Statics Practice

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1.1State the Plane Equilibrium Equations

Exam I | Problem 1.1 |

Equilibrium Equations

Write the three independent equilibrium equations for a rigid body in a plane.

Solution

For a 2D rigid body, the three equilibrium equations are

\sum F_x = 0, \qquad \sum F_y = 0, \qquad \sum M_O = 0

The moment may be taken about any convenient point $O$.

1.2Identify a Particle Model

Exam I | Problem 1.2 |

Modeling Assumptions

A body is modeled as having mass but no size, so only translation matters. What idealization is this?

Solution

This is the particle idealization.

A particle has mass but no dimensions, so rotation is ignored and only force balance matters.

1.3Compute a Force Moment

Exam I | Problem 1.3 |

Moments

A 40 N force acts perpendicular to a wrench 0.6 m from point $O$. Find the moment magnitude about $O$.

Solution

Use the perpendicular-distance formula:

$$ M_O = Fd = 40(0.6) = 24 $$

So the moment magnitude is

24\ \text{N} \cdot \text{m}

1.4Compute a Couple Moment

Exam I | Problem 1.4 |

Two equal and opposite 25 N forces are separated by 0.30 m. Find the couple moment magnitude.

Solution

For a couple,

$$ M = Fd $$

$$ M = 25(0.30) = 7.5 $$

The couple moment magnitude is

7.5\ \text{N} \cdot \text{m}

1.5Identify Pin Reactions

Exam I | Problem 1.5 |

Support Reactions

In 2D, what reaction components can a pin or hinge supply?

Solution

A pin or hinge can supply two force components:

R_x \quad \text{and} \quad R_y

It does not supply a reaction moment.

1.6Identify Roller Reactions

Exam I | Problem 1.6 |

Support Reactions

A roller sits on a smooth horizontal surface. What reaction does it supply?

Solution

A roller on a smooth horizontal surface supplies one vertical normal reaction.

It allows horizontal motion and does not resist a moment.

1.7Replace a Uniform Load

Exam I | Problem 1.7 |

Replace a uniform load of 5 kN/m acting over 4 m with a single equivalent force. Give the magnitude and location.

Solution

The resultant of a uniform load is

$$ R = wL = 5(4) = 20 $$

So the equivalent force is

20\ \text{kN}

It acts at the midpoint of the loaded region, so the line of action is 2 m from either end.

1.8Replace a Triangular Load

Exam I | Problem 1.8 |

Replace a triangular load that increases from 0 to 9 kN/m over 6 m with a single equivalent force. Give the magnitude and its location from the zero-load end.

Solution

The resultant equals the area of the triangle:

R = \frac{1}{2}(9)(6) = 27

The centroid of a triangular load lies two-thirds of the way from the zero end, so the line of action is

\frac{2}{3}(6) = 4\ \text{m}

from the zero-load end.

1.9Find Maximum Static Friction

Exam I | Problem 1.9 |

Friction

If $\mu_s = 0.35$ and $N = 400$ N, what is the maximum static friction?

Solution

At impending motion,

f_s^{\max} = \mu_s N

f_s^{\max} = 0.35(400) = 140

The maximum static friction is

140\ \text{N}

1.10Locate the Centroid of a Rectangle

Exam I | Problem 1.10 |

A 8 cm by 10 cm rectangular plate is measured from a lower-left corner. Where is its centroid?

Solution

For a rectangle, the centroid is at its geometric center.

So the coordinates are

\bar{x} = \frac{8}{2} = 4, \qquad \bar{y} = \frac{10}{2} = 5

The centroid is at

(4,\ 5)\ \text{cm}

Difficulty

Integrated

2.1Find Beam Reactions for a Center Load

Exam II | Problem 2.1 |

A 6 m simply supported beam carries a 12 kN point load at midspan. Find the support reactions at $A$ and $B$.

Solution

Because the load is centered, the reactions are equal.

Using equilibrium:

$$ R_A + R_B = 12 $$

Taking moments about $A$:

$$ 6R_B - 12(3) = 0 $$

$$ R_B = 6 $$

Then

$$ R_A = 12 - 6 = 6 $$

So the reactions are

R_A = 6\ \text{kN}, \qquad R_B = 6\ \text{kN}

2.2Find Reactions Under a Uniform Load

Exam II | Problem 2.2 |

A 5 m simply supported beam carries a uniform load of 2 kN/m over its entire span. Find the support reactions.

