Question #99512
A uniform 5.0 m long ladder of mass 25 kg leans against a frictionless wall at an angle of 62 degrees with the floor. Determine the minimum coefficient of static friction needed on the floor for the ladder not to slide.
1
Expert's answer
2019-11-28T08:35:44-0500
NF=mg,NW=Ffriction=μNF=μmgN_F=mg, N_W=F_{friction}=\mu N_F=\mu mg

NwLsinϕ=mg(0.5L)cosϕN_wL\sin{\phi}=mg(0.5L)\cos{\phi}

So,


μmgLsinϕ=mg(0.5L)cosϕ\mu mgL\sin{\phi}=mg(0.5L)\cos{\phi}

tanϕ=12μ\tan{\phi}=\frac{1}{2\mu}

tan62=12μ\tan{62}=\frac{1}{2\mu}

μ=0.27\mu=0.27


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Comments

Assignment Expert
25.03.21, 20:13

Dear Rachel, \mu = 1/(2*tan(62)) = 0.27

Rachel
25.03.21, 08:11

I tried doing the calculations myself but couldn’t get 0.27 so I was wondering how to get to that answer?

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