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A 60 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 18° above the horizontal. (a) If the coefficient of static friction is 0.51, what minimum force magnitude is required from the rope to start the crate moving? (b) If μk = 0.26, what is the magnitude of the initial acceleration (m/s^2) of the crate?

Respuesta :

Answer:

Explanation:

Given

mass of crate=60 kg

inclination[tex]=18^{\circ}[/tex]

[tex]\mu _=0.51[/tex]

Suppose Force applied by rope is F

[tex]F-mg\sin \theta -f_r=0[/tex]

[tex]f_r=\mu _smg\cos \theta [/tex]

[tex]F=60\times 9.8\times \sin 18+\mu _s\times 60\times 9.8\times \cos 18[/tex]

F=181.70+285.20=466.9 N

(b)[tex]\mu _k[/tex]=0.26

[tex]F-mg\sin \theta -f_r=m\times a[/tex]

here [tex]f_r=\mu _kmg\cos \theta =0.26\times 60\times 9.8\times \cos 18=145.39 N[/tex]

[tex]466.9-187.701-145.39=60\times a[/tex]

[tex]466.9-33.09=60\times a[/tex]

[tex]a=\frac{133.809}{60}=2.23 m/s^2[/tex]

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