex pulley. Taking $g = 9.8\text{ m/s}^2$:
(a) Determine the direction and magnitude of acceleration of the blocks.
(b) Find the tension $T$ in the string.
(c) Calculate the magnitude of the force exerted by the string on the pulley.
Solution:
(a) Downward gravity components along the slopes:
$F_{g1} = m_1 g \sin(30^\circ) = 4.0(9.8)(0.5) = 19.6\text{ N}$.
$F_{g2} = m_2 g \sin(60^\circ) = 8.0(9.8)\left(\frac{\sqrt{3}}{2}\right) = 39.2\sqrt{3} \approx 67.89\text{ N}$.
Since $F_{g2} > F_{g1}$, block $m_2$ moves down and $m_1$ moves up the incline.
$a = \frac{m_2 g \sin(60^\circ) – m_1 g \sin(30^\circ)}{m_1 + m_2} = \frac{67.89 – 19.6}{12.0} = \frac{48.29}{12.0} \approx 4.024\text{ m/s}^2$.
(b) Tension: $T = m_1(g\sin 30^\circ + a) = 19.6 + 4.0(4.024) = 35.7\text{ N}$.
(c) Angle between the two string segments at apex: $\alpha = 180^\circ – (30^\circ + 60^\circ) = 90^\circ$.
$F_{\text{pulley}} = \sqrt{T^2 + T^2} = T\sqrt{2} = 35.7\sqrt{2} \approx 50.49\text{ N}$.
Example 4 (Heavy Rope with Non-Uniform Continuous Tension):
A uniform rope of mass $M = 2.0\text{ kg}$ and length $L = 1.0\text{ m}$ is tied between a block of mass $m_1 = 5.0\text{ kg}$ at its top and a block of mass $m_2 = 3.0\text{ kg}$ at its bottom. A vertical upward force $F = 150.0\text{ N}$ pulls the upper block. Taking $g = 10\text{ m/s}^2$:
(a) Find the upward acceleration of the system.
(b) Derive an expression for the tension $T(y)$ in the rope at a distance $y$ below the top block.
(c) Calculate the tension at the top ($y = 0$), midpoint ($y = 0.5\text{ m}$), and bottom ($y = 1.0\text{ m}$) of the rope.
Solution:
(a) Total mass: $M_{\text{tot}} = 5.0 + 2.0 + 3.0 = 10.0\text{ kg}$.
$F – M_{\text{tot}} g = M_{\text{tot}} a \implies 150.0 – 100.0 = 10.0 a \implies a = 5.0\text{ m/s}^2$.
Effective acceleration: $g_{\text{eff}} = g + a = 10 + 5.0 = 15.0\text{ m/s}^2$.
(b) Subsystem below cross-section $y$ has mass: $m_{\text{sub}}(y) = m_2 + M\left(\frac{L – y}{L}\right) = 3.0 + 2.0(1 – y) = 5.0 – 2.0y\text{ kg}$.
$T(y) = m_{\text{sub}}(y)(g + a) = (5.0 – 2.0y)(15.0) = 75.0 – 30.0y\text{ N}$.
(c) Specific values:
– At top ($y = 0$): $T(0) = 75.0\text{ N}$.
– At midpoint ($y = 0.5\text{ m}$): $T(0.5) = 75.0 – 30.0(0.5) = 60.0\text{ N}$.
– At bottom ($y = 1.0\text{ m}$): $T(1.0) = 75.0 – 30.0(1.0) = 45.0\text{ N}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
Two blocks of masses $m_1 = 3\text{ kg}$ and $m_2 = 2\text{ kg}$ are in contact on a frictionless horizontal table. A horizontal force $F = 20\text{ N}$ is applied on $m_1$ pushing it towards $m_2$. The contact force between the two blocks is:
(A) $8\text{ N}$
(B) $12\text{ N}$
(C) $20\text{ N}$
(D) $10\text{ N}$
Problem 2 (JEE Main – Single Correct):
Three blocks of masses $m_1 = 1\text{ kg}, m_2 = 2\text{ kg}$, and $m_3 = 3\text{ kg}$ are connected by light inextensible strings and pulled by a horizontal force $F = 18\text{ N}$ on a smooth horizontal floor applied to $m_3$. The tension in the string between $m_1$ and $m_2$ is:
(A) $3\text{ N}$
(B) $6\text{ N}$
(C) $9\text{ N}$
(D) $12\text{ N}$
Problem 3 (JEE Main – Single Correct):
A uniform rope of mass $M$ and length $L$ is pulled on a smooth horizontal surface by a constant horizontal force $F$ applied at one end. The tension in the rope at a distance $x$ from the pulled end is:
(A) $F\left(\frac{x}{L}\right)$
(B) $F\left(1 – \frac{x}{L}\right)$
(C) $F$
(D) $F\left(1 + \frac{x}{L}\right)$
Problem 4 (JEE Main – Single Correct):
Two blocks of masses $m_1$ and $m_2$ are connected by a light string on a smooth inclined plane of angle $\theta$. A force $F$ pulls $m_2$ up along the incline. The tension in the string connecting the blocks is:
(A) $\frac{m_1 F}{m_1 + m_2}$
(B) $\frac{m_2 F}{m_1 + m_2}$
(C) $\frac{m_1 F}{m_1 + m_2} + m_1 g\sin\theta$
(D) $\frac{m_1 F}{m_1 + m_2} – m_1 g\sin\theta$
Problem 5 (JEE Main – Single Correct):
A block of mass $m$ is suspended by a string from a rigid ceiling. A horizontal force $F$ is applied to the block, keeping it in equilibrium with the string inclined at an angle $\theta$ to the vertical. The tension in the string is:
(A) $\sqrt{F^2 + (mg)^2}$
(B) $F + mg$
(C) $mg\cos\theta$
(D) $\frac{mg}{\sin\theta}$
Problem 6 (JEE Advanced – One or More Correct):
Which of the following principles must be strictly obeyed when constructing a Free Body Diagram (FBD)?
