Physics_Chapt6

Physics_Chapt6 - 1 Chapter 6 1 We do not consider the...

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1 Chapter 6 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 ~ F in the + x direction). Applying Newton’s second law to the x and y axes, we obtain F - f s, max = ma N - mg = 0 respectively. The second equation yields the normal force N = mg , whereupon the maximum static friction is found to be (from Eq. 6-1) f s, max = μ s mg . Thus, the Frst equation becomes F - μ s mg = ma = 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 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 signiFcant Fgures, 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 F = 1 . 2 × 10 2 N. 2. 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 = tan - 1 μ s = tan - 1 0 . 04 2 . 3. The free-body diagram for the player is shown below. ~ N is the normal force of the ground on the player, m~g is the force of gravity, and ~ f is the force of friction. The force of friction is related to the normal force by f = μ k N . We use Newton’s second law applied
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to the vertical axis to fnd the nor- mal Force. The vertical compo- nent oF the acceleration is zero, so we obtain N - mg = 0; thus, N = mg . Consequently, μ k = f N = 470 N (79 kg) ± 9 . 8 m / s 2 ² = 0 . 61 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... ~ N m~g ~ f 4. To maintain the stone’s motion, a horizontal Force (in the + x direction) is needed that cancels the retarding e±ect due to kinetic Friction. Applying Newtons’ second to the x and y axes, we obtain F - f k = ma N - mg = 0 respectively. The second equation yields the normal Force N = mg , so that (using Eq. 6-2) the kinetic Friction becomes f k = μ k mg . Thus, the frst equation becomes F - μ k mg = ma = 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 fnd F = 1 . 6 × 10 2 N. 5. We denote ~ F as the horizontal Force oF the person exerted on the crate (in the + x direction), ~ f k is the Force oF kinetic Friction (in the - x direction), ~ N is the vertical normal Force exerted by the ²oor (in the + y direction), and m~g is the Force oF gravity. The magnitude oF the Force oF Friction is given by f k = μ k N (Eq. 6-2). Applying Newtons’ second to the x and y axes, we obtain F - f k = ma N - mg = 0 respectively.
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(a) The second equation yields the normal force N = mg , so that the friction is f k = μ k mg = (0 . 35)(55 kg) ± 9 . 8 m / s 2 ² = 1 . 9 × 10 2 N . (b) The Frst equation becomes F - μ k mg = ma which (with F = 220 N) we solve to Fnd a = F m - μ k g = 0 . 56 m / s 2 .
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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Physics_Chapt6 - 1 Chapter 6 1 We do not consider the...

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