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**Velocity v= v0 + at Position x = x0 + v0t + ½ at2 FORCES**

Acceleration a=F/m Velocity v= v0 + at Position x = x0 + v0t + ½ at2 NET FORCE

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FREE BODY DIAGRAM FBD

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

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**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 PHYS2053 4. 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

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y x positive y positive x

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y x negative y positive x

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y x x negative y negative

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y x x positive y negative

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

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y q x q

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**Horizontal plane 1) Horizontal force Fa**

A horizontal force Fa 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

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**FBD Normal force N Fa applied force y weight W**

Normal force: mg = 6kg* 9.81 m/s2 (along y) Weight: -mg (along y) Applied force = 12 N along x 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 Fx = 12 N and Fy = 0 N

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**F = m a 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 Fx = 12 N and Fy = 0 N Applying the second Newton/s law F = m a Fx = max ax = Fx/m ax = 12N/6kg ax = 2 m/s2 Velocity after it travels 0.4 m from the origin: The block was originally at rest: vx0 = 0m/s vx2 = vx02 + 2ax(x-x0) vx = √2ax(x-x0) vx = 1.26 m/s

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**Horizontal plane 2) Force at an angle**

q A force Fa of 15 N making an angle of 35o 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

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**FBD Normal force N y There is no motion in the y**

Direction (the block does not jump !!!) Fy = 0 N Hence: Normal force N = w+Fsin(q) Motion along x: Fa,x = m ax ax = 15N cos (35o)/6kg ax = 2.05 m/s2 vx = vo + axt vx = 0m/s + (2.05m/s2)(5s) vx= m/s Fa,x m q Fa,y applied force Fa W weight x Fx = Fcos(q) Fy = Fsin(q)

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**Horizontal plane 3) Force at an angle + friction**

q A force Fa of 15 N making an angle of 35o from the horizontal is applied to a block with mass m=6kg, on a table with friction force Ff 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

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**FBD Normal force N y There is no motion in the y**

Direction (the block does not jump !!!) Fy = 0 N Hence: Normal force N = w+Fsin(q) Motion along x: Fa,x – Ff = m ax ax = (15N cos (35o) -5.2N)/6kg ax = 1.18 m/s2 x = xo + voxt + (1/2) axt2 x = (1/2)(1.85m/s2)(5s) x = m m Fa,x Ff q Fa,y applied force Fa W weight x Fx = Fcos(q) Fy = Fsin(q)

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y q x q

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

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**Inclined plane 1) Only gravitational force M d= h/sin(q) h q**

M =25 kg θ= 25o h= 2 m Question: What is the velocity of the block at the bottom of the frictionless incline?

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FBD y q W N x Wx Wy θ Wx = W sin θ Wy = W cos θ N =-Wy

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**y-component of the net force is zero!**

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22o M h Fa 37o Ff A box with mass 45 kg is at rest when a force Fa (20N)making an angle of 370 with the inclined plane. The inclined plane makes an angle of 220 with the horizontal plane. When the box is moving on the inclined Plane, there is a friction force Ff 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.

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**The conditions for a particle to be in equilibrium**

Necessary conditions for an object to settle into equilibrium SF = 0 SFx = 0 and SFy = 0

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**Both x and y forces must be considered separately.**

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

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Chapter 4 Changes in Motion Objectives

Chapter 4 Changes in Motion Objectives

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