Forces in Equilibrium & Motion along an Incline
Chapter 7.1
EquilibriumNewton’s 1st Law of Motion When the forces on an object are
balance, it is said to be in equilibrium. When an object is in equilibrium, it is
not accelerating. An object that is not accelerating is
stationary or moving at constant speed in a straight line.
You balanced forces during the force table lab.
Ex. 1: EquilibriumA 100 N sign is hung by two wires as seen below. What is the tension in the wires?
Physics is Fun
Fg = 100 N
FA FB
= 15°
Diagram the Problem
Physics is Fun
Fg = 100 N
FA FB
y
x
System
= 15°
y
x
FBx
FByFAy
FAx
Fg = 100 N
FA FB
State the Known & Unknown
What is known? Fg = 100N θ = 15°What is not known? FA FB
Perform CalculationsIsolate the x and y components separately.Since the sign is not moving, Fnet = ma = 0 in both the x and y directions.x – direction:
-FAx + FBx = 0
-FA cosθ + FB cosθ = 0 FA cosθ = FB cosθ
y – direction: FA sinθ + FB sinθ – Fg = 0 2FA sinθ = Fg FA = (100N)/(sin(15°)(2) FA = 193 N
Motion on an InclineWhen objects are not on a flat level surface: A portion of the
gravitational force is directed along the surface.
The normal force is not equal to the weight of the object.
Choose a coordinate system such that the x-axis is directed parallel to the slope.
FN
Fg
Fgx
Fgy
x
y
Ex. 2: Motion on an InclineDetermine the rate of acceleration of a 25 kg wooden crate as it slides down a wooden ramp with a coefficient of friction, = 0.2. The angle the ramp makes with the horizontal is 30.What is known?
m = 25 kg = 0.2 θ = 30 vi = 0 m/s
What is not known? a = ?
System
FN
Fg
Fgx
Fgy
Diagram and Solve the Problem
y-direction: Fnet,y = FN – Fgy Since Fnet,y = 0 FN = Fgy = mgcosθ
x-direction: Fnet,x = Fgx – Ff ma = mgsinθ – μFN ma = mgsinθ - μmgcosθ a = g(sinθ – μcosθ) a = 9.8 m/s2(sin 30° - μcos 30°) a = 3.2 m/s2
Fg(y)
Ff
FN
Fg
Fg(x)
y-axisx-axis
Key Ideas Equilibrium: When an object is at rest, or when an
object is in motion at a constant speed in a straight line.