3.1 Draw consistent free-body and kinetic diagrams | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Particle Kinetics: Force and Acceleration

EN PH 131 Engineering Mechanics: Dynamics — Study topic 3.1

A free-body diagram and a kinetic diagram answer different questions. The free-body diagram shows the external forces acting on a chosen system. The kinetic diagram shows the direction and magnitude of the system’s mass multiplied by its acceleration. Together, they help organize Newton’s second law without treating acceleration as another applied force. In this lesson, the systems are particles or objects treated as particles. Before drawing, state what is included in the system, who observes it, the reference frame, and the positive coordinate directions. Then place only appropriate forces on the free-body diagram, place the acceleration term on the kinetic diagram, and check that the two agree through the equations of motion.

What you will learn

Define the system and the observer’s reference frame before drawing a diagram.
Separate external forces on a system from the mass–acceleration term used in a kinetic diagram.
Choose and label axes so the free-body diagram, kinetic diagram, and equations use consistent signs.
Use Newton’s second law to check that a proposed diagram is consistent with the motion.

1. Define the system and set up the diagrams

The system is the object or group of objects whose motion you are studying. The observer measures its motion relative to a reference frame. In these examples, the observer is stationary relative to the room or ground, and the frame is treated as inertial. An inertial frame is one in which Newton’s second law can be applied in its usual form.

Choose coordinates that make the forces and motion easy to describe. State which direction is positive on each axis. A positive coordinate direction does not have to match the object’s motion; it is a sign choice. Keep that choice unchanged in the diagram and equations.

To draw a free-body diagram, isolate the chosen system and represent it simply, often as a dot or small block. Add every external force exerted on it by its surroundings. Typical forces are weight, contact normal force, friction, tension, and an applied push or pull. A force exerted by the system on something else does not belong on this diagram.

A kinetic diagram is paired with the free-body diagram. For a particle of mass

m

, show the vector

m\mathbf{a}

in the direction of its acceleration. It is a representation of the left side of Newton’s second law, not an additional physical force. In a coordinate equation, its components are

ma_x

and

ma_y

.

\sum \mathbf{F}_{\mathrm{ext}}=m\mathbf{a}

State the system, observer, reference frame, axes, and positive directions.
The free-body diagram contains external forces on the system only.
The kinetic diagram represents mass times acceleration, not an interaction force.

2. Make signs, force directions, and acceleration agree

A contact force must match the contact. A surface’s normal force is perpendicular to the surface and pushes away from it. Friction, when present, acts along the surface and opposes the relative sliding or impending sliding at the contact. Weight acts vertically downward near Earth’s surface.

Resolve forces into components using the axes you selected. If a force points opposite a positive axis, its component is negative. The acceleration component can also be negative: that means acceleration points opposite the chosen positive direction, not that the calculation failed.

A useful consistency check is to compare the net force direction with the acceleration direction. Since mass is positive, Newton’s second law requires them to point in the same direction. If the diagram’s arrows and the calculated signs contradict this, revisit the system, force directions, axes, or algebra.

For a planar particle, draw the free-body and kinetic diagrams separately or clearly label their parts. Do not put

m\mathbf{a}

among the applied forces. If acceleration is unknown, choose a trial positive direction; a negative answer reports the opposite direction.

\sum F_x=ma_x,\qquad \sum F_y=ma_y

Normal force is perpendicular to the contact; friction is parallel to it.
A negative component indicates direction relative to the selected axes.
A valid diagram and Newton’s second law must predict the same acceleration direction.

3. A practical drawing and solving routine

First define the system and the observer’s frame. Next sketch the physical situation only far enough to identify contacts and directions. Isolate the system, draw and label each external force, then make the kinetic diagram with

m\mathbf{a}

. List known quantities and the requested quantity before writing equations.

Choose coordinates that reduce unnecessary components. For a horizontal surface, horizontal and vertical axes are usually convenient. For a straight incline, axes parallel and perpendicular to the surface often simplify the equations. The axes are mathematical directions; they do not change the actual forces.

Write Newton’s second law component by component. Use the vertical equation to determine a normal force only when the acceleration in that direction is known or constrained. For instance, an object remaining on a level surface has zero vertical acceleration. Do not assume that every normal force equals the object’s weight; other vertical forces or accelerations can change it.

Finally, check units and direction. Force and

ma

must both have units of newtons. Confirm that the sign of the result matches the diagram and the chosen axes. If the acceleration is zero, the net force must also be zero; this is a useful limiting check.

