Chapter 5Forces in Two Dimensions

Vectors Again!How do we find the resultant in multiple dimensions?1. Pythagorean Theorem- if the two vectors are at

right anglesR2= A2 + B2

2. At an angle other than 90°a. Law of Cosines

R2 = A2 +B2 –2AB cos b. Law of Sines

R = A = B sin sin a sin b

Examples

1. A car is driven 125.0 km due west, then 65.0 km due south. What is the magnitude of its displacement?

2. Find the magnitude of the sum of two forces, one 20.0 N and the other 7.0 N, when the angle between them is 30.0 °.

Does a vector have components?

A = Ax + Ay

A Ay

Ax

Vector Resolution

The process of breaking a vector into its components

How can we determine the vector components?

By using Trigonometry

Trig. Functions

1. SOH sin = opposite/hyp.

2. CAH cos = adj./ hyp.

3. TOA tan = opp./adj.

Example

As the 60-Newton tension force acts upward and rightward on Fido at an angle of 40 degrees, the components of this force can be determined using trigonometric functions.

  

Adding 2/more Vectors

By resolving each vector into its x and y components then add the x components to form the x component of the resultant then add the y components to form the y component of the resultantRx= Ax + Bx + Cx

Ry= Ay + By + Cy

ThenR2 = Rx

2 + Ry2

How do we find the angle of the resultant?

Use the Angle of Resultant Vector

= tan-1 (Ry/Rx)

Example: A hiker travels 4.0 m South then 7.3 m Northwest. Find the displacement and angle of the hiker.

What is Friction?

A force opposing motion

2 Types of Friction1. Kinetic Friction(Ffk)- friction

created between moving surfaces2. Static Friction(Ffs)-force between

2 nonmoving surfaces

What does Frictional Force depend upon?

1. Surface materials- depends on the nature of the surfacesCoefficient of Friction- value describing the nature of the surfaces in contact

2. Normal force- perpendicular contact force exerted by a surface on an object

How is it Determined?

1. Kinetic Friction(Ffk)

Ffk= kFN

2. Static Friction(Ffs)

Ffs sFN

Example Problem 3 p.128

You push a 25.0 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. How much force do you exert on the box?

What do we know?M= 25.0 kg Fapp.=?V= 1.0 m/s a= 0.0m/s/s= .20(Table 5-1)

Practice Problem p.130, #22

A 1.4 kg block slides across a rough surface that it slows down with an acceleration of 1.25 m/s/s. What is the coefficient of friction between the block and the surface?

Motion Along an Inclined Plane

A crate weighing 562 N is resting on a plane inclined 30.0° above the horizontal. Find the components of the weight forces that are parallel and perpendicular to the plane.

Example Problem #6, p.134

A 62 kg person on skis is going down a hill sloped at 37°. The coefficient of kinetic friction between the skis and the snow is 0.15. How fast is the skier going after 5.0 s after starting from rest?