The initial force it takes to move the crate is given by μs*mg
where μs is the coefficient of static friction.
Similarly, the coefficient of kinetic friction can be calculated from the force it takes to maintain motion.
Post for an answer check if you wish.
A dockworker loading crates on a ship finds that a 24 kg crate, initially at rest on a horizontal surface, requires a 79 N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 57 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.
2 answers
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Sum F = ma ( because it get loading the crates )
Fp - F(fr) = ma
Fp = ma + F(fr)
Fp = ma + u(k) mg
=> u(k) = (Fp - ma) / mg ( with ma is 0 )
so u(k) = 57/ ( 24*9.8) ( because kenetic friction always moving )
The u(s) you guys can do the same thing , just replace another force number to the equation.
Sum F = ma ( because it get loading the crates )
Fp - F(fr) = ma
Fp = ma + F(fr)
Fp = ma + u(k) mg
=> u(k) = (Fp - ma) / mg ( with ma is 0 )
so u(k) = 57/ ( 24*9.8) ( because kenetic friction always moving )
The u(s) you guys can do the same thing , just replace another force number to the equation.