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

1. Resolving Forces Into Vector ComponentsPhysicsMontwood High SchoolR. Casao

2. 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.

3. 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

4. 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.

5. FN m·g 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

6. Resolving Weight • For  > 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.

7. Resolving Weight • Fx 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.

8. Resolving Weight • Fy 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

9. FN Fx = m·g·sin  Fy = m·g·cos   Fw = m·g  Normal Force on Ramp • The normal force is perpendicular to inclined ramps as well. It’s always equal to the component of weight perpendicular to the surface. FN = m·g·cos 

10. 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.

11. Resolving Weight • Complete a right triangle:

12. Resolving Weight • General equations (if N are given): • General equations (if kg are given): • Accelerating force: Fx=m·a

13. How Does the Incline Affect the Components? m mgsin m mgsin mgcos mgcos mg mg 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. Extreme cases: When  = 0, the ramp is flat; red= mg; blue= 0.When  = 90, the ramp is vertical; red= 0;blue= mg.

14. Inclined Plane: Normal Force 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)orgreen (mg cos  upward). FN> mgcos  would mean the box is jumping off the ramp. FN < mgcos  would mean that the ramp is being crushed. FN = mgcos mgsin m  mgcos mg

15. FN Fx = m·g·sin  Fy = m·g·cos   Fw = m·g  Acceleration on a Ramp What will the acceleration be in this situation? SF= m·a Fx= m·a m·g·sin q = m·a g·sin q= a FN = m·g·cos 

16. FN F Fx = m·g·sin  Fy = m·g·cos   Fw = m·g  Acceleration on a Ramp • 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 = m·g·cos 

17. 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.

18. Pulling an Object at an Angle wrt the Horizontal

19. 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

20. 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.

21. Pushing an Object at an Angle wrt the Horizontal

22. 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

23. FT FT FN m1·g m2·g Pulley Problems Tension is determined by examining one block or the other. SF = (m1+m2)·a m2·g – FT + FT – m1·g·sin q = (m1 + m2)·a m1 m2 q

24. FT FT FN m1·g m2·g Pulley Problems Tension is determined by examining one block or the other. m2·g - FT = m2·a FT- m1·g·sin q= m1·a m1 m2 q