(A) Only external forces acting ON the isolated body are included.
(B) Newton’s third law reaction forces exerted BY the body on other objects must NOT be included.
(C) If the body is analyzed from a non-inertial reference frame, an appropriate pseudo-force must be added to the FBD.
(D) Internal forces between constituents within the isolated body must always be shown.
Problem 7 (JEE Advanced – One or More Correct):
Two blocks $A$ ($m_A = 2\text{ kg}$) and $B$ ($m_B = 4\text{ kg}$) on a smooth horizontal table are connected by a light spring of stiffness $k = 100\text{ N/m}$. A constant force $F = 18\text{ N}$ is suddenly applied horizontally to block $B$ pulling it away from $A$. At the initial instant before the spring stretches:
(A) The acceleration of block $A$ is zero.
(B) The acceleration of block $B$ is $4.5\text{ m/s}^2$.
(C) When the spring achieves maximum elongation, both blocks have the same acceleration of $3.0\text{ m/s}^2$.
(D) The tension in the spring is always $18\text{ N}$.
Problem 8 (JEE Advanced – One or More Correct):
Two blocks $m_1$ and $m_2$ ($m_1 > m_2$) are tied to the ends of a light inextensible cord passing over a fixed frictionless pulley (Atwood machine). If the system is released from rest:
(A) The acceleration of each block is $\frac{m_1 – m_2}{m_1 + m_2}g$.
(B) The tension in the cord is $\frac{2m_1 m_2}{m_1 + m_2}g$.
(C) The force exerted on the pulley support is $\frac{4m_1 m_2}{m_1 + m_2}g$.
(D) The acceleration of the center of mass of the system is directed downwards with magnitude $\left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2 g$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Three blocks $m_1 = 2.0\text{ kg}, m_2 = 3.0\text{ kg}, m_3 = 5.0\text{ kg}$ are in contact on a frictionless table. A horizontal force $F = 30.0\text{ N}$ is applied on $m_1$ towards $m_2$. The contact force between $m_2$ and $m_3$ in Newtons is $N$. Find $N$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
Two masses $m_1 = 4.0\text{ kg}$ and $m_2 = 6.0\text{ kg}$ are connected by a light string on a frictionless horizontal floor. A horizontal pulling force $F = 40.0\text{ N}$ acts on $m_2$ away from $m_1$. The tension in the connecting string in Newtons is $T$. Find $T$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (A) | 3. (B) | 4. (A) | 5. (A) | 6. (A, B, C) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 15 | 10. 16
Solution 1:
$a = \frac{F}{m_1 + m_2} = \frac{20}{3 + 2} = 4\text{ m/s}^2$.
Contact force pushing $m_2$: $N_c = m_2 a = 2 \times 4 = 8\text{ N}$.
Correct Answer: (A)
Solution 2:
Common acceleration $a = \frac{18}{1 + 2 + 3} = \frac{18}{6} = 3\text{ m/s}^2$.
Tension $T_1$ accelerates $m_1$: $T_1 = m_1 a = 1 \times 3 = 3\text{ N}$.
Correct Answer: (A)
Solution 3:
Acceleration $a = \frac{F}{M}$. The mass behind cross-section $x$ is $m(x) = M(1 – x/L)$.
Tension $T(x) = m(x) a = F(1 – x/L)$.
Correct Answer: (B)
Solution 4:
$a = \frac{F}{m_1 + m_2} – g\sin\theta$.
For $m_1$: $T – m_1 g\sin\theta = m_1 a \implies T = m_1(a + g\sin\theta) = \frac{m_1 F}{m_1 + m_2}$.
Correct Answer: (A)
Solution 5:
$T\sin\theta = F, \quad T\cos\theta = mg \implies T = \sqrt{F^2 + (mg)^2}$.
Correct Answer: (A)
Solution 6:
– (A) True: only forces on the isolated system belong on an FBD.
– (B) True: reaction forces on other objects must never be included.
– (C) True: non-inertial frames require $-m\vec{a}_{\text{frame}}$.
– (D) False: internal forces cancel out and are not shown.
Correct Answer: (A, B, C)
Solution 7:
– (A) Initially $x = 0 \implies F_s = 0 \implies a_A = 0$ (True).
– (B) On block $B$, $a_B = \frac{18}{4} = 4.5\text{ m/s}^2$ (True).
– (C) At maximum stretch, relative velocity is zero and $a_A = a_B = a_{\text{cm}} = \frac{18}{6} = 3.0\text{ m/s}^2$ (True).
– (D) Spring force increases continuously as stretch grows (False).
Correct Answer: (A, B, C)
Solution 8:
All statements (A, B, C, D) are standard verified results for the ideal Atwood machine.
Correct Answer: (A, B, C, D)
Solution 9:
$a = \frac{30.0}{2.0 + 3.0 + 5.0} = 3.0\text{ m/s}^2$.
$N = m_3 a = 5.0 \times 3.0 = 15.0\text{ N}$.
Correct Answer: 15
Solution 10:
$a = \frac{40.0}{4.0 + 6.0} = 4.0\text{ m/s}^2$.
$T = m_1 a = 4.0 \times 4.0 = 16.0\text{ N}$.
Correct Answer: 16