1\ \mathrm{N}=1\ \mathrm{kg\,m/s^2}

Use one system and one set of axes consistently from diagram through equations.
Apply a motion constraint only when the physical situation supports it.
Check dimensions, signs, and the net-force direction.

4. What the paired diagrams do—and do not—say

A free-body diagram records the external interactions relevant to the chosen system. It does not show velocity, displacement, or the path unless those are added to a separate motion sketch. A kinetic diagram records acceleration through

m\mathbf{a}

; it does not show the forces that caused that acceleration.

For an object moving at constant velocity, acceleration is zero, so the kinetic diagram has no nonzero

m\mathbf{a}

vector. The free-body diagram may still contain several forces, but their vector sum must be zero. Conversely, a nonzero acceleration requires a nonzero net external force.

The diagram is a model, so it must match the stated physical situation. For example, do not draw a normal force if there is no contact surface, and do not draw friction at an ideal smooth contact. Distinguish mass, measured in kilograms, from weight, a force measured in newtons.

\mathbf{a}=\mathbf{0}\ \Longrightarrow\ \sum\mathbf{F}_{\mathrm{ext}}=\mathbf{0}

Velocity is not acceleration: constant velocity corresponds to zero acceleration.
Zero acceleration means zero net external force, not necessarily no forces.
Include only interactions that actually act on the selected system.

Worked example

A crate pulled across a level floor

A

12\ \mathrm{kg}

crate is pulled horizontally to the right by a

50\ \mathrm{N}

force. The floor exerts a

14\ \mathrm{N}

friction force to the left. Find the acceleration and normal force. Treat the crate as a particle and use a ground-fixed frame.

Crate free-body diagram

Free-body diagram; pair it with a kinetic diagram showing ma to the right.

Define the system and axes
The system is the crate; the observer is fixed to the floor. Take positive $x$ to the right and positive $y$ upward. The crate remains in contact with the level floor, so its vertical acceleration is zero.
Draw the paired diagrams
The free-body diagram contains the pull, friction, normal force, and weight. The kinetic diagram contains $m\mathbf{a}$ to the right if the net horizontal force is positive. Weight is $mg=(12)(9.81)=117.72\ \mathrm{N}$ downward.
Apply Newton’s second law
Horizontally, the pull is positive and friction is negative. Vertically, zero acceleration means the normal force balances the weight.
$50-14=12a_x,\qquad N-117.72=0$
Solve and check
The acceleration is positive, agreeing with the rightward kinetic vector. The normal force is upward. Both sides of each force equation have units of newtons; dividing by kilograms gives acceleration in metres per second squared.
$a_x=3.00\ \mathrm{m/s^2},\qquad N=117.72\ \mathrm{N}$

Answer: The crate accelerates at

3.00\ \mathrm{m/s^2}

to the right. The normal force is

117.72\ \mathrm{N}

upward.

Check: The horizontal net force is

50-14=36\ \mathrm{N}

to the right, and

ma=(12)(3.00)=36\ \mathrm{N}

. Vertically, normal force and weight cancel.

Worked example

A block on a smooth incline

A

5.0\ \mathrm{kg}

block is released from rest on a smooth straight incline at

30^\circ

above the horizontal. Find its acceleration and the normal force. Use a ground-fixed frame and treat the block as a particle.

Incline free-body diagram

Axes are parallel and perpendicular to the incline; no friction acts.

Define the system and directions
The system is the block, observed from the stationary ground frame. Take positive $s$ down the incline and positive $n$ perpendicular away from the surface. The block starts from rest, but its acceleration is not zero.
Draw force and kinetic diagrams
The smooth incline exerts a normal force perpendicular to its surface and no friction force. Weight acts vertically downward. Its component along the incline points in positive $s$ ; the kinetic diagram therefore has $ma_s$ down the slope.
Resolve weight and apply the equations
The components of weight are $mg\sin 30^\circ$ down the slope and $mg\cos 30^\circ$ into the surface. There is no acceleration normal to the incline because the block stays in contact with it.
$mg\sin 30^\circ=ma_s,\qquad N-mg\cos 30^\circ=0$
Calculate and verify
Cancel the positive mass in the along-slope equation. The positive result confirms the assumed downhill direction. The normal force is less than the full weight, as expected for this inclined contact.
$a_s=4.905\ \mathrm{m/s^2},\qquad N=42.5\ \mathrm{N}$

Answer: The acceleration is

4.905\ \mathrm{m/s^2}

down the incline, and the normal force is approximately

42.5\ \mathrm{N}

away from the incline.

