Question #157191
A uniform ladder AB of wieght W and length 2l rests with the end A against a smooth vertical wall and the end B on a rough horizontal ground. A man of weight equal to that of the ladder stands at the point C on the ladder, where BC = 5l/3 . The coefficient of friction between the ladder and the ground is 1/3. Given that the ladder is in limiting equilibrium when it makes an angle s to the horizontal;
(a) Show that tans = 2
(b) Find the magnitude of the reaction force at the wall and on the ground.
1
Expert's answer
2021-02-08T18:42:15-0500

let O middle of ladder\text {let }O\text{ middle of ladder}

AO=OB=l;AC=l3;BC=5l3AO=OB= l;AC=\frac{l}{3};BC=\frac{5l}{3}

sangle s to the horizontal;s-\text{angle s to the horizontal;}

Ffrfriction force at point BF_{fr}-\text{friction force at point }B

TAwall reaction forceT_A-\text{wall reaction force}

TBreaction force at point BT_B-\text{reaction force at point B}

mgthe gravity of the ladder at a point Omg - \text{the gravity of the ladder at a point }O

mgthe gravity of the man at a point Cmg - \text{the gravity of the man at a point }C

forces along the X-axis (horizontal)\text{forces along the X-axis (horizontal)}

TAFfr=0T_A-F_{fr}=0

forces along the Y-axis (vertical)\text{forces along the Y-axis (vertical)}

TBmgmg=0;TB2mg=0;T_B-mg-mg=0;T_B-2mg=0;

equilibrium condition about point B\text{equilibrium condition about point }B

TAsin(s)mgcos(s)ACmgcos(s)AO=0T_A\sin{(s)}-mg*\cos(s)*AC-mg*\cos(s)*AO=0

TAsin(s)mgcos(s)13mgcos(s)l=0T_A\sin{(s)}-mg*\cos(s)*\frac{1}{3}-mg*\cos(s)*l=0

TA=43cot(s)mgT_A=\frac{4}{3}\cot(s)mg

Ffr=43cot(s)mgF_{fr}=\frac{4}{3}\cot(s)mg

a)friction static of the ladder and the man13mg+13mg=23mga)\text{friction static of the ladder and the man} - \frac{1}{3}mg+\frac{1}{3}mg=\frac{2}{3}mg

 limiting equilibriumFfrequal to static friction\text{ limiting equilibrium} F_{fr} \text{equal to static friction}

Ffr=43cot(s)mgF_{fr}=\frac{4}{3}\cot(s)mg

43cot(s)mg=23mg\frac{4}{3}\cot(s)mg=\frac{2}{3}mg

cot(s)=12\cot(s)=\frac{1}{2}

tan(s)=2\tan(s)=2

b)b) TA=Ffr=43cot(s)mgT_A=F_{fr}=\frac{4}{3}\cot(s)mg

Tgreaction force on the ground.T_g-\text{reaction force on the ground.}

Tg=TB2+Ffr2T_g= \sqrt{T_B^2+F_{fr}^2}

TB=2mgT_B=2mg

Tg=(4m2g2+169cot2(s)m2g2)T_g= \sqrt{(4m^2g^2+\frac{16}{9}}\cot^2(s)m^2g^2)


Answer:a)tan(s)=2\tan(s)=2 b)TA=43cot(s)mgT_A=\frac{4}{3}\cot(s)mg ;Tg=(4m2g2+169cot2(s)m2g2)T_g= \sqrt{(4m^2g^2+\frac{16}{9}}\cot^2(s)m^2g^2)













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