7.7 Rolling Resistance

Transcription

1 7.7 Rolling Resistance

2 7.7 Rolling Resistance Example 1, page 1 of 4 1. The fertilizer spreader and the fertilizer it contains have a combined mass of 40 kg and a center of gravity located at point G. If the coefficient of rolling resistance for the tires is 5 mm, determine the resultant force that must be applied to the handle to move the spreader at a constant speed. 250 mm Px 100 mm P y G 850 mm Radius = 120 mm

3 7.7 Rolling Resistance Example 1, page 2 of 4 1 Locate the point where the resisting force from the ground acts on the wheel. 4 For later use in calculating moments, find the vertical distance between and. Rotation of wheel Direction of motion of center of wheel 2 s the wheel rolls to the left, the ground exerts a force, R, opposing the motion. R 5 mm ( = coefficient of rolling resistance) 100 mm 5 mm (100 mm) 2 (5 mm) 2 = mm 3 R acts at a point on the circumference 5 mm from a vertical line through the center of the wheel.

4 Rolling Resistance Example 1, page 3 of 4 5 Free-body diagram of spreader. 250 mm Px Weight = 40 kg 9.81 m/s 2 = N 100 mm G P y 850 mm 6 Equilibrium equations Fx = 0: 2Rx Px = 0 (1) F y = 0: 2R y + P y N = 0 (2) M = 0 : Px (850 mm mm) + P y (250 mm + 5 mm) (392.4 N)(100 mm + 5 mm) = 0 (3) 2Rx mm Three equations, four unknowns; an additional free body is needed. Components of force from ground acting on two wheels 2R y 5 mm

5 + 7.7 Rolling Resistance Example 1, page 4 of 4 7 Free body diagram of a wheel and axle 8 Moment equation for wheel Forces from spreader acting on axle M = 0: Rx( mm) R y (5 mm) = 0 (4) Solving equations 1-4 simultaneously gives x y Rx = N R y = N Px = N Rx 100 mm mm P y = N The resultant force acting on the handle is 5 mm (Rx 2 + R y 2 ) = (7.102 N) 2 + ( N) 2 R y = 142 N ns.

6 7.7 Rolling Resistance Example 2, page 1 of 4 2. The mass of the upright piano is 200 kg. The coefficient of rolling resistance of the casters is 0.2 mm. Determine the value of the horizontal force P required to move the piano at constant speed. Make the simplifying assumption that the resisting forces are the same at all four casters. P Radius of the casters = 14 mm

7 7.7 Rolling Resistance Example 2, page 2 of 4 1 Locate the point where the resisting force from the floor acts on the wheel. Direction of motion of center of wheel Rotation of wheel 14 mm 2 R s the wheel rolls to the left, the floor exerts a force, R, opposing the motion. 0.2 mm ( = coefficient of rolling resistance) 0.2 mm (14 mm) 2 (0.2 mm) 2 = mm 3 R acts at a point on the circumference 0.2 mm from a vertical line through the center of the wheel. 4 For later use in calculating moments, find the vertical distance between and.

8 + 7.7 Rolling Resistance Example 2, page 3 of 4 5 Free-body diagram of piano Weight = (200 kg)(9.81 m/s 2 ) = 1962 N P 6 Equilibrium equations. + Fx = 0: 4Rx P = 0 (1) Rx Rx Rx Rx F y = 0: 4R y 1962 N = 0 (2) R y R y R y R y Two equations, three unknowns; an additional free body is needed. Resisting forces from floor acting on caster are same at each caster

9 + 7.7 Rolling Resistance Example 2, page 4 of 4 7 Free-body diagram of a caster and axle Forces from piano acting on axle 8 Moment equation for caster M = 0 : Rx( mm) R y (0.2 mm) = 0 (3) Solving Eqs.1-3 simultaneously gives mm Rx = 7.01 N Rx R y 0.2 mm R y = N P = 28.0 N ns.

