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Physics
Chapter 7: Forces & Motion in 2-D
Day 57-66
Inclined Plane
The motion along an inclined plane causes some variations in equations.
The normal force is perpendicular to the incline. The weight force is toward the center of
the earth.
The force of friction will either speed up or slow down the object. The direction of the
velocity and the acceleration will be parallel to the hill along the x-axis.
The y-axis is perpendicular to the hill and is the normal force.
Diagram
Fgx = -Fg sin Ǿ and Fgy = -Fg cos Ǿ
If a box with a weight of 562 N rests on an incline at an angle of 30 degrees, find the Fgx
and Fgy forces. Remember both will be negative because of their direction in relationship
to the coordinate system.
A skier (mass of 62 kg) is going down a hill that has a slope of 37 degrees. What are the
horizontal and vertical forces on the skier?
Find weight. Find Fx = -Fg sin Ǿ and Fy = -Fg cos Ǿ
The skier will accelerate down the hill. The coefficient of friction of the snow is 0.15.
What velocity will the skier have after 5.0 seconds of skiing?
The acceleration occurs in the x direction. There must be a net force.
Fnet x = ma but ma = Fx –Ff
ma = Fg sin Ǿ - µ FN or mg sin Ǿ - µ FN
Remember that FN = Fy = mg cos Ǿ
So ma = mg sin Ǿ - µmg cos Ǿ
Divide both sides by m, factor and
a = g (sin Ǿ - µ cos Ǿ)
Solve for a then to find v = vo + at
V = 24 m/s
Lab on Inclined Plane
Worksheet: two problems, section 7.1