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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 == =
Background image of page 4
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 .
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 111

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online