LON-CAPA Copper rod Secure I https ohio edu hrw/9e/chapter 28/28-01 w.problem?sy

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LON-CAPA Copper rod Secure I https ohio edu hrw/9e/chapter 28/28-01 w.problem?symb uploaded%2fohiou%2f8x43014f1bbMa580boucapa2%2fdefault 1455 /5858089185145/665255 esequence 38 wiley Mail-Wilson, Timoth Chegg Study I Guided Welcom LON-CAPM The Leam b CE2010. mothy i Timothy Wilson Student section: 101) Physics 2052/ Spring 2017 New Messages Courses Help Logout Contents Grades Main Menu O Ilmer Notes d Evaluate Feedback Pnnt info s week's orrlework due Fl day Copper rod just slides Course Contents copper rod just slides Due in 7 hnurs, 9 minutes A 1.12 kg copper rod rests nn two horizontal rails 0.62 m apart and carries a cument of 55 A from one rail to the other. The coefficient of static friction between rod and rails is 0.60. What is the magnitude of the smallest magnetic field that puts the rod on the verge of ding? 220 Submit Answer Incorrect. Tries 1/10 nes What is its angle (relative tn the vertical)? Submit Arlswel es 0/10 Threaded Vlew Chronological Vlew Other Vlews My general preferences on what ls marked as NEW Mark NEW posts no longer new Export NEW Part 1 Anonymous 1 R (Thu Mar 16 08:11:48 am 2017 (CDT I found force using the coefficient ot friction and such, but when I tried using the equation for torce due to the magnetic tield to tind the smallest magnetic tield it didn't work. Anyone have any ideas as to how to wo s problem out? NEW Re: Part 1 Anonymous 2 Reply (Fri Mar 17 12:24:36 am 2017 (HDT)) s am having a similar problem. I found the correct angle using arota ction rorce/Normal roroe) but when I solve for B using r-iUDsin(0): r being the foroe of friction, but I cannot get the correct magnetic field. Does anyone know what I'm doing wrong here? NEW Re: Re: Part 1 Alexandra Semposki Reply (Fr Mar 17 12:44 6 am 2017 (EDT) haven't tried this yet, but could it be a resultant force of the two forces that we're supposed to use? NEW Re: Re: Re: Part 1 Alexandra Semposki Reply (Tri Mar 17 12:45:26 a 2017 (CDT)) But ot course they're at 90 deg to each other aren't they? 420 PM Ask me any 3/1/20

Explanation / Answer

m = 1.12 kg, L = 0.62 m , I =55 A

u =0.6

Fx = ILBx , Fy = ILBy

forces along vertical direction

FN = mg - Fy

FN = mg - ILBy

f = u*FN

f = u*(mg - ILBy)

Net force along horizontal diection

Fx - f =0

ILBx = u(mg - ILBy)

Bx = Bcos(theta) , By = B sin(theta)

ILBcos(theta) = u(mg - ILBsin(theta))

B = u*mg/(IL(cos(theta)+usin(theta))

if we differentiate and equal to zer.

Then we get

theta = arctan(u)

theta = actan(0.6) = 31 degrees

Bmin = (0.6*1.12*9.8)/(55*0.62*(cos(31)+ 0.6*sin(31))

Bmin = 0.166 T

theta = 31 degrees

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