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Unformatted text preview: 217 Chapter 6 1. The greatest deceleration (of magnitude a ) is provided by the maximum friction force (Eq. 6-1, with F N = mg in this case). Using Newton’s second law, we find a = f s,max / m = s g . Eq. 2-16 then gives the shortest distance to stop: | x | = v 2 /2 a = 36 m. In this calculation, it is important to first convert v to 13 m/s. 2. Applying Newton’s second law to the horizontal motion, we have F k m g = ma , where we have used Eq. 6-2, assuming that F N = mg (which is equivalent to assuming that the vertical force from the broom is negligible). Eq. 2-16 relates the distance traveled and the final speed to the acceleration: v 2 = 2 a x . This gives a = 1.3 m/s 2 . Returning to the force equation, we find (with F = 25 N and m = 3.5 kg) that k = 0.60. 3. The free-body diagram for the bureau is shown to the right. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion problem (with the person’s push F in the + x direction). Applying Newton’s second law to the x and y axes, we obtain , max s N F f ma F mg respectively. The second equation yields the normal force F N = mg , whereupon the maximum static friction is found to be (from Eq. 6-1) f mg s s ,max . Thus, the first equation becomes F mg ma s where we have set a = 0 to be consistent with the fact that the static friction is still (just barely) able to prevent the bureau from moving. (a) With s 0 45 . and m = 45 kg, the equation above leads to F = 198 N. To bring the bureau into a state of motion, the person should push with any force greater than this value. Rounding to two significant figures, we can therefore say the minimum required push is F = 2.0 10 2 N. CHAPTER 6 218 (b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly 2 1.2 10 N F . Note: The values found above represent the minimum force required to overcome the friction. Applying a force greater than s mg results in a net force in the + x-direction, and hence, non-zero acceleration. 4. We first analyze the forces on the pig of mass m . The incline angle is . The + x direction is “downhill.’’ Application of Newton’s second law to the x- and y-axes leads to sin cos 0. k N mg f ma F mg Solving these along with Eq. 6-2 ( f k = k F N ) produces the following result for the pig’s downhill acceleration: sin cos . k a g To compute the time to slide from rest through a downhill distance , we use Eq. 2-15: v t at t a 2 1 2 2 . We denote the frictionless ( k = 0) case with a prime and set up a ratio: t t a a a a 2 2 / / which leads us to conclude that if t / t' = 2 then a' = 4 a . Putting in what we found out above about the accelerations, we have sin 4 sin cos ....
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