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A belt/pulley system has tight side of 1000N, a slack side of 100N and a wrap angle of 500 degrees. The belt is just on the point of slipping. What is the cofficient of friction?

User Breavyn
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1 Answer

3 votes

Answer:

0.26386

Step-by-step explanation:

T₁ = Tight side tension = 1000 N

T₂ = Slack side tension = 100 N

θ = Wrap angle = 500° = 500π/180 radians

μ = Coefficient of friction

Belt tension theory


(T_1)/(T_2)=e^(\mu\theta)\\\Rightarrow (1000)/(100)=e^{\mu* 500* (\pi)/(180)}\\\Rightarrow 10=e^{\mu* 500* (\pi)/(180)}

Taking natural logs on both sides


ln10=\mu* 500* (\pi)/(180)\\\Rightarrow \mu=(ln10)/(500* (\pi)/(180))=0.26386

∴ Coefficient of friction is 0.26386

User ClaraU
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