10 7.7 Rolling Resistance Example 3, page 1 of 3 3. If the coefficient of rolling resistance is 0.3 in., determine the magnitude of the horizontal force F required to push the 300-lb drum up the inclined plane. 8 in. F 1 Locate the point where the resisting force from the inclined plane acts on the drum. 15 Rotation of drum Direction of motion of center of drum 3 R acts at a point on the circumference 0.3 in. from the center of the drum. Note that the 0.3 in. is measured parallel to the inclined plane. 15 R 0.3 in. ( = coefficient of rolling resistance) 2 s the wheel rolls up the plane, the plane exerts a force, R, opposing the motion.

11 + 7.7 Rolling Resistance Example 3, page 2 of 3 4 For use in calculating moments, find the distance from to in the direction perpendicular to the inclined plane. D (8 in.) 2 (0.3 in.) 2 = in. C 8 in. 0.3 in. F in. 15 Reference line perpendicular to inclined plane 5 Free-body diagram of drum D C 300 lb (Weight of drum) 8 in. R y 0.3 in. Rx y x Components of force from inclined plane acting on wheel 6 Moment equilibrium equation M = 0: (300 cos lb)(0.3 in.) + (300 sin lb)(7.994 in.) (F cos )(7.994 in.) + (F sin )(0.3 in) = 0 (1)

12 7.7 Rolling Resistance Example 3, page 3 of 3 7 Calculate D 8 = = 15 Inclined plane C = 75 9 Substitute = 15 into Eq. 1: (300 cos lb)(0.3 in.) + (300 sin lb)(7.994 in.) (F cos )(7.994 in.) + (F sin )(0.3 in) = 0 (Eq. 1 repeated) Solving gives F = 92.6 lb ns.

13 7.7 Rolling Resistance Example 4, page 1 of lb steel-rim wheel of 24-in. diameter rolls at constant velocity down the inclined plane. If the coefficient of rolling resistance is 0.08 in., determine the angle. 1 Locate the point where the resisting force from the inclined plane acts as the wheel. Rotation of wheel Direction of motion of the center of the wheel 0.08 in. ( = coefficient of rolling resistance) R 2 s the wheel rolls down the plane, the plane exerts a force, R, opposing the motion. 3 R acts at a point on the circumference 0.08 in. from the center of the wheel, measured parallel to the inclined plane.

14 Rolling Resistance Example 4, page 2 of 3 4 For use in calculating moments, find the distance from to in the direction perpendicular to the inclined plane. D 5 Free-body diagram of wheel D y (12 in.) 2 (0.08 in.) in. x = in. C 0.08 in. Radius = 24/2 in. = 12 in. C 0.08 in. R y Rx Components of force from inclined plane acting on wheel 6 + Equilibrium equations Fx = 0: Rx (100 lb) sin = 0 (1) F y = 0: R y (100 lb) cos = 0 (2) M = 0 : R y (0.08 in.) Rx( in.) = 0 (3) Solving gives Rx = lb R y = lb = (4)

15 7.7 Rolling Resistance Example 4, page 3 of 3 7 Geometry D ut = 0.382, by Eq. 4 so, = ns. C 8 We could have saved same work by noticing that the wheel is a two-force body. Thus because the line of action of the weight is a vertical line through the center, the line of action of the resisting force R must also be a vertical line passing through the center. The angle can be found from geometry, as shown below. Weight Complement of in. = sin 0.08 in. 12 in. = in. R

16 7.7 Rolling Resistance Example 5, page 1 of 4 5. If the coefficient of rolling resistance at the top of the cylinder is 0.4 mm and at bottom of the cylinder is 0.8 mm, determine the horizontal force P required to start the block moving to the left. The weight of the cylinder is negligible compared to the weight of the block. 100 kg P 80 mm

17 7.7 Rolling Resistance Example 5, page 2 of 4 1 Locate the points where the resisting forces act on the cylinder. 0.4 mm (= coefficient of rolling resistance) Direction of motion of center of the wheel relative to the ground 100 kg P 2 The resisting force from the block opposes the rolling of the cylinder on the block. Rotation of wheel C R block Direction of motion of center of the wheel relative to the block R ground 0.8 mm (= coefficient of rolling resistance) 3 The resisting force from the ground opposes the rolling of the cylinder on the ground.

