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7.4 · Analyze wedges with dry friction

Learn to analyze wedges with dry friction through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Friction

Contact forces, impending motion, and equilibrium in simple wedge systems

A wedge has a sloped surface that transfers force through contact with another body. The central challenge is choosing the friction direction correctly. Friction opposes the tendency of one contacting surface to slide relative to the other; it does not always oppose the motion of the wedge itself. Isolate the wedge and the supported body, show the contact forces on each, then apply planar equilibrium. In the examples, the bodies are idealized as particles: their force resultants are treated as concurrent at the contact point. This makes their moment balance automatic, while keeping the calculations focused on wedge friction.

What you will learn

  • Draw and interpret free-body diagrams for a wedge and the body it contacts.
  • Resolve normal and friction forces using clear axes and signs.
  • Apply the dry-friction model to wedge contacts and determine the impending-motion direction.
  • Solve wedge equilibrium problems and independently check force and moment balance.

1. Contact forces and dry friction

At a contact, the normal force NN acts perpendicular to the touching surfaces and pushes the bodies apart. The friction force FF acts along the surfaces. For a body that remains at rest, static friction adjusts as needed up to a limiting value: 0≤F≤μsN0\leq F\leq\mu_sN, where μs\mu_s is the coefficient of static friction.
When relative sliding is impending, friction is at its limiting value, F=μsNF=\mu_sN. Determine its direction by considering how the surfaces would slide relative to each other if friction were absent. The friction forces on the two bodies at the same contact are equal in magnitude and opposite in direction.
For a surface rising to the right at angle α\alpha, the up-slope direction is (cos⁡α,sin⁡α)(\cos\alpha,\sin\alpha). A normal force on a body resting on the surface points up and left, with components (−Nsin⁡α,Ncos⁡α)(-N\sin\alpha,N\cos\alpha). A down-slope friction force on that body has components (−Fcos⁡α,−Fsin⁡α)(-F\cos\alpha,-F\sin\alpha). Use the reverse directions when the actual tendency calls for them.
0≤F≤μsN0\leq F\leq\mu_sN
  • Use F=μsNF=\mu_sN at impending slip; otherwise use 0≤F≤μsN0\leq F\leq\mu_sN.
  • Determine friction direction from relative sliding tendency at each contact.
  • Contact forces on the two bodies are opposite, not identical.

2. Free-body diagrams and equilibrium

Define the system before solving. In a wedge problem, it is often useful to isolate the supported body and the wedge separately. Show each body's weight, applied force, guide or floor reactions, and both the normal and friction forces at every rough contact. A smooth contact supplies a normal force only; a rough contact may also supply friction.
Choose horizontal xx to the right and vertical yy upward. Take counterclockwise moments as positive. For each isolated body in planar equilibrium, the horizontal and vertical force sums and the moment sum about any point must each be zero. A particle model has concurrent forces, so its moment sum about the common point is zero. For a rigid-body model with separated lines of action, write the moment equation using the actual force locations instead.
Resolve inclined forces into the chosen axes before combining them. Keep force units, such as newtons, distinct from moment units, such as newton-metres. If a solved force is negative, its actual direction is opposite the assumed direction; check that the result still matches the stated motion tendency.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad\sum F_y=0,\quad\sum M_O=0
  • Draw separate free-body diagrams so the action-reaction pair is clear.
  • Include floor friction when the wedge base is rough.
  • Check both force components and moment equilibrium; do not assume moment balance for separated force lines.

3. Solving and checking wedge problems

For impending motion, first state the direction of relative slip at each rough contact. This sets the friction directions and the limiting relation. Then write equilibrium for the supported body, which often gives the sloped-contact normal force. Use the equal-and-opposite contact forces on the wedge to find the required applied force. If the base is rough, use a separate normal force and friction relation for that contact.
A useful check is to return to each isolated body after solving. Add horizontal and vertical force components independently. For a rigid body, also sum moments using the actual perpendicular distances to the force lines of action. For the particle idealization used here, the force lines are concurrent at the idealized contact point, so the moments about that point are zero. Verify that normal forces are nonnegative and that any friction not at impending slip remains within its limiting value.
Fmax⁡=μsNF_{\max}=\mu_sN
  • Do not confuse the friction coefficient or limiting friction with the actual friction force at every contact.
  • Use the supported body's equilibrium to determine contact-force components, then apply their opposite forces to the wedge.
  • State the particle or rigid-body idealization and check its moment implications.

