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Physics Subject Area Test MECHANICS: DYNAMICS. Newtons First Law.

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Presentation on theme: "Physics Subject Area Test MECHANICS: DYNAMICS. Newtons First Law."— Presentation transcript:

1 Physics Subject Area Test MECHANICS: DYNAMICS

2 Newtons First Law

3 Dynamics – connection between force and motion Force – any kind of push or pull required to cause a change in motion (acceleration) measured in Newtons (N)

4 Newtons First Law of Motion

5 First Law – Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it. First Law – (Common) An object at rest remains at rest, and a object in motion, remains in motion unless acted upon by an outside force.

6 Newtons Laws are only valid in an Inertial Frame of Reference For example, if your frame of reference is an accelerating car – a cup in that car will slide with no apparent force being applied

7 An inertial frame of reference is one where if the first law is valid Inertia – resistance to change in motion

8 Mass

9 Mass – a measurement of inertia A larger mass requires more force to accelerate it Weight – is a force, the force of gravity on a specific mass

10 Newtons Second Law

11 Second Law – acceleration is directly proportional to the net force acting on it, and inversely proportional to its mass. -the direction is in the direction of the net force Easier to see as an equation more commonly written

12 F – the vector sum of the forces In one dimension this is simply adding or subtracting forces.

13 Free Body Diagram The most important step in solving problems involving Newtons Laws is to draw the free body diagram Be sure to include only the forces acting on the object of interest Include any field forces acting on the object Do not assume the normal force equals the weight F table on book F Earth on book


15 Objects in Equilibrium Objects that are either at rest or moving with constant velocity are said to be in equilibrium Acceleration of an object can be modeled as zero: Mathematically, the net force acting on the object is zero Equivalent to the set of component equations given by

16 A lamp is suspended from a chain of negligible mass The forces acting on the lamp are the downward force of gravity the upward tension in the chain Applying equilibrium gives

17 A traffic light weighing 100 N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper cables make angles of 37 ° and 53° with the horizontal. Find the tension in each of the three cables. Conceptualize the traffic light Assume cables dont break Nothing is moving Categorize as an equilibrium problem No movement, so acceleration is zero Model as an object in equilibrium

18 Need 2 free-body diagrams Apply equilibrium equation to light Apply equilibrium equations to knot

19 Suppose a block with a mass of 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle can the ramp make with the horizontal before the block starts to slip down?

20 Newton 2nd law: Then So

21 A block of mass m1 on a rough, horizontal surface is connected to a ball of mass m2 by a lightweight cord over a lightweight, frictionless pulley as shown in figure. A force of magnitude F at an angle θ with the horizontal is applied to the block as shown and the block slides to the right. The coefficient of kinetic friction between the block and surface is μ k. Find the magnitude of acceleration of the two objects.

22 We all remember the fun see-saw of our youth. But what happens if...

23 Moral Both the masses and their positions affect whether or not the see saw balances.

24 Need: M 1 d 1 = M 2 d 2 M1M1 M2M2 d1d1 d2d2

25 The great Greek mathematician Archimedes said, give me a place to stand and I will move the Earth, meaning that if he had a lever long enough he could lift the Earth by his own effort.

26 We can think of leaving the masses in place and moving the fulcrum. It would have to be a pretty long see-saw in order to balance the school bus and the race car, though!

27 (We still) need: M 1 d 1 = M 2 d 2 M2M2 d1d1 d2d2 M1M1

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