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 GFin the +xdirection). Applying Newton’s second law to the xand yaxes, we obtain , max0sNFfmFmga−=−=respectively. The second equation yields the normal force FN= mg, whereupon the maximum static friction is found to be (from Eq. 6-1) fss,maxmg=μ. Thus, the first equation becomes Fmgmas−==μ0where 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 μs=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 ×102N. (b) Replacing m= 45 kg with m= 28 kg, the reasoning above leads to roughly . 21.210NF=×3. We denote as the horizontal force of the person exerted on the crate (in the +xdirection), Gis the force of kinetic friction (in the –xdirection), is the vertical normal force exerted by the floor (in the +ydirection), and GFfkNFmgGis the force of gravity. The magnitude of the force of friction is given by fk= μkFN(Eq. 6-2). Applying Newton’s second law to the xand yaxes, we obtain 0kNFfmaFmg−=−=respectively. (a) The second equation yields the normal force FN= mg, so that the friction is ()()220.3555 kg (9.8 m/s )1.910N .kkfmgμ===×(b) The first equation becomes Fmgmka−=μ
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