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 569 lb crate rests on a 24° 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 = 569·cos 24° = 519.8 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.55(519.8)=285.9lb$$

5. Driving component — W·sin θ = 569·sin 24° = 231.4 lb

6. Compare — 231.4 < 285.9 → stays in place

7. Push to move up — P = W·sin θ + μN = 517.3 lb

Why the other options are there

• 313.0 lb (weight used as the normal force)
• -54 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Belt Friction

A screw jack with a mean thread diameter of 1.50 in and a lead of 0.35 in per turn raises a 20,000 lb load. The coefficient of friction on the screw thread is 0.11. Find the required torque.

Start with the thinking

• The screw thread is an inclined plane wrapped around a cylinder — the lead angle comes from the lead over the mean circumference.
• Raising adds the friction angle to the lead angle; lowering subtracts it.

Step-by-step solution

1. Formula

   $$\tan \alpha =L/(\pi d_m)$$

2. Substituting

   $$\tan \alpha =0.35/(\pi \times 1.50)=0.0743,\alpha =4.25^{\circ }$$

3. Friction angle

   $$\varphi =\arctan \mu =\arctan 0.11=6.28^{\circ }$$

4. Formula

   $$T=W(d_m/2)\tan (\alpha +\varphi )$$

5. Substituting

   $$T=20000(0.750)\tan (10.52^{\circ })$$

6. Evaluate — T = 2,787 in·lb = 232.2 ft·lb

Why the other options are there

• 1,114 in·lb (friction ignored)
• 531.6 in·lb (lowering torque)

Reference: FE Reference Handbook — Statics → Belt Friction

A 691 lb crate rests on a 21° 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 = 691·cos 21° = 645.1 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.40(645.1)=258.0lb$$

5. Driving component — W·sin θ = 691·sin 21° = 247.6 lb

6. Compare — 247.6 < 258.0 → stays in place

7. Push to move up — P = W·sin θ + μN = 505.7 lb

Why the other options are there

• 276.4 lb (weight used as the normal force)
• -10 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Belt Friction

A screw jack with a mean thread diameter of 1.00 in and a lead of 0.30 in per turn raises a 3,000 lb load. The coefficient of friction on the screw thread is 0.19. Find the required torque.

Start with the thinking

• The screw thread is an inclined plane wrapped around a cylinder — the lead angle comes from the lead over the mean circumference.
• Raising adds the friction angle to the lead angle; lowering subtracts it.

Step-by-step solution

1. Formula

   $$\tan \alpha =L/(\pi d_m)$$

2. Substituting

   $$\tan \alpha =0.30/(\pi \times 1.00)=0.0955,\alpha =5.45^{\circ }$$

3. Friction angle

   $$\varphi =\arctan \mu =\arctan 0.19=10.76^{\circ }$$

4. Formula

   $$T=W(d_m/2)\tan (\alpha +\varphi )$$

5. Substituting

   $$T=3000(0.500)\tan (16.21^{\circ })$$

6. Evaluate — T = 436.2 in·lb = 36.3 ft·lb

Why the other options are there

• 143.2 in·lb (friction ignored)
• 139.2 in·lb (lowering torque)

Reference: FE Reference Handbook — Statics → Belt Friction

A 834 lb crate rests on a 28° incline with μ_s = 0.25. 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 = 834·cos 28° = 736.4 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.25(736.4)=184.1lb$$

5. Driving component — W·sin θ = 834·sin 28° = 391.5 lb

6. Compare — 391.5 > 184.1 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 575.6 lb

Why the other options are there

• 208.5 lb (weight used as the normal force)
• 207.4 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Belt Friction

A screw jack with a mean thread diameter of 2.50 in and a lead of 0.30 in per turn raises a 20,000 lb load. The coefficient of friction on the screw thread is 0.12. Find the required torque.

Start with the thinking

• The screw thread is an inclined plane wrapped around a cylinder — the lead angle comes from the lead over the mean circumference.
• Raising adds the friction angle to the lead angle; lowering subtracts it.

Step-by-step solution

1. Formula

   $$\tan \alpha =L/(\pi d_m)$$

2. Substituting

   $$\tan \alpha =0.30/(\pi \times 2.50)=0.0382,\alpha =2.19^{\circ }$$

3. Friction angle

   $$\varphi =\arctan \mu =\arctan 0.12=6.84^{\circ }$$

4. Formula

   $$T=W(d_m/2)\tan (\alpha +\varphi )$$

5. Substituting

   $$T=20000(1.250)\tan (9.03^{\circ })$$

6. Evaluate — T = 3,973 in·lb = 331.1 ft·lb

Why the other options are there

• 954.9 in·lb (friction ignored)
• 2,036 in·lb (lowering torque)

Reference: FE Reference Handbook — Statics → Belt Friction

A 371 lb crate rests on a 25° 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 = 371·cos 25° = 336.2 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.35(336.2)=117.7lb$$

5. Driving component — W·sin θ = 371·sin 25° = 156.8 lb

6. Compare — 156.8 > 117.7 → slips without restraint

7. Push to move up — P = W·sin θ + μN = 274.5 lb

Why the other options are there

• 129.9 lb (weight used as the normal force)
• 39 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Belt Friction

A screw jack with a mean thread diameter of 1.75 in and a lead of 0.40 in per turn raises a 6,000 lb load. The coefficient of friction on the screw thread is 0.24. Find the required torque.

Start with the thinking

• The screw thread is an inclined plane wrapped around a cylinder — the lead angle comes from the lead over the mean circumference.
• Raising adds the friction angle to the lead angle; lowering subtracts it.

Step-by-step solution

1. Formula

   $$\tan \alpha =L/(\pi d_m)$$

2. Substituting

   $$\tan \alpha =0.40/(\pi \times 1.75)=0.0728,\alpha =4.16^{\circ }$$

3. Friction angle

   $$\varphi =\arctan \mu =\arctan 0.24=13.50^{\circ }$$

4. Formula

   $$T=W(d_m/2)\tan (\alpha +\varphi )$$

5. Substituting

   $$T=6000(0.875)\tan (17.66^{\circ })$$

6. Evaluate — T = 1,671 in·lb = 139.3 ft·lb

Why the other options are there

• 382.0 in·lb (friction ignored)
• 863.0 in·lb (lowering torque)

Reference: FE Reference Handbook — Statics → Belt Friction

A 339 lb crate rests on a 10° 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 = 339·cos 10° = 333.8 lb

3. Maximum friction — F_max = μ_s·N

4. Substituting

   $$F_{m}ax=0.35(333.8)=116.8lb$$

5. Driving component — W·sin θ = 339·sin 10° = 58.9 lb

6. Compare — 58.9 < 116.8 → stays in place

7. Push to move up — P = W·sin θ + μN = 175.7 lb

Why the other options are there

• 118.6 lb (weight used as the normal force)
• -58 lb (friction sign reversed)

Reference: FE Reference Handbook — Statics → Belt Friction