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Acceleration a=F/m Velocity v= v 0 + at Position x = x 0 + v 0 t + ½ at 2 FORCES NET FORCE.

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Presentation on theme: "Acceleration a=F/m Velocity v= v 0 + at Position x = x 0 + v 0 t + ½ at 2 FORCES NET FORCE."— Presentation transcript:

1 Acceleration a=F/m Velocity v= v 0 + at Position x = x 0 + v 0 t + ½ at 2 FORCES NET FORCE

2 FREE BODY DIAGRAM FBD

3 From Simple to Complex 1) Horizontal plane 2) Inclined plane

4 Drawing a FBD of forces on an object (on, not by) 1. Choose the object to analyze. Draw it as a dot. 2. What forces physically touch this object? This object, not some other 3. What “action at a distance” forces act on the object? Gravity is the only one for this PHYS Draw these forces as arrows with tails at the dot (object). 5. Forces only! No accelerations, velocities, … Get components of Newton’s 2nd Law Choose a convenient xy coordinate system Find the x and y components of each force in the FBD Add the x and y components separately

5 x y x positive y positive

6 x negative y positive x y

7 x y x negative y negative

8 x y x positive y negative

9 x y ?????? x positive y positive

10 x y  

11 Horizontal plane 1) Horizontal force A horizontal force F a of 12 N is applied To a block with mass m=6kg, on a frictionless table. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its velocity after it travels 0.4m from the origin FaFa

12 FBD m FaFa N W Normal force weight applied force x y Normal force: mg = 6kg* 9.81 m/s 2 (along y) Weight: -mg (along y) Applied force = 12 N along x The net force F : the y component is zero because the normal Force and the weight cancel. The x component is the applied Force. Hence F x = 12 N and F y = 0 N

13 The net force F : the y component is zero because the normal Force and the weight cancel. The x component is the applied Force. Hence F x = 12 N and F y = 0 N Applying the second Newton/s law F = m a F x = ma x a x = F x /m a x = 12N/6kg a x = 2 m/s 2 Velocity after it travels 0.4 m from the origin: The block was originally at rest: v x0 = 0m/s v x 2 = v x a x (x-x 0 ) v x = √ 2a x (x-x 0 ) v x = 1.26 m/s

14 Horizontal plane 2) Force at an angle A force F a of 15 N making an angle of 35 o from the horizontal is applied to a block with mass m=6kg, on a frictionless table. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its velocity after it travels for 5 seconds from the origin 

15 FBD m FaFa N W Normal force weight applied force x y  F a, y F a,x There is no motion in the y Direction (the block does not jump !!!) F y = 0 N Hence: Normal force N = w+Fsin(   Motion along x: F a,x = m a x a x = 15N cos (35 o )/6kg a x = 2.05 m/s 2 v x = v o + a x t v x = 0m/s + (2.05m/s 2 )(5s) v x = m/s F x = Fcos(   F y = Fsin(  

16 Horizontal plane 3) Force at an angle + friction A force F a of 15 N making an angle of 35 o from the horizontal is applied to a block with mass m=6kg, on a table with friction force F f opposing the motion of the block of 5.2 N magnitude. The block was originally at rest when the force was applied. Draw a FBD and find the acceleration of the block and its position after it travels for 5 seconds from the origin 

17 FBD m FaFa N W Normal force weight applied force x y  F a, y F a,x There is no motion in the y Direction (the block does not jump !!!) F y = 0 N Hence: Normal force N = w+Fsin(   Motion along x: F a,x – F f = m a x a x = (15N cos (35 o ) -5.2N)/6kg a x = 1.18 m/s 2 x = x o + v ox t + (1/2) a x t 2 x = (1/2)(1.85m/s 2 )(5s) 2 x = m F x = Fcos(   F y = Fsin(   FfFf

18 x y  

19  α β ??? β = 90 o - α β = 90 o - (90 o -  ) β =  α = 90 o - 

20 Inclined plane 1) Only gravitational force M =25 kg θ= 25 o h= 2 m  M h Question: What is the velocity of the block at the bottom of the frictionless incline? d d= h/sin( 

21 FBD W x = W sin θ W y = W cos θ N =-W y y  W N x WxWx WyWy θ

22 y-component of the net force is zero!

23   M h FaFa   FfFf A box with mass 45 kg is at rest when a force Fa (20N)making an angle of 37 0 with the inclined plane. The inclined plane makes an angle of 22 0 with the horizontal plane. When the box is moving on the inclined Plane, there is a friction force F f of 5 N opposing the motion. The box was Originally at a height h from the ground (h=4 m). Draw the free body diagram for the box. Determine the components x, y for all the forces acting on the box. Find the net force and acceleration Of the box.

24 The conditions for a particle to be in equilibrium Necessary conditions for an object to settle into equilibrium –  F = 0 –  F x = 0 and  F y = 0

25 Both x and y forces must be considered separately.

26 Homework from chapter 4 3, 9, 14, 20, 21, 23, 30, 34, 41, 52

27


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