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

Step-by-step 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* 9.8* \sin 18+\mu _s* 60* 9.8* \cos 18`

F=181.70+285.20=466.9 N

(b)
`\mu _k`=0.26

`F-mg\sin \theta -f_r=m* a`

here
`f_r=\mu _kmg\cos \theta =0.26* 60* 9.8* \cos 18=145.39 N`

`466.9-187.701-145.39=60* a`

`466.9-33.09=60* a`

`a=(133.809)/(60)=2.23 m/s^2`