06_InstSolManual_PDF_Part2

06_InstSolManual_PDF_Part2 - Friction is static friction...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Figure 6.2 (b) gives 6.3. Set Up: Each hand travels in a circle of radius 0.750 m and has mass and weight 6.4 N. The period for each hand is Let be toward the center of the circular path. (a) The free-body diagram for one hand is given in Figure 6.3. is the force exerted on the hand by the wrist. This force has both horizontal and vertical components. Figure 6.3 (b) gives (c) The horizontal force from the wrist is 12 times the weight of the hand. Reflect: The wrist must also exert a vertical force on the hand equal to its weight. 6.4. Set Up: Since the car travels in an arc of a circle, it has acceleration directed toward the center of the arc. The only horizontal force on the car is the static friction force exerted by the roadway. To calculate the minimum coefficient of friction that is required, set the static friction force equal to its maximum value,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Friction is static friction because the car is not sliding in the radial direction. Solve: (a) The free-body diagram for the car is given in Figure 6.4. The diagram assumes the center of the curve is to the left of the car. Figure 6.4 y n mg x f s 5 m s n a rad f s 5 m s n . a rad 5 v 2 / R , F w 5 77 N 6.4 N 5 12. F x 5 ma rad 5 1 0.65 kg 21 118 m / s 2 2 5 77 N g F x 5 ma x a rad 5 4 p 2 R T 2 5 4 p 2 1 0.750 m 2 1 0.50 s 2 2 5 118 m / s 2 w x y F F x F y a rad F S 1 x T 5 1 1.0 s 2 / 1 2.0 2 5 0.50 s. 1 0.0125 21 52 kg 2 5 0.65 kg v 5 TR m 5 1 60.0 N 21 0.90 m 2 0.80 kg 5 8.22 m / s. T 5 m v 2 R . g F x 5 ma x y x mg T n a rad 6-2 Chapter 6...
View Full Document

Ask a homework question - tutors are online