Chapter06

# Chapter06 - 1. We do not consider the possibility that the...

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1. 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 G F in the + x direction). Applying Newton’s second law to the x and y axes, we obtain ,max 0 s N Ff m a Fm g −= respectively. The second equation yields the normal force F N = mg , whereupon the maximum static friction is found to be (from Eq. 6-1) fm g ss ,max = μ . Thus, the first equation becomes g m a s = 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. (a) With s = 045 . 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 .

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2. 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 Ff m a Fm g −= 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 g m a k = 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 k = 0.80, we find F = 1.6 × 10 2 N.
3. We denote G F as the horizontal force of the person exerted on the crate (in the + x direction), G 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 G 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 Ff m a Fm g −= respectively. (a) The second equation yields the normal force F N = mg , so that the friction is () ( ) 22 0.35 55 kg (9.8 m/s ) 1.9 10 N . kk fm g == = × (b) The first equation becomes g m a k which (with F = 220 N) we solve to find a F m g k =− = 056 2 .. m/s

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4. The free-body diagram for the player is shown next. N F G is the normal force of the ground on the player, mg G is the force of gravity, and G 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, () 2 470 N 0.61. 79 kg 9.8m/s k N f F == =
5. 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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## This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter06 - 1. We do not consider the possibility that the...

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