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

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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 fm Fm g 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 ss ,max m g = 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 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 = 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 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 g == = × (b) The first equation becomes g m k a =
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which (with F = 220 N) we solve to find a F m g k =− = μ 056 2 .. m/s 13. (a) The free-body diagram for the crate is shown on the right. G T is the tension force of the rope on the crate, is the normal force of the floor on the crate, mg N F G G is the force of gravity, and is the force of friction. We take the + x direction to be horizontal to the right and the + y direction to be up. We assume the crate is motionless. The equations for the x and the y components of the force according to Newton’s second law are: G f T cos θ f = 0 sin 0 N TF m g +− = where = 15° is the angle between the rope and the horizontal. The first equation gives f = T cos and the second gives F N = mg T sin . If the crate is to remain at rest, f must be less than s F N , or T cos < s ( mg T sin ). When the tension force is sufficient to just start the crate moving, we must have T cos = s ( mg T sin ). We solve for the tension: ( ) ( ) ( ) 2 2 0.50 68 kg 9.8 m/s 304 N 3.0 10 N. cos sin cos 15 0.50 sin 15 s s mg T θμ == = + ° × ) (b) The second law equations for the moving crate are T cos f = ma F N + T sin mg = 0. Now f = k F N , and the second equation gives F N = mg T sin , which yields (s i n k fm g T . This expression is substituted for f in the first equation to obtain T cos k ( mg T sin ) = ma , so the acceleration is
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( ) cos sin k k T ag m θμ θ μ + =− .
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This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

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

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