Force and free-body diagram

Force is a vector quantity describing the interaction of bodies; in a free-body diagram (FBD) you draw only the forces acting on the chosen body, not the forces it exerts on other bodies. To construct an FBD, mentally isolate the body, replace it with a point, and draw arrows for external forces: weight $\vec{W}=m\vec{g}$, normal force $\vec{N}$, tension $\vec{T}$, friction $\vec{f}$, and others. The arrow lengths show relative magnitudes, and labels show the source of each force. Then choose coordinate axes and write Newton's second law as $\sum \vec{F}=m\vec{a}$ or in components: $\sum F_x=ma_x$, $\sum F_y=ma_y$. For a body in equilibrium $\sum \vec{F}=0$, so unknown forces can be found from balance conditions. An FBD does not include acceleration as a separate force: acceleration is the result of the sum of forces, and it may be marked separately, not drawn as a force arrow.

A block of mass $2.0$ kg rests motionless on a horizontal table. Draw its free-body diagram and find the magnitude of the normal force from the table. Take $g=9.8$ m/s$^2$.

A box of mass $5.0$ kg is pulled along a frictionless horizontal surface by a rope with force $T=20$ N at an angle $30^\circ$ above the horizontal. Draw the FBD, and find the box's acceleration and the normal force.

A block of mass $m_1=3.0$ kg lies on a smooth incline of angle $\theta=25^\circ$. It is connected by an inextensible string over an ideal pulley to a hanging mass $m_2=2.0$ kg. Draw an FBD for each body, and find the system's acceleration and the string tension. Neglect friction, $g=9.8$ m/s$^2$.

Safe self-study investigation: open the PhET simulation “Forces and Motion: Basics” (or a similar one) and choose the mode with a crate on a horizontal surface or an incline. For three cases — rest, uniform motion, and accelerated motion — draw an FBD, label the forces, set axes, and write $\sum F_x=ma_x$, $\sum F_y=ma_y$. Vary the applied force and mass, record the acceleration from the simulation, and compare it with your calculation. Do not use real inclined planes with heavy loads without supervision; for household force measurements, a light spring scale and a book on a table are sufficient.

Show that an FBD is the isolation of one body and the replacement of all external interactions by force vectors, not a drawing of the whole scene.

Diagram: the body is drawn as a point; arrows emerge from it for $\vec{W}$ (downward), $\vec{N}$ (perpendicular to the support), $\vec{T}$ (along the string), and $\vec{f}$ (opposing relative motion); next to it are the $x,y$ axes and the label “only forces on the body.”

A ball of mass $0.44$ kg flies at $20$ m/s after being kicked. Gravity $F_g = mg = 0.44 \cdot 9.8 \approx 4.3$ N acts straight down, and air drag of about $0.4$ N acts opposite to the velocity. The player's foot no longer touches the ball: the only other bodies acting on it are the Earth and the air.

A free-body diagram starts by choosing one body and listing only the forces acting on it. Here there are exactly two: $\vec F_g$ and $\vec F_{drag}$. Their vector sum $\vec F_{net} = \vec F_g + \vec F_{drag}$ sets the acceleration $\vec a = \vec F_{net}/m$: vertically $a_y = 4.3/0.44 \approx 9.8$ m/s² downward, and horizontally the ball slows at $a_x \approx 0.4/0.44 \approx 0.9$ m/s².

A car of mass $1200$ kg brakes on a level road from $20$ m/s ($72$ km/h). The forces on it are: weight $F_g = 1200 \cdot 9.8 = 11\,760$ N downward, the road's normal force $N = 11\,760$ N upward, and a friction force $F_{fr} = 6000$ N directed backwards.

The free-body diagram is four arrows drawn from one point. Along the vertical axis $F_g$ and $N$ are equal and opposite, so there is no acceleration on that axis. Along the horizontal axis only one force is left unbalanced, and Newton's second law gives $a = F_{fr}/m = 6000/1200 = 5.0$ m/s². Hence the stopping time $t = v/a = 20/5.0 = 4.0$ s and the stopping distance $s = v^2/(2a) = 400/10 = 40$ m.

A lamp of mass $12$ kg hangs under a footbridge on two symmetric cables, each tilted $30^\circ$ from the vertical. The lamp's weight $F_g = 12 \cdot 9.8 \approx 118$ N points downward; the cables pull the junction upward and sideways.

The free-body diagram of the junction holds three arrows: $\vec T_1$, $\vec T_2$ and $\vec F_g$. The lamp hangs at rest, so $\vec T_1 + \vec T_2 + \vec F_g = 0$. In components: vertically $2T\cos 30^\circ = F_g$, giving $T = 118/(2 \cdot 0.866) \approx 68$ N; horizontally the $T\sin 30^\circ$ components are equal in magnitude and opposite in direction, so they cancel.