Lesson 2 · 30 min

Free-Body and Kinetic Diagrams

Most mistakes in kinetics are made before any algebra: a force left off, a force invented, an acceleration in the wrong direction. Two drawings prevent them. The free-body diagram shows every force on the body; the kinetic diagram shows the body's mass times its acceleration. Newton's second law says the two drawings are equivalent, and that turns into one equation per direction.

Learning objectives

Draw a free-body diagram that shows every force on an isolated particle, and a kinetic diagram that shows \(m\avec\).
Apply a four-step procedure: diagrams, axes, equations of motion, kinematics.
Handle unknown forces and accelerations by assuming a direction and reading the sign of the answer.
Find the force a support or a floor must exert on an accelerating body, as in an elevator.

Two diagrams, one equation

Isolate the body from everything around it. On the free-body diagram, draw each force that its surroundings exert on it: its weight, and a force at every point where something touches it, such as a floor, a rope, a hand or a spring. On the kinetic diagram, draw the single vector \(m\avec\), in the direction of the acceleration. Newton's second law says the resultant of the first diagram equals the second.

Figure 2.1 The free-body diagram (left) and the kinetic diagram (right) of a crate pulled across a rough floor by a rope at \(30^\circ\). The weight \(\colW{W}\) ends at the center of gravity, the floor's push \(N\) ends on the bottom face, and friction \(\colF{F_f}\) and the rope's tension \(T\) start where they act. The kinetic diagram has one vector, \(\colMA{m\avec}\), along the floor: the crate cannot accelerate up or down, so \(a_y = 0\).

The kinetic diagram is not a force. Never put \(m\avec\) on the free-body diagram: that counts it twice and makes the equation read \(\sum\Fvec - m\avec = m\avec\).

Procedure for analysis

Four steps

Free-body diagram. Isolate the particle and draw every force on it. Choose axes; one along the acceleration, if you know its direction, saves work.
Kinetic diagram. Draw \(m\avec\), or its components \(ma_x\) and \(ma_y\). If you do not know a direction, assume one.
Equations of motion. \(\sum F_x = ma_x\) and \(\sum F_y = ma_y\), one per direction. Add friction \(F_f = \mu_k N\) if the body slides, and spring forces \(F_s = ks\).
Kinematics. If the question asks for a speed, a distance or a time, use the acceleration you found.

Assumed directions. Draw an unknown force or acceleration in whichever direction you like and solve. A negative answer means it points the other way; keep the sign when you substitute it into another equation.
Directions you know. Kinetic friction opposes the sliding. A cord can only pull. A surface can only push, along its normal.
Directions in which nothing moves. A crate on a floor, a car on a level road: \(a_y = 0\), so the forces in that direction balance. That is usually how you find the normal force.

Example 2.1 — A crate pulled by a rope

The crate in Figure 2.1 has a mass of \(20\ \text{kg}\), the rope pulls with \(T = 100\ \text{N}\) at \(30^\circ\) above the horizontal, and the coefficient of kinetic friction is \(0.25\). The crate is sliding. Find its acceleration.

Show solution

Vertical, \(a_y = 0\):

\[ \textstyle\sum F_y = 0: \quad N + 100\sin 30^\circ - 20(9.81) = 0 \quad\Rightarrow\quad N = 146.2\ \text{N} \]

The rope lifts part of the weight, so the floor pushes less than \(mg\). Friction is \(\colF{F_f = \mu_k N = 36.55\ \text{N}}\), against the motion.

Horizontal, in the direction of motion:

\[ \textstyle\sum F_x = ma_x: \quad 100\cos 30^\circ - 36.55 = 20\,a \quad\Rightarrow\quad a = 2.50\ \text{m/s}^2 \]

What a scale reads

A person stands on a bathroom scale in an elevator. The scale reads the force \(N\) it pushes up with. Take up as positive: the person's free-body diagram has \(N\) up and \(\colW{W = mg}\) down, and the kinetic diagram has \(\colMA{ma}\), where \(a\) is the elevator's acceleration.

\[ N - mg = ma \quad\Rightarrow\quad N = m(g + a) \]

Figure 2.2 A \(70\ \text{kg}\) passenger on a scale in an elevator. Left: the elevator and its acceleration. Middle: the passenger's free-body diagram, with \(N\) drawn to scale. Right: the kinetic diagram. Speeding up on the way up, or slowing down on the way down (\(a\) upward), the scale reads more than the weight; the reverse reads less; in free fall it reads zero.

The scale's reading depends only on the acceleration, not on the velocity: at a steady \(5\ \text{m/s}\), up or down, it reads exactly the weight. Slowing down on the way down needs an upward acceleration, so it feels the same as speeding up on the way up.

Example 2.2 — The tension in an elevator cable

An \(800\ \text{kg}\) elevator car hangs from one cable. Find the tension (a) while the car speeds up at \(2\ \text{m/s}^2\) going up, (b) while it slows down at \(1.5\ \text{m/s}^2\) going down, and (c) while it moves at a steady speed.

Show solution

Up positive. The free-body diagram has \(T\) up and \(mg = 7848\ \text{N}\) down; the kinetic diagram, \(ma\), with \(a\) up positive. So \(T = m(g + a)\).

(a) Speeding up going up: \(a = +2\), so \(T = 800(11.81) = 9448\ \text{N}\).
(b) Slowing down going down: the acceleration points up, \(a = +1.5\), so \(T = 800(11.31) = 9048\ \text{N}\).
(c) Steady speed: \(a = 0\), so \(T = 7848\ \text{N}\), the weight.

The direction of the velocity never entered: only the acceleration does.

Common mistakes

Missing a contact force. Walk round the boundary of the isolated body: every place it touches something has a force, and only those places (plus the weight).
Forces on the wrong body. A free-body diagram shows forces on the body, not the forces it exerts on others.
"Force of motion". A moving body does not carry a force forward. If nothing pushes it, there is no forward force; it keeps moving because of the first law.
\(N = mg\) by habit. Find \(N\) from the direction in which nothing moves. It is \(mg\) only on a level surface with no other vertical force and no vertical acceleration.

Check your understanding

Key takeaways

The free-body diagram shows every force on the isolated body; the kinetic diagram shows \(m\avec\). \(\sum\Fvec = m\avec\) says they are equivalent.
Procedure: free-body diagram, kinetic diagram, one equation per direction, then kinematics.
Assume unknown directions and read the sign; find normal forces from the direction with no acceleration.
A support pushes with \(m(g + a)\) on a body accelerating upward at \(a\): scales and cables feel accelerations, not velocities.
Next, Lesson 3 applies the procedure in rectangular coordinates, with friction and inclines.