{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ssm_ch06

# ssm_ch06 - Chapter 6 Student Solutions Manual 1 We do not...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 6 – Student Solutions Manual 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 F f m F mg a = = 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 s s ,max mg = 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. (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 = × 3. We denote as the horizontal force of the person exerted on the crate (in the + x direction), G is the force of kinetic friction (in the – x direction), is the vertical normal force exerted by the floor (in the + y direction), and G F f k N F 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 F f ma F mg = = respectively. (a) The second equation yields the normal force F N = mg , so that the friction is ( )( ) 2 2 0.35 55 kg (9.8 m/s ) 1.9 10 N . k k f mg μ = = = × (b) The first equation becomes F mg m k a = μ

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

View Full Document