PHYSICS HELP!

Alpine3

Alpine3

Active member
i have a question that i cannot solve.... and need some help...

a steel ramp is to be built for sliding blocks of ice from a refrigeration plant down to ground level. if coefficient of friction =.05, find the angle with the horizontal at which the ice will slide at constant velocity.

anyone have any ideas?
 
Use the mass of the ice to figure out when Mu (Coeff. of friction) allows it go at a constant velocity.
 
The trick is to solve on a plain X axis. On this new axis, the force down the slope is: height x sin (theta)*angle of slope, this times the coefficient factor will give you the force down the slope, from that find the speed

 
don't hate on physics to much theres some really interesting parts in it, just stick it out and you'll see...
 
Ff=µ*Fn

F=M*A

Ff=Fg

Fg=9.8*cos(θ)*M

Ff=.05*Fn

Fn=9.8*sin(θ)*M

9.8*cos(θ)*M=.05*9.8*sin(θ)*M

9.8*cos(θ)=.05*9.8*sin(θ)

9.8/(.05*9.8)=sin(θ)/cos(θ)=tan(θ)

arctan(1/.05)=θ

θ≈1.521

YAY, PHYSICS!!!
 
sorry I messed that one up. I was thinking that θ was with respect to the horizontal, also that was radians, so...

Ff=µ*Fn

F=M*A

Ff=Fg

Fg=9.8*sin(θ)*M

Ff=.05*Fn

Fn=9.8*cos(θ)*M

9.8*sin(θ)*M=.05*9.8*cos(θ)*M

9.8*sin(θ)=.05*9.8*cos(θ)

(.05*9.8)/9.8=sin(θ)/cos(θ)=tan(θ)

arctan(.05)=θ

θ≈2.862º or 0.050 radians.

 
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