Problem 11.66 Part A One end of a uniform meter säck is placed against a vertica
ID: 1506903 • Letter: P
Question
Problem 11.66 Part A One end of a uniform meter säck is placed against a vertical wall Figure 1). The other end is held by a lightweight oord that makes an angle wth te stak. The ccefficient of static friction between the end of the mener what is the maximum value the angle can have if the stick is to rernan in equabrium? Express your answer using two significant figures My Answers Give Up Correct Part B Let the ange between the cord and the stick is 18g. Ablock ofthe same weight as meter sta is suspended born the stick, as shown at Elure2 , at adstance z from the wat what is he minmum value orfor which the sick wil remain regAbnm? Express your answer using two significant figures. r0.396 Incorrect; Try Again; 5 attempts remaining Faure 1 | 2 Part C when causing 18 g , how large must the coemcient of static tiction be sothat the block can be atached 11 CLnfrom the left end ofthe stck without to sap? Express your answer using two significant figures My Answers Give UpExplanation / Answer
Let length of the meter stick is L.,apply torque about the right end of the stick ,
fs= W/2
apply torque about the point where the cord is attatched to the wall ,
Fn tan (L) = W/2
Fn tan (1.0 m) = fs
fs /Fn = tan = 0.40
= arc tan(0.40) = 21.8 degree or 22 degree
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Apply torques as in above parts ,
fsL = W(L/2) + W (L - x)
FnL tan = W(L/2) + W (x)
FnL = [W(L/2) + W (x)] / tan
take ration fsL / FnL
fs / Fn = {W(L/2) + W (L - x) } / {[W(L/2) + W (x)] / tan }
the above equation can be write as in terms coeefficient of static friction ,
s > fs / Fn = (tan){[(L/2) +(L - x)] /[(L/2) + (x)]}
s > fs / Fn = (tan){[(3L/2) - x)] /[(L/2) + (x)]}
s > fs / Fn = (tan){[(3L - 2x)] /[(L+ 2x)]}
solve for x is
x = (1/2){[3tan18 - 0.40] / [0.40+ tan18]}
=39.64 cm
(c)
s = (tan18){[(3(100 cm) - 2(11cm))] /[(100 cm+ 2(11 cm))] =0.74