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• An inclined plane is a type of simple machine
• An inclined plane is a large and flat object that is tilted so that one end is higher than the other
What is an Inclined Plane?
Real World Applications of Inclined Planes
Hadley Canal in Massachusetts used Inclined Plane engineering around 1800’s to raise and lower boats over Great Falls
• The greater the angle of the inclined surface, the faster an object will slide down the incline
• There are always at least 2 forces acting on an object on an inclined plane Fgrav and Fnorm
Background Information
The normal force is always perpendicular to the inclined surface
The gravitational force (WEIGHT) is always in downward direction
2. Deduce the net force and solve for other unknowns including acceleration and µ
What is the net force for the
Inclined Plane diagram to the
right?
Example to Do Together!
Solve for all unknowns listed. The crate has a mass of 100 kg and the coefficient of friction between the crate and the incline is 0.3
F║ = Ffrict = a =
F┴ = Fnet =
1. Break down Fgrav into its components
F║ = mgsinΘF║ = (100)(9.8)sin30°F║ = 490 N
F┴ = mgcosΘF┴ = (100)(9.8)cos30°F┴ = 849 N
Fgrav = mgFgrav = (100)(9.8)Fgrav = 980 N
2. This example has friction, therefore let’s solve for Ffrict next
Ffrict = µ•Fnorm
Ffrict = 0.3•849
Ffrict = 255 N
F ║ = 490 N
F ┴ = 849 N
Remember,
Fnorm is
ALWAYS equal and
opposite F┴
849
980
3. Now, from our results we can deduce Fnet
Fnet = F║ - Ffrict
Fnet = 490 – 255
Fnet = 235 NF ║ = 490 N
F ┴ = 849 N
849
255
980Remember, when there is
NO FRICTION Fnet = F║
4. Lastly, we need to solve for the acceleration
Fnet = ma
a = Fnet ÷ m
a = 235 ÷ 100
a = 2.35 m/s2
F ║ = 490 N
F ┴ = 849 N
849
255
980Fnet = 235 N
Answers
1. F║ = mgsinΘ
F║ = (100)(9.8)sin30°
F║ = 490 N
2. F┴ = mgcosΘ
F┴ = (100)(9.8)cos30°
F┴ = 849 N
3. F┴ = Fnorm therefore Fnorm = 849 N