How Is Static Friction Affecting the Boy and the Crate's Movement?

[marissa12]

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip. You start to wonder how heavy the crate is. You call to ask the boy his mass, and he replies " 51.0 kg." From your recent physics class you estimate that the static and kinetic coefficients of friction are 0.900 and 0.4 for the boy's shoes, and 0.5 and 0.2 for the crate.

what is the mass of the crate?

<<Forces acting on the boy>>

X= F(static friction)-F(box on boy)=ma=0 F(static f)=F(boy on box)

y= Fg=Fn

<<Forces acting on box>>

X= F(boy on box)- F(kinetic friction)=ma=0 F(boy on b)= F(kinetic friction)

y: Fn=Fg

and the F(boy on b)= -F(box on boy)= F


i can't understand why this isn't working if i plug in all my numbers and manipulate the equations.

 

[Doc Al]

<<Forces acting on the boy>>

X= F(static friction)-F(box on boy)=ma=0 F(static f)=F(boy on box)

Good. (I assume you meant F(box on boy).)

<<Forces acting on box>>

X= F(boy on box)- F(kinetic friction)=ma=0 F(boy on b)= F(kinetic friction)

Good.

and the F(boy on b)= -F(box on boy)= F

Good. In your earlier equations F(boy on box) & F(box on boy) represent the magnitudes of those forces, so F(boy on box) = F(box on boy).

Now put it all together: If F = A and F = B, then A = B. (That's a hint.)

 

[marissa12]

hah yay.. i had it right the whole time.. its just i wrote my work soo sloppy i messed up the numbers. thank you !