Chapter 5 Forces in Two Dimensions Vectors Again! How do we find the resultant in multiple dimensions? 1. Pythagorean Theorem- if the two vectors are at right angles R 2 = A 2 + B 2 2. At an angle other than 90° a. Law of Cosines R 2 = A 2 +B 2 –2AB cos  b. Law of Sines R = A = B sin  sin a sin b

Examples 1. A car is driven 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 = A x + A y A  A y A x

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 resultant R x = A x + B x + C x R y = A y + B y + C y Then R 2 = R x 2 + R y 2

How do we find the angle of the resultant? Use the Angle of Resultant Vector  = tan -1 (R y /R x ) 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 Friction 1. Kinetic Friction(F fk )- friction created between moving surfaces 2. Static Friction(F fs )-force between 2 nonmoving surfaces

What does Frictional Force depend upon? 1. Surface materials- depends on the nature of the surfaces  Coefficient 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(F fk ) F fk =  k F N 2. Static Friction(F fs ) F fs   s F N

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 kgF app. =? V= 1.0 m/sa= 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 How fast is the skier going after 5.0 s after starting from rest?