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7.2 · Determine impending motion and friction direction

Learn to determine impending motion and friction direction through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Friction

How to identify the likely slip direction and use static friction correctly

Friction direction is not chosen by a memorized rule such as “friction opposes the applied force.” It is chosen by asking how the surfaces would move relative to one another if friction did not hold them in place. In a statics problem, a body can remain at rest while static friction adjusts as needed. At impending motion—the instant the available static friction is just at its limit—the friction magnitude equals the limiting value. This lesson develops that reasoning using free-body diagrams and equilibrium.

What you will learn

  • Identify the direction a body would begin to slide if friction could not prevent it.
  • Place static friction on a free-body diagram opposite the impending relative motion at the contact.
  • Use the limiting static-friction relation only when motion is impending.
  • Solve planar equilibrium problems and check both force and moment balance.

1. Decide which way motion is impending

Define the body you are isolating and the contact surface that may slip. Impending motion means the body is on the verge of sliding relative to that surface. It is still treated as being in equilibrium for the calculation; the phrase identifies the direction of the possible slip, not motion that has already begun.
To choose friction direction, temporarily imagine that the contact offers no friction. Use the other forces and the geometry to decide which way the body would tend to slide. Static friction acts along the contact surface in the opposite direction. The two surfaces exert equal and opposite friction forces on one another, so the direction depends on which body is in the free-body diagram.
For example, a block on a slope tends to slide downhill under gravity if no other force holds it. The friction force on the block therefore points uphill. If a force instead makes the block tend to move uphill, friction on the block points downhill. It is the impending relative slip at the contact—not simply the direction of gravity or the applied force—that determines the answer.
  • First identify the isolated body and the contact.
  • Determine the direction of relative slip that would occur without friction.
  • Draw friction on the isolated body opposite that impending slip.

2. Static friction and its limiting value

At a contact, the normal force acts perpendicular to the surface and static friction acts parallel to it. The normal force NN is a force exerted by the surface. The coefficient of static friction, written μs\mu_s, describes the limiting ratio for the contact in the model used here; it has no units.
Static friction is not automatically at its maximum. While the body remains at rest, its actual value is whatever equilibrium requires, up to the limiting value. When the body is just about to slip, use equality. If the equilibrium equations require more friction than the limit allows, the assumed resting condition is not possible under the stated model.
For a rigid body in planar statics, choose convenient horizontal and vertical axes and use force balance in both directions and moment balance about a chosen point. For a block, moments often confirm that the force directions and locations are consistent, even when the force equations alone determine the unknown magnitudes.
0≤Fs≤μsN;Fs=μsN at impending slip0 \leq F_s \leq \mu_s N; \quad F_s=\mu_s N\text{ at impending slip}
  • Before impending motion, static friction can be below its limit.
  • At impending slip, friction has limiting magnitude.
  • Use force and moment equilibrium for the isolated body.

3. A reliable solution sequence

Start by stating the body and the contact whose possible slip matters. Draw its free-body diagram with the weight, applied forces, normal reactions, and friction forces. Show friction in the direction opposite the suspected impending slip. If that direction is uncertain, assume one direction, solve, and interpret a negative result as evidence that the actual direction is opposite.
Choose axes that simplify the geometry. On a horizontal surface, horizontal and vertical axes are natural. On a slope, axes parallel and perpendicular to the slope are often convenient. State the sign convention so that positive force components and positive moments are unambiguous.
Write equilibrium before substituting numbers. Resolve angled forces into components when needed. Use the static-friction inequality if the problem only asks whether rest is possible; use the limiting equality when it states or implies impending motion. Then solve, retain units, and verify force and moment balance independently.
Do not set friction equal to its limiting value merely because a coefficient is supplied. The coefficient gives the maximum available static friction; impending motion is what justifies equality.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Draw the FBD before writing equations.
  • Choose friction direction from impending relative motion.
  • Use limiting equality only for an impending-motion condition.

Worked example

A block pulled across a level floor

A 20 kg block rests on a horizontal floor. A horizontal force pulls it to the right. The coefficient of static friction is 0.30. Find the force required for impending motion to the right. Use g=9.81 m/s2g=9.81\,\text{m/s}^2.
  1. Isolate and choose directions
    The body is the block. Without friction, the pull would make it slip right, so friction on the block points left. The floor's normal force points up; weight points down.
  2. Apply limiting friction and equilibrium
    At impending slip, the friction magnitude is its limiting value. Vertical equilibrium makes the normal force equal to the weight. Horizontal equilibrium then makes the pull equal to friction.
    N=mg,P=Fs=μsNN=mg,\quad P=F_s=\mu_sN
  3. Calculate the force
    The weight and normal force are each 196.2 N196.2\,\text{N}. The limiting friction, and thus the required pull, is 58.9 N58.9\,\text{N} to the right.
    P=0.30(20 kg)(9.81 m/s2)=58.9 NP=0.30(20\,\text{kg})(9.81\,\text{m/s}^2)=58.9\,\text{N}
  4. Verify equilibrium
    The vertical forces cancel, and the horizontal pull is balanced by friction. The forces act through the block's centre in this idealized diagram, so their net moment about that point is zero.
    ∑Fx=58.9−58.9=0,∑Fy=196.2−196.2=0,∑M=0\sum F_x=58.9-58.9=0,\quad \sum F_y=196.2-196.2=0,\quad \sum M=0
Answer: The force required for impending rightward motion is 58.9 N58.9\,\text{N}.
Check: The friction value is at its limit: 0.30(196.2 N)=58.9 N0.30(196.2\,\text{N})=58.9\,\text{N}.

