2D Forces: Pulling & Pushing
Review: Free Body Diagrams
But what if some of the forces are not in the x or y plane?
Forces at Angles
Part 1: Pulling
FN
Fg
Ff
Fpx
Fp = pulling force
Fp
Fpy
motion
y component
x component
(if constant velocity)
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(a) FN (b) Ff (c) Fnetx
FIRST: Draw your FBD
SECOND: List your givens (go ahead and find Fpx and Fpy)
FN
Fg
Fpx
Fp = 70N
Fpy
motion
x components y components both
μ = 0.4 Fnety = 0N θ = 35°
m = 15 kg
Fp = 70N
Ff
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(a) FN
FN
Fg
Fpx
Fp = 70N
Fpy
motion
Ff
In the previous unit, FN = Fg, but because the applied force is in both the x and y direction, FN ≠ Fg
Fg = mg
Fg = (15)(-9.8)
Fg = -147N
Fpy = Fpsinθ
Fpv = 70sin35
Fpv = 40N
0 = FN + 40 + (-147)
FN = 107N
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(a) FN (b) Ff (c) Fnetx
UPDATED GIVENS:
FN
Fg
Fpx
Fp = 70N
Fpy
motion
x components y components both
μ = 0.4 Fnety = 0N θ = 35°
Fg = -147N m = 15 kg
FN = 107N Fp = 70N
Fpy = 40N
Ff
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(b) Ff
FN
Fg
Fpx
Fp = 70N
Fpy
motion
Ff
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(a) FN (b) Ff (c) Fnetx
UPDATED GIVENS:
FN
Fg
Fpx
Fp = 70N
Fpy
motion
x components y components both
μ = 0.4 Fnety = 0N θ = 35°
Ff = 42.8N Fg = -147N m = 15 kg
FN = 107N Fp = 70N
Fpy = 40N
Ff
Sample Problem 1
A passenger is pulling on the strap of a 15 kg suitcase with a force of 70N. The strap makes a 35° angle above the horizontal. The coefficient of friction between the wheels of the suitcase and the surface is 0.4. Calculate the following quantities:
(c) Fnetx
FN
Fg
Fpx
Fp = 70N
Fpy
motion
Ff
Fpx = Fpcosθ
Fpx = 70cos35
Fpx = 57N
Sample Problem 2
(a) FN (b) Ff (c) Fnetx (d) a
Lesson Check 4.3
Part 2: Pushing
FN
Fg
Ff
Fpx
Fp = pushing force
Fp
Fpy
motion
y component
x component
(if constant velocity)
Fpy is negative due to the down direction
What’s the Difference?
Sample Problem 1
A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate
FIRST: Draw your FBD
SECOND: List your givens (go ahead and find Fpx and Fpy)
Fg
Fpx
Fp = 88N
Fpy
motion
Ff
FN
x components y components both
Fnetx = 0N Fnety = 0N θ = 45°
m = 14 kg
Fp = 88N
Sample Problem 1
A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate
Fnety = FN + Fpy + Fg
Fnety = 0
Need to find Fpy and Fg (both down, both negative)
0 = FN + -62 + (-137.2)
FN = 199.2N
Fg
Fpx
Fp = 88N
Fpy
motion
Ff
FN
Fpy = 88sin45
Fpy = -62N
Fg = (14)(-9.8)
Fg = -137.2N
Sample Problem 1
A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate
UPDATED GIVENS:
Fg
Fpx
Fp = 88N
Fpy
motion
Ff
FN
x components y components both
Fnetx = 0N Fnety = 0N θ = 45°
Fpy = 62N m = 14 kg
Fg = -137.2N Fp = 88N
FN = 199.2N
Sample Problem 1
A person pushes a 14 kg lawn mower at constant speed with a force of F = 88N directed along the handle, which is at an angle of 45° to the horizontal. Calculate
(b) the horizontal friction force on the mower
Fnetx = Fpx + Ff
Fnetx = 0
Need to find Fpx
0 = 62 + Ff
Ff = -62N
Fg
Fpx
Fp = 88N
Fpy
motion
Ff
FN
Fpx = 88cos45
Fpx = 62N