Statics Practice

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

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

This is the particle idealization.

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

Use the perpendicular-distance formula:

So the moment magnitude is

For a couple,

So

The couple moment magnitude is

A pin or hinge can supply two force components:

It does not supply a reaction moment.

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

It allows horizontal motion and does not resist a moment.

The resultant of a uniform load is

So the equivalent force is

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

The resultant equals the area of the triangle:

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

from the zero-load end.

At impending motion,

So

The maximum static friction is

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

So the coordinates are

The centroid is at

Because the load is centered, the reactions are equal.

Using equilibrium:

Taking moments about $A$:

Then

So the reactions are

Replace the uniform load with a single resultant:

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

So

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

This follows directly from joint equilibrium.

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:

From horizontal equilibrium:

So both members are in tension:

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

The maximum static friction is

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

It acts opposite the push.

First find the support reactions. By symmetry,

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

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

Taking moments about the cut, with counterclockwise positive:

So the internal resultants are

Treat the hole as negative area.

Big rectangle:

Hole:

Total area:

Centroid coordinates:

So the centroid is

Count the unknown reactions:

• Pin at $A$: 2 components

• Roller at $B$: 1 component

• Cable at $C$: 1 tension

That gives a total of

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

Additional compatibility or geometry information would be needed.

Replace the triangular load with an equivalent force:

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

Now write equilibrium:

Take moments about $A$:

Then

So the reactions are

On a level floor, the normal force is

At impending motion,

Since the push is 300 N,

so

The force creates a clockwise moment about $A$:

The couple adds directly:

So the net moment is

That is a moment of

clockwise.

Use the center of gravity formula for discrete masses:

Substitute the values:

So

First find the reactions.

Taking moments about the left support:

Then

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

Vertical equilibrium with positive shear upward on the cut face gives

Take moments about the cut, with counterclockwise positive:

So the internal resultants are

Replace the uniform load with a single force:

It acts 1 m from $A$.

Now apply equilibrium:

Take moments about $A$, with counterclockwise positive:

Then

So the reactions are

Use negative area for the removed square.

Large square:

Removed square:

Total area:

Centroid coordinates:

So the centroid is

Count the unknown reactions:

• Pin at $A$: 2

• Roller at $B$: 1

• Cable at $C$: 1

So there are

unknowns in total.

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

There are no horizontal forces, so

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

Now take moments about the cut, with counterclockwise positive:

So the internal resultants are