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CH-06 - Chapter 6 1 An excellent discussion and equation...

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Chapter 6 1. An excellent discussion and equation development related to this problem is given in Sample Problem 6-3. We merely quote (and apply) their main result (Eq. 6-13) θ μ = = ° - - tan tan . . 1 1 0 04 2 s 2. The free-body diagram for the player is shown next. N F r is the normal force of the ground on the player, mg r is the force of gravity, and r f is the force of friction. The force of friction is related to the normal force by f = μ k F N . We use Newton’s second law applied to the vertical axis to find the normal force. The vertical component of the acceleration is zero, so we obtain F N mg = 0; thus, F N = mg . Consequently, ( 29 ( 29 2 470 N 79 kg 9.8 m/s 0.61. k N f F μ = = = 3. 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 r F in the + x direction). Applying Newton’s second law to the x and y axes, we obtain , max 0 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 = μ . Thus, the first equation becomes F mg ma s - = = μ 0 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. 223
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CHAPTER 6 (a) With μ 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. (b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly 2 1.2 10 N F = . 4. To maintain the stone’s motion, a horizontal force (in the + x direction) is needed that cancels the retarding effect due to kinetic friction. Applying Newton’s second to the x and y axes, we obtain 0 k N F f ma F mg - = - = respectively. The second equation yields the normal force F N = mg , so that (using Eq. 6- 2) the kinetic friction becomes f k = μ k mg . Thus, the first equation becomes F mg ma k - = = μ 0 where we have set a = 0 to be consistent with the idea that the horizontal velocity of the stone should remain constant. With m = 20 kg and μ k = 0.80, we find F = 1.6 × 10 2 N. 5. We denote r F as the horizontal force of the person exerted on the crate (in the + x direction), r f k is the force of kinetic friction (in the – x direction), N F is the vertical normal force exerted by the floor (in the + y direction), and mg r is the force of gravity. The magnitude of the force of friction is given by f k = μ k F N (Eq. 6-2). Applying Newton’s second law to the x and y axes, we obtain 0 k N F f ma F mg - = - = respectively. (a) The second equation yields the normal force F N = mg , so that the friction is f mg k k = = = × μ 0 35 55 9 8 19 10 2 2 . . . . b gb gc h kg m / s N (b) The first equation becomes F mg ma k - = μ which (with F = 220 N) we solve to find 224
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a F m g k = - = μ 056 2 . . m / s 6. 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 = μ s g .
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