Worked example

1. Force to raise a guided load

A 30∘30^\circ wedge face rises to the right. A 200 N200\,\mathrm{N} load rests on it and is guided so it can move vertically. The wedge is on a smooth horizontal floor. The coefficient of static friction at the sloped contact is 0.200.20. Find the magnitude and direction of the horizontal force that is just sufficient to drive the wedge left and raise the load. Use the particle idealization: represent the load and wedge contact resultants as concurrent at the contact point; the load's weight and guide force are idealized as passing through that point.
  1. Set directions and isolate the load
    As the wedge moves left, the load tends to rise relative to the wedge. Friction on the load therefore acts down the slope. The normal on the load points up-left. At impending slip, friction has magnitude 0.20N0.20N.
    F=0.20NF=0.20N
  2. Use vertical equilibrium of the load
    The upward normal component and the downward-slope friction component balance the 200 N200\,\mathrm{N} weight. This gives the sloped-contact normal force.
    N(cos⁡30∘−0.20sin⁡30∘)=200 NN(\cos30^\circ-0.20\sin30^\circ)=200\,\mathrm{N}
  3. Find the horizontal force
    The load's net contact force on the wedge is to the right. Its horizontal component is N(sin⁡30∘+0.20cos⁡30∘)N(\sin30^\circ+0.20\cos30^\circ). The applied force must act left to balance it; the smooth floor has no horizontal reaction.
    P=N(sin⁡30∘+0.20cos⁡30∘)P=N(\sin30^\circ+0.20\cos30^\circ)
  4. Calculate and check
    The normal force is about 247.4 N247.4\,\mathrm{N}. The required applied force is about 165.1 N165.1\,\mathrm{N} left. The load's guide reaction acts right with magnitude N(sin⁡30∘+0.20cos⁡30∘)N(\sin30^\circ+0.20\cos30^\circ), balancing its horizontal contact force. Vertically, the load's normal and friction components balance its weight. On the wedge, the contact's horizontal component balances PP and the floor normal balances the downward contact component. With the stated concurrent-force idealization, moments about the contact point are zero for each particle.
    N≈247.4 N,P≈165.1 NN\approx247.4\,\mathrm{N},\quad P\approx165.1\,\mathrm{N}
Answer: Apply approximately 165.1 N165.1\,\mathrm{N} to the left.
Check: For the load, Ncos⁡30∘−Fsin⁡30∘−200 N=0N\cos30^\circ-F\sin30^\circ-200\,\mathrm{N}=0 and the guide balances the horizontal contact component. For the wedge, PP balances the rightward contact component, and the floor normal balances the downward component. Both particle moment sums about the common point are zero.

Worked example

2. Raising a load with a rough wedge base

A 120 N120\,\mathrm{N} load rests on a wedge face rising right at 25∘25^\circ. A smooth guide keeps the load moving vertically. The wedge base is rough, with coefficient of static friction 0.250.25; the sloped contact has coefficient 0.150.15. Find the horizontal force needed to drive the wedge left and raise the load at impending slip at both rough contacts. Use the particle idealization, with the load weight and guide reaction passing through the sloped-contact point.
  1. Determine the sloped-contact normal
    The load tends to rise relative to the left-moving wedge, so friction on the load acts down-slope. Vertical equilibrium gives the normal force using F=0.15NF=0.15N.
    N(cos⁡25∘−0.15sin⁡25∘)=120 NN(\cos25^\circ-0.15\sin25^\circ)=120\,\mathrm{N}
  2. Determine the base resistance
    The wedge's vertical equilibrium gives a floor normal of 120 N120\,\mathrm{N}: its vertical contact load equals the load's weight. At impending base slip, floor friction has magnitude 0.250.25 times this normal and acts right, opposite the wedge's leftward tendency.
    Ng=120 N,Fg=0.25Ng=30 NN_g=120\,\mathrm{N},\quad F_g=0.25N_g=30\,\mathrm{N}
  3. Balance horizontal forces on the wedge
    The sloped contact pushes the wedge right with component N(sin⁡25∘+0.15cos⁡25∘)N(\sin25^\circ+0.15\cos25^\circ). Base friction also acts right. The applied force acts left and balances both contributions.
    P=N(sin⁡25∘+0.15cos⁡25∘)+30 NP=N(\sin25^\circ+0.15\cos25^\circ)+30\,\mathrm{N}
  4. Calculate and verify
    The normal is approximately 139.3 N139.3\,\mathrm{N}, giving an applied force of approximately 125.8 N125.8\,\mathrm{N} left. On the load, the normal and down-slope friction components balance its weight vertically, and the guide balances the horizontal component. On the wedge, the contact force and base friction balance the applied force horizontally; the floor normal balances the vertical contact load. The particle moments about the common contact point are zero.
    N≈139.3 N,P≈125.8 NN\approx139.3\,\mathrm{N},\quad P\approx125.8\,\mathrm{N}
Answer: The required force is approximately 125.8 N125.8\,\mathrm{N} to the left.
Check: The base friction is 30 N30\,\mathrm{N}, within its limiting value of 0.25(120 N)0.25(120\,\mathrm{N}). The sloped friction is 0.15N0.15N. Horizontal and vertical force sums are zero for both bodies by the equations used, and their particle moment sums about the common point are zero.

