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Force Applied at an Angle
Force Applied at an Angle
A force may be applied at an angle to a
horizontal surface:
Fapp
Force Applied at an Angle
Force Applied at an Angle
Once the parallel applied force is found, we
may use our equations for a flat surface:
=
Fnet = Fapp + Ff
θ
We must use trigonometry to
find the parallel component:
Fapp
AND
Fnet = ma
Friction calculations are done like normal,
including coefficient of friction (µ)
θ
Fapp
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
=
=
Fapp = Fapp x cosθ
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
Specific Outcome:
i. I can explain, qualitatively and quantitatively, static and kinetic forces of
friction acting on an object.
ii. I can apply Newton’s laws of motion to solve, algebraically, linear motion
problems in horizontal, vertical and inclined planes near the surface of
Earth, ignoring air resistance.
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
1
Force Applied at an Angle
ex. A desk of mass 30.0 kg is pushed forward at an
angle of 30° to the floor at a constant speed by a
force of 50 N. Find the coefficient of kinetic friction
of the surface.
=
Fapp = (50 N) cos30 = 43.301…N [forward]
ex. A desk of mass 30.0 kg is pushed forward at an
angle of 30° to the floor at a constant speed by a
force of 50 N. Find the coefficient of kinetic friction
of the surface.
FN = |Fg| = |mg| = |(30.0 kg)(-9.81 m/s2)|
= 294.3 N
=
Fnet = Fapp + Ff
Force Applied at an Angle
=
FF = Fnet – Fapp = 0 N – 43.301…N = -43.301…N
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
µK =
Ff
43.301…N
=
= 0.15
FN
294.3 N
Dulku – Physics 20 – Unit 2 (Dynamics) – Topic L
2