Question-227607

Question Number 227607 by MWSuSon last updated on 10/Feb/26

Answered by MWSuSon last updated on 10/Feb/26

i got (((√3)W)/3) for the reaction at C and μ= (1/( (√3))) The pronlem is μ=(√3) in the textbook. What did i do wrong?

$${i}\:{got}\:\frac{\sqrt{\mathrm{3}}{W}}{\mathrm{3}}\:{for}\:{the}\:{reaction}\:{at}\:{C}\:{and}\:\mu=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\: \\ $$$${The}\:{pronlem}\:{is}\:\mu=\sqrt{\mathrm{3}}\:{in}\:{the}\:{textbook}.\:{What}\:{did} \\ $$$${i}\:{do}\:{wrong}? \\ $$

Answered by mr W last updated on 10/Feb/26

Commented by mr W last updated on 10/Feb/26

MC=CB=a, say ⇒AM=MB=2a SC=(√3)a μ_A =tan φ=((AC)/(SC))=((3a)/( (√3)a))=(√3) ✓

$${MC}={CB}={a},\:{say} \\ $$$$\Rightarrow{AM}={MB}=\mathrm{2}{a} \\ $$$${SC}=\sqrt{\mathrm{3}}{a} \\ $$$$\mu_{{A}} =\mathrm{tan}\:\phi=\frac{{AC}}{{SC}}=\frac{\mathrm{3}{a}}{\:\sqrt{\mathrm{3}}{a}}=\sqrt{\mathrm{3}}\:\checkmark \\ $$

Commented by MWSuSon last updated on 10/Feb/26

R is the resultant of the reaction at A and the limiting friction right? also is the dotted line for the reaction? since the angle of friction is to that dotted line.

Commented by mr W last updated on 10/Feb/26

$${yes} \\ $$

Commented by MWSuSon last updated on 10/Feb/26

Commented by mr W last updated on 10/Feb/26

R=(((√3)W)/6) μR=(W/2) ⇒μ=((W×6)/(2×(√3)W))=(√3) ✓

$${R}=\frac{\sqrt{\mathrm{3}}{W}}{\mathrm{6}} \\ $$$$\mu{R}=\frac{{W}}{\mathrm{2}} \\ $$$$\Rightarrow\mu=\frac{{W}×\mathrm{6}}{\mathrm{2}×\sqrt{\mathrm{3}}{W}}=\sqrt{\mathrm{3}}\:\checkmark \\ $$

Commented by MWSuSon last updated on 10/Feb/26

i tried using moment and equating opposite forces, but still couldn't get √3. I will like to show you what I did and if possible, help me point out where it went wrong.

Commented by MWSuSon last updated on 10/Feb/26

I also prefer the geometric way. any recommendations on textbook or ebook I can learn more?

Commented by MWSuSon last updated on 10/Feb/26

Thanks. I see I somehow changed √3/6 to √3/2. Such a clumsy mistake. Thank you so much.

Commented by mr W last updated on 10/Feb/26

Commented by mr W last updated on 10/Feb/26

$${what}'{s}\:{the}\:{answer}\:{in}\:{this}\:{case}? \\ $$

Commented by MWSuSon last updated on 10/Feb/26

Care to share what you use for your illustration drawings?

Commented by MWSuSon last updated on 10/Feb/26

Looks like the same diagram. Is there something different? I can't see the difference in the diagrams.

Commented by mr W last updated on 10/Feb/26

Commented by mr W last updated on 10/Feb/26

$${drawn}\:{with}\:{the}\:{app}\:{INKredible} \\ $$

Commented by mr W last updated on 11/Feb/26

Commented by mr W last updated on 11/Feb/26

μ_A =tan φ=((2a×((√3)/2))/(2a×(1/2)+a×2))=((√3)/3)=(1/( (√3))) ✓

$$\mu_{{A}} =\mathrm{tan}\:\phi=\frac{\mathrm{2}{a}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{2}{a}×\frac{\mathrm{1}}{\mathrm{2}}+{a}×\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\checkmark \\ $$

Commented by MWSuSon last updated on 11/Feb/26

if I understand this correctly when it is just the edge of the rod touching A, the reaction becomes vertical through A?

Commented by mr W last updated on 11/Feb/26

$${yes}.\:{in}\:{first}\:{case}\:{the}\:{reaction}\:{force} \\ $$$${is}\:{orthogonal}\:{to}\:{the}\:{surface}\:{of}\:{rod}.\: \\ $$$${in}\:{second}\:{case}\:{it}\:{is}\:{orthogonal}\:{to}\:{the}\: \\ $$$${surface}\:{of}\:{the}\:{peg}\:{A}. \\ $$

Commented by fantastic2 last updated on 11/Feb/26