Worked example

A block pushed up a slope

A 12 kg block rests on a slope that rises to the right at 20∘20^\circ. A force parallel to the slope pushes the block uphill. If μs=0.35\mu_s=0.35, find the push required for impending uphill motion. Use g=9.81 m/s2g=9.81\,\text{m/s}^2.
  1. Set the impending direction
    The isolated body is the block. Since it is about to move uphill, friction on it points downhill. Choose the positive axis uphill and the perpendicular positive axis outward from the slope.
  2. Resolve weight and use equilibrium
    Weight has a component down the slope and a component into the slope. There is no acceleration in this statics model, so the normal force balances the into-slope component. At impending uphill slip, friction is limiting and acts downhill.
    N=mgcos⁡20∘,P−mgsin⁡20∘−μsN=0N=mg\cos 20^\circ,\quad P-mg\sin 20^\circ-\mu_sN=0
  3. Substitute values
    The normal force is about 110.5 N110.5\,\text{N}. The uphill push must balance both the downhill weight component and limiting friction, giving about 79.7 N79.7\,\text{N}.
    P=12(9.81)(sin⁡20∘+0.35cos⁡20∘)=79.7 NP=12(9.81)(\sin 20^\circ+0.35\cos 20^\circ)=79.7\,\text{N}
  4. Check force and moment balance
    In the slope-parallel direction, the push balances the two downhill contributions. In the normal direction, the normal force balances the weight component into the surface. In the idealized point-contact representation, the forces are concurrent at the block's centre, so moment balance there is also satisfied.
    ∑F∥=79.7−40.2−39.5=0,∑F⊥=110.5−110.5=0,∑M=0\sum F_{\parallel}=79.7-40.2-39.5=0,\quad \sum F_{\perp}=110.5-110.5=0,\quad \sum M=0
Answer: The push for impending uphill motion is 79.7 N79.7\,\text{N} uphill.
Check: The friction direction is downhill, opposite the specified impending uphill slip.

Worked example

A block pressed against a vertical wall

A 240 N block is pressed horizontally against a vertical wall. The coefficient of static friction is 0.50. Find the minimum horizontal push that can keep the block at rest when it is on the verge of slipping downward.
  1. Identify contact forces
    The block is the isolated body. It would slip downward without friction, so friction from the wall on the block points up. The push presses the block into the wall, and the wall's normal force points left.
  2. Use limiting friction
    At the stated threshold, upward friction balances the 240 N weight. The horizontal push balances the normal force. The limiting relation connects the friction and normal forces.
    Fs=240 N,Fs=0.50N,P=NF_s=240\,\text{N},\quad F_s=0.50N,\quad P=N
  3. Solve and verify
    The normal force must be 480 N480\,\text{N}, so the required push is also 480 N480\,\text{N}. Vertical forces balance, as do horizontal forces. Taking moments about the block's centre gives zero because the idealized forces pass through it.
    N=P=240 N0.50=480 N;∑Fx=480−480=0,∑Fy=240−240=0,∑M=0N=P=\frac{240\,\text{N}}{0.50}=480\,\text{N};\quad \sum F_x=480-480=0,\quad \sum F_y=240-240=0,\quad \sum M=0
Answer: The minimum horizontal push for equilibrium at impending downward slip is 480 N480\,\text{N}.
Check: At the threshold, the wall can supply at most 0.50(480 N)=240 N0.50(480\,\text{N})=240\,\text{N} of upward friction.

Common mistakes and how to avoid them

Always drawing friction opposite the applied force.
Correction: Determine the impending relative slip at the contact. Friction on the chosen body points opposite that slip.
Setting static friction equal to μsN\mu_sN in every resting problem.
Correction: Use the inequality for ordinary rest. Use equality only when the body is at the threshold of slipping.
Drawing friction perpendicular to the surface.
Correction: Friction acts parallel to the contact surface; the normal force acts perpendicular to it.
Treating a negative assumed friction value as an impossible result.
Correction: A negative result means the actual direction is opposite the direction initially assumed. Redraw or interpret the force accordingly.

Lesson summary

  • Friction direction is found from the impending relative slip, not from a general rule about applied forces.
  • Static friction adjusts as needed up to its limit; at impending slip, its magnitude is μsN\mu_sN.
  • Use a free-body diagram, stated axes, equilibrium equations, and force and moment checks.

Check your understanding

Question 1

A block on a rough horizontal floor is about to slide left. Which way does friction on the block act?
  1. Left
  2. Right
  3. Up
  4. Down
Show answer and explanation
Right
Friction on the block opposes its impending relative slip, so it acts to the right.

Question 2

A body is at rest, and equilibrium requires a friction force smaller than μsN\mu_sN. What is the correct conclusion?
  1. The body cannot remain at rest because friction must equal μsN\mu_sN.
  2. Static friction can take the required smaller value.
  3. The normal force must be zero.
  4. Friction must point in the same direction as impending slip.
Show answer and explanation
Static friction can take the required smaller value.
Static friction can adjust below its limiting value. Equality is used only at impending motion.

Question 3

A block is about to slide downhill on a slope. Which direction is friction on the block?
  1. Down the slope
  2. Up the slope
  3. Perpendicular into the slope
  4. Vertically downward
Show answer and explanation
Up the slope
The impending slip is downhill, so friction acts uphill along the contact.

Key terms

Impending motion
The condition in which a body is just about to slip relative to a contact surface.
Static friction
A contact force parallel to the surface that resists relative slipping while the body remains at rest.
Normal force
The contact force perpendicular to the surface.
Limiting static friction
The greatest static-friction magnitude in the model, equal to μsN\mu_sN at impending slip.

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