Solution

Replace the uniform load with a single resultant:

R = 2(5) = 10\ \text{kN}

The resultant acts at the midpoint of the beam, so the reactions are equal by symmetry:

R_A = R_B = \frac{10}{2} = 5

R_A = 5\ \text{kN}, \qquad R_B = 5\ \text{kN}

2.3Use the Zero-Force Member Rule

Exam II | Problem 2.3 |

At an unloaded truss joint, three members meet. Two are collinear. Which member carries zero force?

Solution

The third member, the one that is not collinear with the other two, is a zero-force member.

This follows directly from joint equilibrium.

2.4Solve a Two-Force Joint

Exam II | Problem 2.4 |

At a truss joint, member $AB$ is horizontal to the left, and member $AC$ rises 3 m for every 4 m of horizontal run to the right. A 12 kN downward load acts at the joint. Find the force in each member, assuming member forces are tensile if they pull away from the joint.

Solution

Let the member forces be positive in tension.

For member $AC$, the direction ratios are $4/5$ in $x$ and $3/5$ in $y$.

From vertical equilibrium:

\frac{3}{5}F_{AC} - 12 = 0

F_{AC} = 20\ \text{kN}

From horizontal equilibrium:

-F_{AB} + \frac{4}{5}(20) = 0

F_{AB} = 16\ \text{kN}

So both members are in tension:

F_{AB} = 16\ \text{kN}, \qquad F_{AC} = 20\ \text{kN}

2.5Find the Friction Force at Impending Slip

Exam II | Problem 2.5 |

A 200 N crate sits on a horizontal floor. A horizontal push of 40 N is applied. If $\mu_s = 0.25$, what friction force acts on the crate?

Solution

First find the normal force. Since the push is horizontal,

N = 200\ \text{N}

The maximum static friction is

f_s^{\max} = \mu_s N = 0.25(200) = 50\ \text{N}

Because the applied push is only 40 N, the crate does not slip. Static friction matches the push:

f_s = 40\ \text{N}

It acts opposite the push.

2.6Find Internal Forces at a Cut

Exam II | Problem 2.6 |

A 4 m simply supported beam carries an 8 kN point load at midspan. Find the internal shear force and bending moment at a section 3 m from the left support, using the left segment.

Solution

First find the support reactions. By symmetry,

R_A = R_B = 4\ \text{kN}

Now cut the beam at $x = 3$ m and isolate the left segment.

Using vertical equilibrium with positive shear upward on the cut face:

$$ 4 - 8 + V = 0 $$

V = 4\ \text{kN}

Taking moments about the cut, with counterclockwise positive:

$$ -4(3) + 8(1) + M = 0 $$

M = 4\ \text{kN} \cdot \text{m}

So the internal resultants are

V = 4\ \text{kN}, \qquad M = 4\ \text{kN} \cdot \text{m}

2.7Find the Centroid of a Composite Area

Exam II | Problem 2.7 |

A 6 cm by 4 cm rectangle has a 2 cm by 2 cm square hole removed from its upper-right corner. Find the centroid of the remaining area relative to the lower-left corner.

Solution

Treat the hole as negative area.

Big rectangle:

A_1 = 24, \qquad (x_1, y_1) = (3, 2)

Hole:

A_2 = 4, \qquad (x_2, y_2) = (5, 3)

Total area:

$$ A = 24 - 4 = 20 $$

Centroid coordinates:

\bar{x} = \frac{24(3) - 4(5)}{20} = \frac{52}{20} = 2.6

\bar{y} = \frac{24(2) - 4(3)}{20} = \frac{36}{20} = 1.8

So the centroid is

(2.6,\ 1.8)\ \text{cm}

2.8Check Static Determinacy

Exam II | Problem 2.8 |

A 2D beam is supported by a pin at $A$, a roller at $B$, and a cable at $C$. How many independent reaction unknowns are there, and is equilibrium alone enough to solve them?

