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The coefficient of static friction between the m = 2.95-kg crate and the 35.0° incline is 0.345. What minimum force must be app...Asked by Anonymous
The coefficient of static friction between the 2.3 kg crate and the 35.0° incline of Figure P4.47 is 0.241. What minimum force, F, must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?
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Answered by
Elena
ΣF(x) = 0: m•g•sinα=F(fr)
ΣF(y) = 0: m•g•cosα +F = N
F(fr)=μ•N= μ• (m•g•cosα +F)
m•g•sinα= μ• (m•g•cosα +F)
F=(m•g/μ) •(sinα -μ• cosα)
ΣF(y) = 0: m•g•cosα +F = N
F(fr)=μ•N= μ• (m•g•cosα +F)
m•g•sinα= μ• (m•g•cosα +F)
F=(m•g/μ) •(sinα -μ• cosα)
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