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.