Solution

Count the unknown reactions:

Pin at $A$: 2 components
Roller at $B$: 1 component
Cable at $C$: 1 tension

That gives a total of

$$ 2 + 1 + 1 = 4 $$

In 2D, rigid-body equilibrium provides only 3 independent equations, so equilibrium alone is not enough.

Additional compatibility or geometry information would be needed.

Difficulty

Applied

3.1Balance a Mixed Beam Load

Final | Problem 3.1 |

A 6 m simply supported beam carries a triangular load that increases from 0 at the left end to 6 kN/m at the right end, plus a 6 kN point load located 2 m from the left support. Find the support reactions.

Solution

Replace the triangular load with an equivalent force:

R_w = \frac{1}{2}(6)(6) = 18\ \text{kN}

Its line of action is 4 m from the left end.

Now write equilibrium:

$$ R_A + R_B = 18 + 6 = 24 $$

Take moments about $A$:

$$ 6R_B - 18(4) - 6(2) = 0 $$

$$ 6R_B = 84 $$

R_B = 14\ \text{kN}

Then

R_A = 24 - 14 = 10\ \text{kN}

So the reactions are

R_A = 10\ \text{kN}, \qquad R_B = 14\ \text{kN}

3.2Required Coefficient of Static Friction

Final | Problem 3.2 |

Friction

A 600 N crate is pushed horizontally with 300 N just before it moves. What minimum coefficient of static friction is required?

Solution

On a level floor, the normal force is

N = 600\ \text{N}

At impending motion,

f_s^{\max} = \mu_s N

Since the push is 300 N,

300 = \mu_s(600)

\mu_s = 0.5

3.3Combine a Force and a Couple

Final | Problem 3.3 |

A 10 kN downward force acts 2 m to the right of point $A$, and a 4 kN*m counterclockwise couple also acts on the body. What is the net moment about $A$? Take counterclockwise as positive.

Solution

The force creates a clockwise moment about $A$:

M_F = -10(2) = -20\ \text{kN} \cdot \text{m}

The couple adds directly:

M_C = +4\ \text{kN} \cdot \text{m}

So the net moment is

M_A = -20 + 4 = -16\ \text{kN} \cdot \text{m}

That is a moment of

16\ \text{kN} \cdot \text{m}

clockwise.

3.4Find the Center of Gravity of Discrete Masses

Final | Problem 3.4 |

Three masses lie on a line at $x = 0$ m, $x = 2$ m, and $x = 5$ m. Their masses are 2 kg, 3 kg, and 5 kg. Find $\bar{x}$.

Solution

Use the center of gravity formula for discrete masses:

\bar{x} = \frac{\sum m_i x_i}{\sum m_i}

Substitute the values:

\bar{x} = \frac{2(0) + 3(2) + 5(5)}{2 + 3 + 5}

\bar{x} = \frac{0 + 6 + 25}{10} = \frac{31}{10} = 3.1

\bar{x} = 3.1\ \text{m}

3.5Find Shear and Moment at a Section

Final | Problem 3.5 |

A 8 m simply supported beam carries a 12 kN point load 2 m from the left support. Find the internal shear force and bending moment at a section 3 m from the left support.

Solution

First find the reactions.

Taking moments about the left support:

$$ 8R_B - 12(2) = 0 $$

R_B = 3\ \text{kN}

Then

R_A = 12 - 3 = 9\ \text{kN}

Now cut the beam at $x = 3$ m and use the left segment.

Vertical equilibrium with positive shear upward on the cut face gives

$$ 9 - 12 + V = 0 $$

V = 3\ \text{kN}

Take moments about the cut, with counterclockwise positive:

$$ -9(3) + 12(1) + M = 0 $$

M = 15\ \text{kN} \cdot \text{m}

So the internal resultants are

V = 3\ \text{kN}, \qquad M = 15\ \text{kN} \cdot \text{m}

Difficulty

Challenge

4.1Load a Beam with Multiple Actions

Final | Problem 4.1 |

A 8 m beam is supported by a pin at $A$ and a roller at $B$. It carries a 4 kN/m uniform load over the first 2 m from $A$, an 8 kN point load 5 m from $A$, and an 8 kN*m counterclockwise couple. Find the reactions at $A$ and $B$.

