Free body diagrams

Engineering · First-year engineering · about 10 min

A free body diagram (FBD) is a sketch of one object with every external force acting on it. It turns a complicated scene into a small set of force equations. A reliable FBD is the first step in questions about supports, ropes, friction and acceleration.

The key idea

For one isolated body, the vector sum of the external forces is

∑F=ma\sum\mathbf{F}=m\mathbf{a}

Here F\mathbf{F} denotes force, mm is mass and a\mathbf{a} is acceleration. Resolve the equation along independent axes, usually ∑Fx=max\sum F_x=ma_x and ∑Fy=may\sum F_y=ma_y.

Show one isolated rectangular crate seen from the side, centred with no floor in contact. Draw an upward vertical arrow starting at the midpoint of its bottom face for the normal force N, a downward vertical arrow from its centre for weight mg, and a sloping arrow from its upper-right corner directed 30 degrees above the horizontal to the right for rope tension T. Draw short positive x and y coordinate arrows to the lower left of the crate, x right and y up. Mark a small 30-degree angle arc between the horizontal ray through the rope attachment and the tension arrow. No acceleration arrow, friction arrow or force components. Make arrow tips and their origins unambiguous.
Only external forces acting on the isolated crate appear in its diagram.

Try it yourself

Move and rotate the force arrows, then compare their components with the resultant.
Move the load and cable angles, then find the cable tensions that keep the point in equilibrium.

Isolate the object

Sketch the object alone, replacing each contact with the force that contact exerts on the object. Gravity acts vertically down with magnitude mgmg. A rope pulls along its length, away from the object. A smooth surface exerts a normal force perpendicular to the surface. Friction, when present, lies along the contact and opposes relative motion or the tendency to slip. Put forces at their relevant contact points if rotation matters; for particle-like translation problems, their directions and magnitudes are the priority. Do not add a force that the object exerts on something else.

Resolve forces into components

Choose axes and mark their positive directions before writing equations. If a pull TT makes an angle θ\theta above the positive horizontal, its components are Tx=Tcos⁡θT_x=T\cos\theta and Ty=Tsin⁡θT_y=T\sin\theta. Components pointing against a chosen axis are negative. Add all components algebraically, rather than adding the magnitudes of slanted forces. For instance, a vertical force cannot directly change horizontal acceleration, although it can change the normal force and therefore the available friction. An FBD should contain the original forces; component arrows can be shown separately if they improve clarity.

Apply the motion condition

For a body at rest or moving with constant velocity, ax=ay=0a_x=a_y=0, so ∑Fx=0\sum F_x=0 and ∑Fy=0\sum F_y=0. For an accelerating body, use the measured or unknown acceleration in ∑F=ma\sum F=ma along each axis. If a crate stays in contact with a horizontal floor, its vertical acceleration is zero even when it accelerates horizontally. The normal reaction is whatever the vertical balance requires, provided it stays non-negative. It is not automatically equal to mgmg, particularly when another force has a vertical component.

Worked example: A crate pulled by an angled rope

A 12.0 kg12.0\,\mathrm{kg} crate is pulled across a smooth horizontal floor by a 50.0 N50.0\,\mathrm{N} rope at 30∘30^\circ above the horizontal. The crate remains on the floor. Take g=9.81 m/s2g=9.81\,\mathrm{m/s^2}. Find its horizontal acceleration and the floor's normal reaction.

Side view of a rectangular crate resting on a long horizontal floor, with a rope attached at the upper-right corner, extending upward to the right at 30 degrees. Put the mass inside the crate. Draw a force arrow pointing along the rope away from the crate, with its given force next to the arrow. Add a short horizontal construction ray from the rope attachment and an arc showing 30 degrees above that ray. Place a downward arrow mg through the crate centre and an upward arrow N from the floor contact. Do not label either unknown acceleration or unknown normal with a numerical value. No friction arrow because the floor is smooth.
The vertical component of the pull reduces the normal reaction.
Show the solution

Choose +x+x right and +y+y up. The floor is smooth, so friction is zero.

  1. The downward weight is
mg=12.0(9.81)=117.72 N.mg=12.0(9.81)=117.72\,\mathrm{N}.
  1. Resolve the rope force:
Tx=50.0cos⁡30∘=43.301 N,Ty=50.0sin⁡30∘=25.0 N.T_x=50.0\cos30^\circ=43.301\,\mathrm{N},\qquad T_y=50.0\sin30^\circ=25.0\,\mathrm{N}.
  1. Use horizontal Newton's second law:
∑Fx=Tx=max⇒ax=43.30112.0=3.608 m/s2.\sum F_x=T_x=ma_x\quad\Rightarrow\quad a_x=\frac{43.301}{12.0}=3.608\,\mathrm{m/s^2}.
  1. As the crate has no vertical acceleration,
∑Fy=N+25.0−117.72=0⇒N=92.72 N.\sum F_y=N+25.0-117.72=0\quad\Rightarrow\quad N=92.72\,\mathrm{N}.

Answer: ax=3.61 m/s2a_x=3.61\,\mathrm{m/s^2} to the right and N=92.72 NN=92.72\,\mathrm{N} upward.

Common mistakes

Drawing forces exerted by the object on other objects

Keep only forces acting on the isolated object. Reaction partners belong on another FBD.

Treating a slanted rope force as entirely horizontal

Resolve it into horizontal and vertical components using the stated angle.

Setting the normal reaction equal to weight without checking

Write the vertical force equation first. Vertical pulls can reduce the normal force.

Using equilibrium equations for an accelerating object

Use ∑F=ma\sum F=ma and set a component to zero only when acceleration in that direction is zero.

Check your understanding

1. What is the weight of a 5.00 kg5.00\,\mathrm{kg} object when g=9.81 m/s2g=9.81\,\mathrm{m/s^2}?

2. A crate remains stationary while a 20 N20\,\mathrm{N} horizontal push acts to the right. What horizontal static-friction force acts on it?

3. Two forces of 30 N30\,\mathrm{N} right and 40 N40\,\mathrm{N} up act on a particle. What is the resultant magnitude?

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