18 7.7 Rolling Resistance Example 5, page 3 of 4 4 For use in calculating moments, find the vertical distances between the center of the wheel and the point of application of the resisting forces. 0.4 mm D (80 mm) 2 (0.4 mm) 2 = mm 80 mm 80 mm (80 mm) 2 (0.8 mm) 2 = mm C 0.8 mm E

19 Rolling Resistance Example 5, page 4 of 4 5 Free-body diagram of block and cylinder Weight = (100 kg)(9.81 m/s 2 ) = 981 N 7 Free-body diagram of the cylinder 0.4 mm R block-y 100 kg F R block-x mm R ground-x C R ground-y 6 Equilibrium equations R ground-x R ground-y C Moment equation E 0.8 mm mm M = 0: R ground-x ( mm mm) R ground-y (0.8 mm mm) = 0 (3) + Fx = 0: R ground-x F = 0 (1) F y = 0: R ground-y 981 N = 0 (2) Solving Eqs. 1-3 gives R groung-x = N Two equations, three unknowns: an additional free-body is needed. R ground-y = 981 N F = 7.36 N ns.

20 7.7 Rolling Resistance Example 6, page 1 of 5 6. The ancient ritons who constructed the prehistoric monument Stonehenge moved massive stones over twenty miles from a quarry to the monument site. One possible way that they might have done this is to roll the stones over logs laid on the ground. ssuming that the coefficient of rolling resistance for the top of a log is 0.01 ft and the coefficient is 0.08 ft for the bottom, estimate the horizontal force required to move the typical stone shown below. Make the simplifying assumptions that the resisting forces are the same for each log. 24 ft Rope 4 ft 100,000 lb Force of men (and perhaps oxen) F Diameter of logs = 1 ft.

21 7.7 Rolling Resistance Example 6, page 2 of 5 1 Locate the points where the resisting forces act on the log. Direction of motion of stone Portion of stone 0.01 ft 2 s the log rolls on the underside of the stone, the stone exerts a force opposing the rolling motion Rotation of log Coefficient of rolling resistance Direction of motion of center of log relative to stone C Direction of motion of center of log relative to ground 0.08 ft Coefficient of rolling resistance R g 3 s the log rolls on the ground, the stone exerts a force opposing the rolling motion.

22 7.7 Rolling Resistance Example 6, page 3 of 5 4 For use in calculating moments, find the vertical distances from the center to the points C and where the resisting forces act ft 0.5 ft (0.5 ft) 2 (0.01 ft) 2 = ft Radius = (1 ft)/2 = 0.5 ft 0.5 ft (0.5 ft) 2 (0.08 ft) 2 = ft C 0.08 ft

23 + 7.7 Rolling Resistance Example 6, page 4 of 5 5 Free-body diagram of stone and logs 100,000 lb F R g-y R g-x The same resisting force acts on each of the ten logs. Equilibrium equations + Fx = 0: F 10R g-x = 0 (1) F y = 0: 10R g-y 100,000 lb = 0 (2) Two equations, three unknowns: an additional free-body is needed.

24 + 7.7 Rolling Resistance Example 6, page 5 of 5 6 Free-body diagram of an individual wheel Rs-x R s-y Force components from the stone ft ft 7 Moment equilibrium M = 0: R g-y (0.01 ft ft) R g-x ( ft ft) = 0 (3) Solving Eqs. 1-3 gives C R g-x R g-x = 906 lb 0.01 ft 0.08 ft R g-y R g-y = 10,000 lb F = 9,060 lb ns. 8 The force F required to move the 100,000-lb stone is large, even when the ground is level. One scholar has estimated that as many as 600 men were needed to move such a stone up one of the slopes lying between the quarry and the monument site.

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