Unit 4 Notes

1. Unit 4 Notes

2. Free Body Diagrams • Show an object and the forces acting on it • The object is represented by a circle (you can write the object’s name inside the circle) • Clearly labeled arrows pointing away from the object represent forces acting on the object • Their length represents the magnitude of the force

3. Good Example

4. Common Forces • General push or pull • Gravity • Normal Force • 90 degrees from a surface • Friction • Resists motion (always in the opposite direction from motion) For most forces, whatever exerts the force is listed first followed by the receiver of the force.

5. Components of Forces When you are given a force and an angle, make a right triangle with the angle inside the triangle

6. SOHCAHTOA is the measure of the angle

7. Fopp = Fsin(ϴ) Fadj= Fcos(ϴ)

8. Find the x- and y- components of the force

11. Example 2

14. Equations of Forces ∑Fy= Upward Forces–Downward Forces ∑Fx= Forces to Right – Forces to Left (∑ means the summation) For this unit, ∑Fx= 0 ∑Fy= 0

15. Equations of Forces • When there is no acceleration, the sum of forces in the x-direction is 0 and the sum of forces in the y-direction is 0. (∑ means the summation) ∑Fy= 0 ∑Fx= 0

16. Example A person pushes a 10kg object to the left with a 20N force. Solve for all forces acting on the object. Step1: Draw a free body diagram.

17. Step 2: Write the force summation equations. ∑Fx=0 ∑Fy=0 Step 3: Insert Forces into the equations. From here it is a 3-step solution. Ff-FP/O=0 FN -Fg=0

18. Fg= 9.8N/kg * 10kg = 98N Ff -FP/O=0 FN - Fg=0 Ff-20N=0 FN– 98N=0 Ff = 20N FN= 98N

19. A person pulls on a 50kg dog’s leash. The dog doesn’t move. • Solve for all forces

21. ∑Fx= 0 ∑Fy= 0 Ff = 38.6N FN+ FLeash/Dog,y- Fg = 0 FLeash/Dog,x-Ff= 0 FN + 10.4N– 490N = 0 38.6N-Ff = 0 FN– 479.6N = 0 FN= 479.6N

22. Ramps • Sometimes it is easier to redefine the x- and y-axes so that they are perpendicular and parallel to a surface

23. A 5kg block is at rest on an incline as shown. Solve for the normal force and the force of friction.

24. Fg = 5kg * 9.8N/kg = 49N Fg,x = 49N * sin(30) = 24.5N Fg,y = 49N * cos(30) = 42.4N

25. ∑Fx= 0 ∑Fy= 0 -Ff+ Fg,x= 0 -Ff + 24.5N= 0 Ff = 24.5N FN - Fg,y= 0 FN– 42.4N= 0