Resolving Forces Into Vector Components Physics Montwood High School R. Casao

Resolving Forces Into Vector

Components

PhysicsMontwood High School

R. Casao

Resolving Weight

The weight vector Fw for a mass can be resolved into an x- and y-component.

The x-component is called the parallel force and is represented by Fx or Fp.

The y-component is called the perpendicular or normal force and is represented by Fy or FN.

Resolving Weight

= 0°

The Fw vector is perpendicular (normal) to the surface: Fw = Fy.

Don’t worry about the sign of Fw or Fy.

Fy represents the amount of the object’s weight supported by the surface.

The normal force: FN – Fw = 0; FN = Fw

Normal Force

The normal force is a force that keeps one object from penetrating into another object.

The normal force is always perpendicular a surface.

The normal exactly cancels out the components of all applied forces that are perpendicular to a surface.

Normal Force on Flat Surface

The normal force is equal to the weight of an object for objects resting on horizontal surfaces.

FN = FW = m·g

m·g

FN

Resolving WeightFor > 0°

The weight vector Fw always points straight down. To make things easier, rotate the x and y axes so that the x-axis is parallel to the surface. The y-axis will be perpendicular to the surface.

Resolving WeightFx is the part of the

weight that causes the mass to slide down the inclined plane.

Fx will be positive because we take the direction of the motion to be positive.

Fx is the accelerating force; Fx = m·a.

Resolving WeightFy is the part of the

weight that presses the mass to the surface.

The surface exerts an equal and opposite upward force to balance Fy – the normal force FN.

The normal force:FN – Fy = 0; FN = Fy

Normal Force on RampThe normal force

is perpendicular to inclined ramps as well. It’s always equal to the component of weight perpendicular to the surface.

Fw = m·g

FN

Fy = m·g·cos

Fx = m·g·sin

FN = m·g·cos

Resolving Weight

In friction problems, FN is the force that presses the mass to the surface.

The angle between Fw (the hypotenuse) and Fy is always equal to the angle of the inclined surface.

Resolving Weight

Complete a right triangle:

w

x

w

y

FF

θsin

F

Fθcos

θsinFF

θcosFF

wx

wy

Resolving WeightGeneral equations (if N are given):

General equations (if kg are given):

Accelerating force: Fx=m·a

θcosFF

θsinFF

wy

wx

θcosgmF

θsingmF

y

x

How Does the Incline Affect the Components?

m

mg

mg cos

mg sin

The steeper the incline, the greater is, and the greater sin is. Thus, a steep incline means a large parallel component and a small horizontal one. Conversely, a gradual incline means a large horizontal component and a small vertical one.

m

mg

mg

cos

mg sin

Extreme cases: When = 0, the ramp is flat; red = mg; blue = 0.When = 90, the ramp is vertical; red = 0; blue = mg.

Inclined Plane: Normal Force

m

mgmg cos

mg sin

FN = mg cos

Recall normal force is perpen-dicular to the contact surface. As long as the ramp itself isn’t accelerating and no other forces are lifting the box off the ramp or pushing it into the ramp, N matches the perpendicular component of the weight. This must be the case, otherwise the box would be accelerating in the direction of red (mg cos downward) or green (mg cos upward).

FN > mg cos would mean the box is jumping off the ramp. FN < mg cos would mean that the ramp is being crushed.

Acceleration on a Ramp

Fw = m·g

What will the acceleration be in this situation?F = m·aFx = m·am·g·sin = m·ag·sin = a

FN

Fy = m·g·cos

Fx = m·g·sin

FN = m·g·cos

Acceleration on a Ramp

Fw = m·g

How could you keep the block from accelerating?Supply a pulling force F that is equal in magnitude to Fx and is opposite in direction to Fx.

FN

Fy = m·g·cos

Fx = m·g·sin

FN = m·g·cos

F

Pulling an Object at an Angle wrt the Horizontal

The pulling force F has an x and y component.

Construct a right triangle to determine Fx and Fy.

Fx is the force that is moving the object forward along the surface.

Fy is the upward pull of the force F.

Pulling an Object at an Angle wrt the Horizontal

cos

sin

cos

sin

x

y

x

y

FFF

FF F

F F

Pulling an Object at an Angle wrt the Horizontal

Fx is the accelerating force: Fx = m·a.

To determine FN, use forces up = forces down.

FN + Fy = Fw, therefore: FN = Fw - Fy

Pushing an Object at an Angle wrt the Horizontal

The pushing force F has an x and y component.

Construct a right triangle to determine Fx and Fy.

Fx is the force that is moving the object forward along the surface.

Fy is the force that is pushing the object to the surface F.

Pushing an Object at an Angle wrt the Horizontal

cos

sin

cos

sin

x

y

x

y

FFF

FF F

F F

Pushing an Object at an Angle wrt the Horizontal

Fx is the accelerating force; Fx = m·a.

To determine FN, use forces up = forces down.

FN= Fy + Fw

Tension is determined by examining one block or the other.

Pulley Problems

m 1 m2

m1·g

FN

FT FT

m2·g

Fm1+m2)·am2·g – FT + FT – m1·g·sin = (m1 + m2)·a

Tension is determined by examining one block or the other.

Pulley Problems

m 1 m2

m1·g

FN

FT FT

m2·g

m2·g FT = m2·aFT m1·g·sin = m1·a