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Friction

Statics · FE Reference Handbook section

Statics
4 formulas
10 exam-style examples
~53 min
All Statics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • The largest frictional force is called the limiting friction.
  • Any further increase in applied forces will cause motion.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Coulomb friction — solve for friction force — Friction

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.7000; normal force (N) = 606.0 lb, determine the friction force (F) in lb.

Given

  • frictioncoefficient(mu)=0.7000friction coefficient (mu) = 0.7000
  • normalforce(N)=606.0lbnormal force (N) = 606.0 lb

Find

friction force (F), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except F is given, so isolate F symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.7000,normalforce(N)=606.0lbList the givens: friction coefficient (mu) = 0.7000, normal force (N) = 606.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    F=424.2 lbF = 424.2\ \text{lb}
  6. Step 6 — Check: returning F = 424.2 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=424.2 lbF = 424.2\ \text{lb}

Why the other options are there

  • 848.4 — kept a factor of two that cancels in the correct rearrangement.
  • 212.1 — dropped that same factor in the other direction.
  • 466.6 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 2
Coulomb friction — solve for friction coefficient — Friction (2)

A statics problem uses Coulomb friction. Given normal force (N) = 457.0 lb; friction force (F) = 1,494 lb, determine the friction coefficient (mu).

Given

  • normalforce(N)=457.0lbnormal force (N) = 457.0 lb
  • frictionforce(F)=1,494lbfriction force (F) = 1,494 lb

Find

friction coefficient (mu)

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except mu is given, so isolate mu symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

  3. Step 3

    Listthegivens:normalforce(N)=457.0lb,frictionforce(F)=1,494lbList the givens: normal force (N) = 457.0 lb, friction force (F) = 1,494 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=3.2691\mu = 3.2691
  6. Step 6 — Check: returning mu = 3.2691 to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
μ=3.2691\mu = 3.2691

Why the other options are there

  • 6.5383 — kept a factor of two that cancels in the correct rearrangement.
  • 1.6346 — dropped that same factor in the other direction.
  • 3.5961 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 3
Coulomb friction — solve for normal force — Friction (3)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.3100; friction force (F) = 983.0 lb, determine the normal force (N) in lb.

Given

  • frictioncoefficient(mu)=0.3100friction coefficient (mu) = 0.3100
  • frictionforce(F)=983.0lbfriction force (F) = 983.0 lb

Find

normal force (N), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that N stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.3100,frictionforce(F)=983.0lbList the givens: friction coefficient (mu) = 0.3100, friction force (F) = 983.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    N=3171 lbN = 3171\ \text{lb}
  6. Step 6 — Check: returning N = 3,171 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
N=3171 lbN = 3171\ \text{lb}

Why the other options are there

  • 6,342 — kept a factor of two that cancels in the correct rearrangement.
  • 1,585 — dropped that same factor in the other direction.
  • 3,488 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 4
Coulomb friction — solve for friction force (case 2) — Friction (4)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.6000; normal force (N) = 1,687 lb, determine the friction force (F) in lb.

Given

  • frictioncoefficient(mu)=0.6000friction coefficient (mu) = 0.6000
  • normalforce(N)=1,687lbnormal force (N) = 1,687 lb

Find

friction force (F), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except F is given, so isolate F symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.6000,normalforce(N)=1,687lbList the givens: friction coefficient (mu) = 0.6000, normal force (N) = 1,687 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    F=1012 lbF = 1012\ \text{lb}
  6. Step 6 — Check: returning F = 1,012 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=1012 lbF = 1012\ \text{lb}

Why the other options are there

  • 2,024 — kept a factor of two that cancels in the correct rearrangement.
  • 506.1 — dropped that same factor in the other direction.
  • 1,113 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 5
Coulomb friction — solve for friction coefficient (case 2) — Friction (5)

A statics problem uses Coulomb friction. Given normal force (N) = 411.0 lb; friction force (F) = 1,309 lb, determine the friction coefficient (mu).

Given

  • normalforce(N)=411.0lbnormal force (N) = 411.0 lb
  • frictionforce(F)=1,309lbfriction force (F) = 1,309 lb

Find

friction coefficient (mu)

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except mu is given, so isolate mu symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

  3. Step 3

    Listthegivens:normalforce(N)=411.0lb,frictionforce(F)=1,309lbList the givens: normal force (N) = 411.0 lb, friction force (F) = 1,309 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=3.1849\mu = 3.1849
  6. Step 6 — Check: returning mu = 3.1849 to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
μ=3.1849\mu = 3.1849

Why the other options are there

  • 6.3698 — kept a factor of two that cancels in the correct rearrangement.
  • 1.5925 — dropped that same factor in the other direction.
  • 3.5034 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 6
Coulomb friction — solve for normal force (case 2) — Friction (6)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.6700; friction force (F) = 736.0 lb, determine the normal force (N) in lb.

