A statics problem uses Moment of a force. Given force (F) = 1,271 lb; moment arm (d) = 10.1000 ft, determine the moment (M) in lb·ft.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• Everything except M is given, so isolate M 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:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that M stands alone on the left-hand side.

3. Step 3

   $$List the givens: force (F) = 1,271 lb, moment arm (d) = 10.1000 ft$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$M = 12837\ \text{lb·ft}$$

6. Step 6 — Check: returning M = 12,837 lb·ft to

   $$M = F \times d$$

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

Why the other options are there

• 25,674 — kept a factor of two that cancels in the correct rearrangement.
• 6,419 — dropped that same factor in the other direction.
• 14,121 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.6100; normal force (N) = 1,006 lb, determine the 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:

   $$F = \mu N$$

2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

3. Step 3

   $$List the givens: friction coefficient (mu) = 0.6100, normal force (N) = 1,006 lb$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F = 613.7\ \text{lb}$$

6. Step 6 — Check: returning F = 613.7 lb to

   $$F = \mu N$$

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

Why the other options are there

• 1,227 — kept a factor of two that cancels in the correct rearrangement.
• 306.8 — dropped that same factor in the other direction.
• 675.0 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread

a wrench applied to an anchor nut Given force (F) = 1,256 N; perpendicular distance (d) = 3.5000 m, determine the moment (M) in N\cdot m.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• Everything except M is given, so isolate M 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for M:

   $$M = M = F d$$

3. Step 3

   $$List the givens: force (F) = 1,256 N, perpendicular distance (d) = 3.5000 m$$

4. Step 4 — Substitute the given values:

   $$M = M = 1256 3.5000$$

5. Step 5 — Evaluate:

   $$M = 4394\ \text{N\cdot m}$$

6. Step 6 — Check: returning M = 4,394 N\cdot m to

   $$M = F d$$

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

Why the other options are there

• 8,789 — kept a factor of two that cancels in the correct rearrangement.
• 2,197 — dropped that same factor in the other direction.
• 4,834 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A statics problem uses Moment of a force. Given moment arm (d) = 18.6000 ft; moment (M) = 7,733 lb·ft, determine the force (F) in lb.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• 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:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that F stands alone on the left-hand side.

3. Step 3 — List the givens: moment arm (d) = 18.6000 ft, moment (M) = 7,733 lb·ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$F = 415.8\ \text{lb}$$

6. Step 6 — Check: returning F = 415.8 lb to

   $$M = F \times d$$

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

Why the other options are there

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

Reference: FE Reference Handbook — Statics → Screw Thread

A statics problem uses Coulomb friction. Given normal force (N) = 1,131 lb; friction force (F) = 196.0 lb, determine the 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:

   $$F = \mu N$$

2. Step 2 — Rearrange the relation so that mu stands alone on the left-hand side.

3. Step 3

   $$List the givens: normal force (N) = 1,131 lb, friction force (F) = 196.0 lb$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$\mu = 0.1733$$

6. Step 6 — Check: returning mu = 0.1733 to

   $$F = \mu N$$

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

Why the other options are there

• 0.3466 — kept a factor of two that cancels in the correct rearrangement.
• 0.0866 — dropped that same factor in the other direction.
• 0.1906 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread

a bracket bolted to a column flange Given moment (M) = 5,304 N\cdot m; perpendicular distance (d) = 1.9000 m, determine the force (F) in N.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for F:

   $$F = F = \dfrac{M}{d}$$

3. Step 3

   $$List the givens: moment (M) = 5,304 N\cdot m, perpendicular distance (d) = 1.9000 m$$

4. Step 4 — Substitute the given values:

   $$F = F = \dfrac{5304}{1.9000}$$

5. Step 5 — Evaluate:

   $$F = 2792\ \text{N}$$

6. Step 6 — Check: returning F = 2,792 N to

   $$M = F d$$

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

Why the other options are there

• 5,583 — kept a factor of two that cancels in the correct rearrangement.
• 1,396 — dropped that same factor in the other direction.
• 3,071 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A statics problem uses Moment of a force. Given force (F) = 235.0 lb; moment (M) = 428.0 lb·ft, determine the moment arm (d) in ft.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• Everything except d is given, so isolate d 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:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that d stands alone on the left-hand side.

3. Step 3 — List the givens: force (F) = 235.0 lb, moment (M) = 428.0 lb·ft.

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$d = 1.8213\ \text{ft}$$

6. Step 6 — Check: returning d = 1.8213 ft to

   $$M = F \times d$$

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

Why the other options are there

• 3.6426 — kept a factor of two that cancels in the correct rearrangement.
• 0.9106 — dropped that same factor in the other direction.
• 2.0034 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread

A statics problem uses Coulomb friction. Given friction coefficient (mu) = 0.3700; friction force (F) = 1,548 lb, determine the 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:

   $$F = \mu N$$

2. Step 2 — Rearrange the relation so that N stands alone on the left-hand side.

3. Step 3

   $$List the givens: friction coefficient (mu) = 0.3700, friction force (F) = 1,548 lb$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$N = 4184\ \text{lb}$$

6. Step 6 — Check: returning N = 4,184 lb to

   $$F = \mu N$$

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

Why the other options are there

• 8,368 — kept a factor of two that cancels in the correct rearrangement.
• 2,092 — dropped that same factor in the other direction.
• 4,602 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread

a cantilevered sign arm Given moment (M) = 6,275 N\cdot m; force (F) = 2,461 N, determine the perpendicular distance (d) in m.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force about a point.
• Everything except d is given, so isolate d 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.
• A force acting at a perpendicular distance d produces a moment.

Step-by-step solution

1. Step 1 — State the governing relation:

   $$M = F d$$

2. Step 2 — Rearrange symbolically for d:

   $$d = d = \dfrac{M}{F}$$

3. Step 3

   $$List the givens: moment (M) = 6,275 N\cdot m, force (F) = 2,461 N$$

4. Step 4 — Substitute the given values:

   $$d = d = \dfrac{6275}{2461}$$

5. Step 5 — Evaluate:

   $$d = 2.5499\ \text{m}$$

6. Step 6 — Check: returning d = 2.5499 m to

   $$M = F d$$

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

Why the other options are there

• 5.0998 — kept a factor of two that cancels in the correct rearrangement.
• 1.2749 — dropped that same factor in the other direction.
• 2.8049 — rounded an intermediate value before the final step.

Reference: FE Handbook — Statics (Moments)

A statics problem uses Moment of a force. Given force (F) = 423.0 lb; moment arm (d) = 1.6000 ft, determine the moment (M) in lb·ft.

Start with the thinking

• The governing relation printed in this handbook section is Moment of a force.
• Everything except M is given, so isolate M 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:

   $$M = F \times d$$

2. Step 2 — Rearrange the relation so that M stands alone on the left-hand side.

3. Step 3

   $$List the givens: force (F) = 423.0 lb, moment arm (d) = 1.6000 ft$$

4. Step 4 — Substitute the given values into the rearranged relation.

5. Step 5 — Evaluate:

   $$M = 676.8\ \text{lb·ft}$$

6. Step 6 — Check: returning M = 676.8 lb·ft to

   $$M = F \times d$$

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

Why the other options are there

• 1,354 — kept a factor of two that cancels in the correct rearrangement.
• 338.4 — dropped that same factor in the other direction.
• 744.5 — rounded an intermediate value before the final step.

Reference: FE Reference Handbook — Statics → Screw Thread