chapter 5 forces in two dimensions
DESCRIPTION
Chapter 5 Forces in Two Dimensions. Vectors Again! How do we find the resultant in multiple dimensions? Pythagorean Theorem- if the two vectors are at right angles R 2 = A 2 + B 2 At an angle other than 90 ° a. Law of Cosines R 2 = A 2 +B 2 –2AB cos - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/1.jpg)
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
![Page 2: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/2.jpg)
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 °.
![Page 3: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/3.jpg)
Does a vector have components?
A = Ax + Ay
A Ay
Ax
![Page 4: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/4.jpg)
Vector Resolution
The process of breaking a vector into its components
How can we determine the vector components?
By using Trigonometry
![Page 5: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/5.jpg)
Trig. Functions
1. SOH sin = opposite/hyp.
2. CAH cos = adj./ hyp.
3. TOA tan = opp./adj.
![Page 6: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/6.jpg)
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.
![Page 7: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/7.jpg)
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
![Page 8: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/8.jpg)
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.
![Page 9: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/9.jpg)
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
![Page 10: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/10.jpg)
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
![Page 11: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/11.jpg)
How is it Determined?
1. Kinetic Friction(Ffk)
Ffk= kFN
2. Static Friction(Ffs)
Ffs sFN
![Page 12: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/12.jpg)
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)
![Page 13: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/13.jpg)
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?
![Page 14: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/14.jpg)
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.
![Page 15: Chapter 5 Forces in Two Dimensions](https://reader036.vdocument.in/reader036/viewer/2022071711/56812d5b550346895d926288/html5/thumbnails/15.jpg)
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?