Question

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?

Answer 1

Answer:

Explanation:

Given

mass of crate=60 kg

inclination$=18^{\circ}$

$\mu _=0.51$

Suppose Force applied by rope is F

$F-mg\sin \theta -f_r=0$

$f_r=\mu _smg\cos \theta$

$F=60\times 9.8\times \sin 18+\mu _s\times 60\times 9.8\times \cos 18$

F=181.70+285.20=466.9 N

(b)$\mu _k$=0.26

$F-mg\sin \theta -f_r=m\times a$

here $f_r=\mu _kmg\cos \theta =0.26\times 60\times 9.8\times \cos 18=145.39 N$

$466.9-187.701-145.39=60\times a$

$466.9-33.09=60\times a$

$a=\frac{133.809}{60}=2.23 m/s^2$