A 4.0 kN crate rests on a floor with μs = 0.35. What horizontal push starts it moving, and what push is needed on a 10° ramp (up-slope)?

Start with the thinking

• On the level, N equals the weight; on a ramp it does not.
• Add the gravity component along the ramp.

Step-by-step solution

1. Level friction — F = μs N = 0.35(4.0) = 1.40 kN

2. Ramp normal

   $$N=W\cos 10^{\circ }=4.0(0.9848)=3.94kN$$

3. Ramp friction

   $$F=0.35(3.94)=1.38kN$$

4. Gravity component

   $$W\sin 10^{\circ }=4.0(0.1736)=0.694kN$$

5. Required push

   $$P=1.38+0.694=2.07kN$$

Why the other options are there

• 1.38 kN (gravity component omitted)
• 2.09 kN (normal force taken as W on the ramp)

Reference: FE Reference Handbook — Statics — Friction

A 455 lb crate rests on a 11° incline with μ_s = 0.55. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 455·cos 11° = 446.6 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.55(446.6)=245.7lb$$

5. Driving component — W·sin θ = 455·sin 11° = 86.8 lb

6. Compare — 86.8 < 245.7 → stays in place

7. Push to move up — P = W·sin θ + μN = 332.5 lb

Why the other options are there

• 250.3 lb (weight used as the normal force)
• -158.8 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 401 lb crate rests on a 19° incline with μ_s = 0.45. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 401·cos 19° = 379.2 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.45(379.2)=170.6lb$$

5. Driving component — W·sin θ = 401·sin 19° = 130.6 lb

6. Compare — 130.6 < 170.6 → stays in place

7. Push to move up — P = W·sin θ + μN = 301.2 lb

Why the other options are there

• 180.5 lb (weight used as the normal force)
• -40 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 883 lb crate rests on a 23° incline with μ_s = 0.40. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 883·cos 23° = 812.8 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.40(812.8)=325.1lb$$

5. Driving component — W·sin θ = 883·sin 23° = 345.0 lb

6. Compare — 345.0 > 325.1 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 670.1 lb

Why the other options are there

• 353.2 lb (weight used as the normal force)
• 20 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 340 lb crate rests on a 12° incline with μ_s = 0.55. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 340·cos 12° = 332.6 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.55(332.6)=182.9lb$$

5. Driving component — W·sin θ = 340·sin 12° = 70.7 lb

6. Compare — 70.7 < 182.9 → stays in place

7. Push to move up — P = W·sin θ + μN = 253.6 lb

Why the other options are there

• 187.0 lb (weight used as the normal force)
• -112.2 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 352 lb crate rests on a 24° incline with μ_s = 0.35. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 352·cos 24° = 321.6 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.35(321.6)=112.5lb$$

5. Driving component — W·sin θ = 352·sin 24° = 143.2 lb

6. Compare — 143.2 > 112.5 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 255.7 lb

Why the other options are there

• 123.2 lb (weight used as the normal force)
• 31 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 227 lb crate rests on a 21° incline with μ_s = 0.30. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 227·cos 21° = 211.9 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.30(211.9)=63.6lb$$

5. Driving component — W·sin θ = 227·sin 21° = 81.3 lb

6. Compare — 81.3 > 63.6 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 144.9 lb

Why the other options are there

• 68 lb (weight used as the normal force)
• 18 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 322 lb crate rests on a 24° incline with μ_s = 0.60. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 322·cos 24° = 294.2 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.60(294.2)=176.5lb$$

5. Driving component — W·sin θ = 322·sin 24° = 131.0 lb

6. Compare — 131.0 < 176.5 → stays in place

7. Push to move up — P = W·sin θ + μN = 307.5 lb

Why the other options are there

• 193.2 lb (weight used as the normal force)
• -46 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 592 lb crate rests on a 13° incline with μ_s = 0.60. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 592·cos 13° = 576.8 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.60(576.8)=346.1lb$$

5. Driving component — W·sin θ = 592·sin 13° = 133.2 lb

6. Compare — 133.2 < 346.1 → stays in place

7. Push to move up — P = W·sin θ + μN = 479.3 lb

Why the other options are there

• 355.2 lb (weight used as the normal force)
• -212.9 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction

A 429 lb crate rests on a 26° incline with μ_s = 0.35. Does it slip, and what force parallel to the incline is required to start it moving up?

Start with the thinking

• Compare the driving component to the maximum available friction.
• Normal force uses the cosine, driving force the sine.

Step-by-step solution

1. Normal force — N = W·cos θ

2. Substituting — N = 429·cos 26° = 385.6 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.35(385.6)=135.0lb$$

5. Driving component — W·sin θ = 429·sin 26° = 188.1 lb

6. Compare — 188.1 > 135.0 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 323.0 lb

Why the other options are there

• 150.1 lb (weight used as the normal force)
• 53 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Friction