Check: The acceleration equals

g\sin 30^\circ

, and the normal force equals

mg\cos 30^\circ

. At a horizontal surface angle of zero, these expressions approach zero along the surface and the full weight normal to it.

Worked example

A suspended load accelerating upward

A

4.0\ \mathrm{kg}

load is lifted vertically by a cable. At one instant its upward acceleration is

2.0\ \mathrm{m/s^2}

. Find the cable tension. Use a ground-fixed frame and treat the load as a particle.

Load free-body diagram

The paired kinetic diagram has ma upward because the specified acceleration is upward.

Set the system and sign convention
The system is the load, and the observer is fixed to the ground. Choose positive $y$ upward. The stated acceleration is $+2.0\ \mathrm{m/s^2}$ ; the load’s instantaneous velocity is not needed to find the tension.
Draw the diagrams
The external forces are cable tension upward and weight downward. The kinetic diagram has $ma_y$ upward. Do not draw the acceleration term as a third force.
Use the vertical force equation
Weight is $mg=(4.0)(9.81)=39.24\ \mathrm{N}$ . With upward positive, tension is positive and weight is negative. Solve the force balance for the unknown tension.
$T-mg=ma_y$
Substitute and check
The result exceeds the weight because the load accelerates upward. If the acceleration were zero, the same equation would give tension equal to weight.
$T=4.0(9.81+2.0)=47.24\ \mathrm{N}$

Answer: The cable tension is

47.24\ \mathrm{N}

upward.

Check: Both

T-mg

and

ma_y

equal

8.0\ \mathrm{N}

, so the net force and acceleration point upward as required.

Common mistakes and how to avoid them

Putting

m\mathbf{a}

on the free-body diagram as if it were an applied force.

Correction: Keep external forces on the free-body diagram and show

m\mathbf{a}

separately on the kinetic diagram.

Drawing friction in whichever direction seems convenient.

Correction: Friction acts along the contact and opposes relative sliding or impending sliding. Use the stated or physically inferred tendency of motion.

Assuming the normal force always equals

mg

.

Correction: Write the force equation perpendicular to the contact. Equality occurs only under suitable conditions, such as a level surface with no other vertical force and zero vertical acceleration.

Treating a negative acceleration result as an impossible answer.

Correction: A negative component means acceleration points opposite to the positive direction chosen for that coordinate.

Lesson summary

Define the system, observer, frame, axes, and positive directions before drawing.
Draw only external forces on the free-body diagram; show $m\mathbf{a}$ on the kinetic diagram.
Resolve forces using the same axes and signs in the equations.
Check that the net force direction matches the acceleration and that the units are consistent.

Check your understanding

Question 1

A particle has zero acceleration. What must be true of its external forces?

There are no external forces.
Their vector sum is zero.
The weight must be zero.
Every external force must point in the positive direction.

Show answer and explanation

Their vector sum is zero.

Newton’s second law gives zero net external force when acceleration is zero. Individual forces may still be present and balance.

Question 2

A block’s positive horizontal axis points right, but its calculated acceleration is negative. What does that mean?

The mass must be negative.
The acceleration points left.
The kinetic diagram should show a force to the left.
The block cannot be accelerating.

Show answer and explanation

The acceleration points left.

A negative component indicates a vector direction opposite the selected positive axis. Acceleration is not an applied force.

Question 3

Which item belongs on a free-body diagram for a crate in contact with a rough floor?

$m\mathbf{a}$
The floor’s friction force on the crate
The crate’s force on the floor
The crate’s velocity

Show answer and explanation

The floor’s friction force on the crate

Friction exerted by the floor is an external force acting on the crate. The mass–acceleration term belongs on the kinetic diagram; the other choices do not represent external forces on the chosen system.

Key terms

System: The object or group of objects selected for analysis.
Reference frame: The coordinate viewpoint relative to which motion is described.
Free-body diagram: A simplified drawing showing external forces acting on the selected system.
Kinetic diagram: A drawing showing the system’s mass–acceleration term in the direction of its acceleration.
Normal force: A contact force perpendicular to the contacting surface.

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3.1 · Draw consistent free-body and kinetic diagrams

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 Engineering Mechanics: Dynamics — Study topic 3.1

What you will learn

1. Define the system and set up the diagrams

2. Make signs, force directions, and acceleration agree

3. A practical drawing and solving routine

4. What the paired diagrams do—and do not—say

A crate pulled across a level floor

A block on a smooth incline

A suspended load accelerating upward

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review