inclined planes presentation
Post on 25-May-2015
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A skier is sliding down a slope.
The gravitational force acting upon the skier is directed _____.
a. straight upward b. straight downward c. perpendicular to the slope d. parallel to the slope
A skier is sliding down a slope.
The gravitational force acting upon the skier is directed _____.
a. straight upward b. straight downward c. perpendicular to the slope d. parallel to the slope
Fg
The normal force acting upon the skier is directed _____.
a. perpendicular to the slope b. parallel to the slopec. straight downward d. straight upward
Fg
The normal force acting upon the skier is directed _____.
a. perpendicular to the slope b. parallel to the slopec. straight downward d. straight upward
Fg
Fn
The friction force acting upon the skier is directed _____.
a. parallel to the slope b. straight upward c. straight downward d. perpendicular to the slope
Fg
Fn
The friction force acting upon the skier is directed _____.
a. parallel to the slope b. straight upward c. straight downward d. perpendicular to the slope
Fg
FnFfr
An object upon an inclined plane is sliding at a constant speed down the incline. Friction is present. Which one of the following diagrams represent the free-body diagram for such an object?
How large is the force of friction?
Fg
FnFfr
How large is the force of friction?
Construct a coordinate system…
Fg
FnFfr
+x
+y
Fg
FnFfr
+x
+y
q
q
How large is the force of friction?
Construct a coordinate system…
Fg
FnFfr
+x
+y
q
q
How large is the force of friction?
Construct a coordinate system…
…now break Fg into its components…
Fperp
Fpar
Fg
FnFfr
+x
+y
q
q
A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine
Fperp
Fpar Fperp
Fpar
Fg= 500 N
FnFfr
+x
+y
30
A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine
Fperp = (500 N) cos 30
= 433 N
Fpar = (500 N) sin 30
= 250 N
Fperp
Fpar
Fg= 500 N
FnFfr
+x
+y
30
A downhill skier has a weight of 500 N,and is skiing down a 30 hill. Determine
433 N = Fperp
250 N = Fpar
Fg= 500 N
FnFfr
+x
+y
30
If the skier is not accelerating, h how large is the force of friction?
How large is the normal force?
433 N = Fperp
250 N = Fpar
20
A 12 kg box slides down a ramp at a constant velocity.
a. Sketch a free body diagram for the box.
20
A 12 kg box slides down a ramp at a constant velocity.
a. Sketch a free body diagram for the box.
Fg
FnFfr
20
A 12 kg box slides down a ramp at a constant velocity.
b. Make a tilted coordinate system and find Fperp and Fpar.
Fg
FnFfr
20
A 12 kg box slides down a ramp at a constant velocity.
b. Make a tilted coordinate system and find Fperp and Fpar.
Fg
FnFfr
+y
+x
20
A 12 kg box slides down a ramp at a constant velocity.
b. Make a tilted coordinate system and find Fperp and Fpar.
Fperp = (117.6 N) cos 20
= 110.5 N
Fpar = (117.6 N) sin 20
= 40.2 N
Fg
FnFfr
+y
+x
20
20
A 12 kg box slides down a ramp at a constant velocity.
c. Determine the force of friction and the normal force on the block.
Ff =
Fn = Fg
FnFfr
+y
+x
2011
0.5
N
40.2 N
20
A 12 kg box slides down a ramp at a constant velocity.
c. Determine the force of friction and the normal force on the block.
Ff = 40.2 N
Fn = 110.5 N
FnFfr
+y
+x110.
5 N
40.2 N
20
A 12 kg box slides down a ramp at a constant velocity.
d. If there was no friction, what would be the acceleration down the ramp?
FNET = m a 40.2 N = (12 kg) a a = 3.35 m/s^2
Fn
+y
+x110.
5 N
40.2 N
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