Worked example

3. Testing a specified leftward force

A 150 N150\,\mathrm{N} load rests on a wedge face rising right at 20∘20^\circ. A smooth guide keeps the load moving vertically, and the wedge rests on a smooth horizontal floor. At the sloped contact, μs=0.30\mu_s=0.30. Determine whether a 40 N40\,\mathrm{N} leftward force can produce impending upward motion of the load. Use the particle idealization, with the load weight and guide reaction passing through the contact point.
  1. Find the limiting normal force
    Assume the load is just about to rise relative to the wedge moving left. Friction on the load acts down-slope and has magnitude 0.30N0.30N. Vertical equilibrium determines the normal force.
    N=150cos⁡20∘−0.30sin⁡20∘ N≈169.6 NN=\frac{150}{\cos20^\circ-0.30\sin20^\circ}\,\mathrm{N}\approx169.6\,\mathrm{N}
  2. Find the force needed at that limit
    The contact force on the wedge has rightward component N(sin⁡20∘+0.30cos⁡20∘)N(\sin20^\circ+0.30\cos20^\circ). This is the leftward applied force required for equilibrium at the upward-impending limit.
    Preq=N(sin⁡20∘+0.30cos⁡20∘)≈106.2 NP_{\mathrm{req}}=N(\sin20^\circ+0.30\cos20^\circ)\approx106.2\,\mathrm{N}
  3. Compare with the available force
    The specified 40 N40\,\mathrm{N} force is smaller than the force needed to reach the upward-impending limit. It cannot produce the stated impending upward motion under this model. The comparison concerns that specified direction of impending motion, not every possible equilibrium state.
    40 N<106.2 N40\,\mathrm{N}<106.2\,\mathrm{N}
Answer: No. Approximately 106.2 N106.2\,\mathrm{N} leftward is required for impending upward motion; 40 N40\,\mathrm{N} is insufficient.
Check: At the upward limit, F=0.30N≈50.9 NF=0.30N\approx50.9\,\mathrm{N}. The load's vertical contact components balance 150 N150\,\mathrm{N}, and its guide balances the horizontal component. For the wedge, the contact's rightward component balances the applied force at the limiting value, and the floor reaction balances the vertical component. The force and moment sums are zero under the stated particle idealization.

Common mistakes and how to avoid them

Setting friction equal to μsN\mu_sN at every static contact.
Correction: Use 0≤F≤μsN0\leq F\leq\mu_sN for ordinary static equilibrium; use equality at impending slip.
Drawing friction in the same direction on both bodies at a contact.
Correction: The contact forces are opposite. Determine the relative sliding tendency first.
Using the surface angle as the normal-force angle.
Correction: The normal is perpendicular to the surface; resolve it using the chosen axes.
Ignoring friction at a rough wedge base.
Correction: Include a separate base friction force, directed opposite the wedge's tendency to slide along the floor.
Claiming moment balance without considering force locations.
Correction: Use actual lines of action for rigid bodies. In a particle idealization, state that the forces are concurrent and check moments about that point.

Lesson summary

  • Isolate the wedge and supported body, and show their contact forces separately.
  • Use normal forces perpendicular to surfaces and friction tangent to them.
  • Choose friction directions from relative impending slip; apply F=μsNF=\mu_sN at the limit.
  • Solve force equilibrium, then verify force and moment balance with the stated model.

Check your understanding

Question 1

A wedge is about to slide left on a rough horizontal floor. Which way does the floor friction on the wedge act?
  1. Left
  2. Right
  3. Vertically upward
  4. Perpendicular to the sloped face
Show answer and explanation
Right
Floor friction opposes the wedge's tendency to move along the floor, so it acts right.

Question 2

At a rough contact that is not known to be at impending slip, which condition applies?
  1. F=μsNF=\mu_sN in every case
  2. F≥μsNF\geq\mu_sN
  3. 0≤F≤μsN0\leq F\leq\mu_sN
  4. F=0F=0 whenever the body is static
Show answer and explanation
0≤F≤μsN0\leq F\leq\mu_sN
Static friction adjusts as needed up to the limiting value. Equality applies at impending slip.

Key terms

Wedge
A rigid body with a sloped surface that transfers force through contact.
Normal force
A contact force perpendicular to the touching surfaces.
Dry friction
A tangential contact force that opposes relative sliding or its tendency.
Impending motion
The limiting condition just before relative sliding begins.

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