Given

  • frictioncoefficient(mu)=0.6700friction coefficient (mu) = 0.6700
  • frictionforce(F)=736.0lbfriction force (F) = 736.0 lb

Find

normal force (N), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that N stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.6700,frictionforce(F)=736.0lbList the givens: friction coefficient (mu) = 0.6700, friction force (F) = 736.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    N=1099 lbN = 1099\ \text{lb}
  6. Step 6 — Check: returning N = 1,099 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
N=1099 lbN = 1099\ \text{lb}

Why the other options are there

  • 2,197 — kept a factor of two that cancels in the correct rearrangement.
  • 549.3 — dropped that same factor in the other direction.
  • 1,208 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 7
Coulomb friction — solve for friction force (case 3) — Friction (7)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.4100; normal force (N) = 1,932 lb, determine the friction force (F) in lb.

Given

  • frictioncoefficient(mu)=0.4100friction coefficient (mu) = 0.4100
  • normalforce(N)=1,932lbnormal force (N) = 1,932 lb

Find

friction force (F), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except F is given, so isolate F symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.4100,normalforce(N)=1,932lbList the givens: friction coefficient (mu) = 0.4100, normal force (N) = 1,932 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    F=792.1 lbF = 792.1\ \text{lb}
  6. Step 6 — Check: returning F = 792.1 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=792.1 lbF = 792.1\ \text{lb}

Why the other options are there

  • 1,584 — kept a factor of two that cancels in the correct rearrangement.
  • 396.1 — dropped that same factor in the other direction.
  • 871.3 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 8
Coulomb friction — solve for friction coefficient (case 3) — Friction (8)

A statics problem uses Coulomb friction. Given normal force (N) = 489.0 lb; friction force (F) = 767.0 lb, determine the friction coefficient (mu).

Given

  • normalforce(N)=489.0lbnormal force (N) = 489.0 lb
  • frictionforce(F)=767.0lbfriction force (F) = 767.0 lb

Find

friction coefficient (mu)

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except mu is given, so isolate mu symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

  3. Step 3

    Listthegivens:normalforce(N)=489.0lb,frictionforce(F)=767.0lbList the givens: normal force (N) = 489.0 lb, friction force (F) = 767.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    μ=1.5685\mu = 1.5685
  6. Step 6 — Check: returning mu = 1.5685 to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
μ=1.5685\mu = 1.5685

Why the other options are there

  • 3.1370 — kept a factor of two that cancels in the correct rearrangement.
  • 0.7843 — dropped that same factor in the other direction.
  • 1.7254 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 9
Coulomb friction — solve for normal force (case 3) — Friction (9)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.1600; friction force (F) = 692.0 lb, determine the normal force (N) in lb.

Given

  • frictioncoefficient(mu)=0.1600friction coefficient (mu) = 0.1600
  • frictionforce(F)=692.0lbfriction force (F) = 692.0 lb

Find

normal force (N), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except N is given, so isolate N symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that N stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.1600,frictionforce(F)=692.0lbList the givens: friction coefficient (mu) = 0.1600, friction force (F) = 692.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    N=4325 lbN = 4325\ \text{lb}
  6. Step 6 — Check: returning N = 4,325 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
N=4325 lbN = 4325\ \text{lb}

Why the other options are there

  • 8,650 — kept a factor of two that cancels in the correct rearrangement.
  • 2,163 — dropped that same factor in the other direction.
  • 4,758 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

Example 10
Coulomb friction — solve for friction force (case 4) — Friction (10)

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.7800; normal force (N) = 403.0 lb, determine the friction force (F) in lb.

Given

  • frictioncoefficient(mu)=0.7800friction coefficient (mu) = 0.7800
  • normalforce(N)=403.0lbnormal force (N) = 403.0 lb

Find

friction force (F), in lb

Start with the thinking

  • The governing relation printed in this handbook section is Coulomb friction.
  • Everything except F is given, so isolate F symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • Statics items reward recognising the unknown before touching a calculator.

Step-by-step solution

  1. Step 1 — State the governing relation:

    F=μNF = \mu N
  2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

  3. Step 3

    Listthegivens:frictioncoefficient(mu)=0.7800,normalforce(N)=403.0lbList the givens: friction coefficient (mu) = 0.7800, normal force (N) = 403.0 lb
  4. Step 4 — Substitute the given values into the rearranged relation.

  5. Step 5 — Evaluate:

    F=314.3 lbF = 314.3\ \text{lb}
  6. Step 6 — Check: returning F = 314.3 lb to

    F=μNF = \mu N

    reproduces the given quantities, and both sides carry the same units.

Answer:
F=314.3 lbF = 314.3\ \text{lb}

Why the other options are there

  • 628.7 — kept a factor of two that cancels in the correct rearrangement.
  • 157.2 — dropped that same factor in the other direction.
  • 345.8 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Friction

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