Solution

Replace the uniform load with a single force:

R_w = 4(2) = 8\ \text{kN}

It acts 1 m from $A$.

Now apply equilibrium:

$$ R_A + R_B = 8 + 8 = 16 $$

Take moments about $A$, with counterclockwise positive:

$$ 8R_B + 8 - 8(1) - 8(5) = 0 $$

$$ 8R_B = 40 $$

R_B = 5\ \text{kN}

Then

R_A = 16 - 5 = 11\ \text{kN}

So the reactions are

R_A = 11\ \text{kN}, \qquad R_B = 5\ \text{kN}

4.2Find the Centroid of an L-Section

Final | Problem 4.2 |

An 8 cm by 8 cm square has a 4 cm by 4 cm square removed from its upper-right corner. Find the centroid of the remaining area relative to the lower-left corner.

Solution

Use negative area for the removed square.

Large square:

A_1 = 64, \qquad (x_1, y_1) = (4, 4)

Removed square:

A_2 = 16, \qquad (x_2, y_2) = (6, 6)

Total area:

$$ A = 64 - 16 = 48 $$

Centroid coordinates:

\bar{x} = \frac{64(4) - 16(6)}{48} = \frac{160}{48} = \frac{10}{3}

\bar{y} = \frac{64(4) - 16(6)}{48} = \frac{10}{3}

So the centroid is

\left(\frac{10}{3},\ \frac{10}{3}\right)\ \text{cm}

4.3Decide Whether Equilibrium Alone Is Enough

Final | Problem 4.3 |

A rigid body in 2D is supported by a pin at $A$, a roller at $B$, and a cable at $C$. How many independent reaction unknowns are present, and can equilibrium alone determine them?

Solution

Count the unknown reactions:

Pin at $A$: 2
Roller at $B$: 1
Cable at $C$: 1

So there are

$$ 2 + 1 + 1 = 4 $$

unknowns in total.

In 2D, a rigid body has only 3 independent equilibrium equations, so equilibrium alone is not enough to solve the problem.

4.4Recover Internal Forces from a Cut Segment

Final | Problem 4.4 |

A beam segment has a 10 kN upward force at the left end, a 6 kN downward force 1 m from the left end, and a 4 kN*m clockwise couple at 2 m from the left end. A cut is made 3 m from the left end. Using the common left-face convention, find $N$, $V$, and $M$ at the cut.

Solution

There are no horizontal forces, so

$$ N = 0 $$

For vertical equilibrium, with positive shear upward on the cut face:

$$ 10 - 6 + V = 0 $$

V = -4\ \text{kN}

Now take moments about the cut, with counterclockwise positive:

$$ -10(3) + 6(2) - 4 + M = 0 $$

M = 22\ \text{kN} \cdot \text{m}

So the internal resultants are

N = 0, \qquad V = -4\ \text{kN}, \qquad M = 22\ \text{kN} \cdot \text{m}

Statics Practice

Selected problems

Quick reference

Direct

1.1State the Plane Equilibrium Equations

1.2Identify a Particle Model

1.3Compute a Force Moment

1.4Compute a Couple Moment

1.5Identify Pin Reactions

1.6Identify Roller Reactions

1.7Replace a Uniform Load

1.8Replace a Triangular Load

1.9Find Maximum Static Friction

1.10Locate the Centroid of a Rectangle

Integrated

2.1Find Beam Reactions for a Center Load

2.2Find Reactions Under a Uniform Load

2.3Use the Zero-Force Member Rule

2.4Solve a Two-Force Joint

2.5Find the Friction Force at Impending Slip

2.6Find Internal Forces at a Cut

2.7Find the Centroid of a Composite Area

2.8Check Static Determinacy

Applied

3.1Balance a Mixed Beam Load

3.2Required Coefficient of Static Friction

3.3Combine a Force and a Couple

3.4Find the Center of Gravity of Discrete Masses

3.5Find Shear and Moment at a Section

Challenge

4.1Load a Beam with Multiple Actions

4.2Find the Centroid of an L-Section

4.3Decide Whether Equilibrium Alone Is Enough

4.4Recover Internal Forces from a Cut Segment