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Forces and the Laws of MotionSection 4 Normal Force Force on an object perpendicular to the surface (F n ) It may equal the weight (F g ), as it does here.

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Presentation on theme: "Forces and the Laws of MotionSection 4 Normal Force Force on an object perpendicular to the surface (F n ) It may equal the weight (F g ), as it does here."— Presentation transcript:

1 Forces and the Laws of MotionSection 4 Normal Force Force on an object perpendicular to the surface (F n ) It may equal the weight (F g ), as it does here. It does not always equal the weight (F g ), as in the second example. F n = mg cos 

2 Forces and the Laws of MotionSection 4 Normal Force An object placed on a tilted surface will often slide down the surface. The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it. In physics, a tilted surface is called an inclined plane.

3 Forces and the Laws of MotionSection 4 Normal Force As shown in the diagram, there are always at least two forces acting upon any object that is positioned on an inclined plane - the force of gravity and the normal force. The force of gravity (also known as weight) acts in a downward directionforce of gravity (also known as weight) The normal force acts in a direction perpendicular to the surface (in fact, normal means "perpendicular").normal force acts in a direction perpendicular to the surface

4 Forces and the Laws of MotionSection 4 Normal Force Up to this point in the course, we have always seen normal forces acting in an upward direction, opposite the direction of the force of gravity. But this is only because the objects were always on horizontal surfaces and never upon inclined planes. The normal forces is not always upwards, but rather that it is directed perpendicular to the surface that the object is on.

5 Forces and the Laws of MotionSection 4 Normal Force The force of gravity will be resolved into two components of force - one directed parallel to the inclined surface and the other directed perpendicular to the inclined surface.

6 Forces and the Laws of MotionSection 4 Normal Force The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force

7 Forces and the Laws of MotionSection 4 Normal Force The parallel component of the force of gravity is not balanced by any other force. This object will subsequently accelerate down the inclined plane due to the presence of an unbalanced force

8 Forces and the Laws of MotionSection 4 Normal Force It is the parallel component of the force of gravity that causes this acceleration. The parallel component of the force of gravity is the net force.

9 Forces and the Laws of MotionSection 4 Normal Force The task of determining the magnitude of the two components of the force of gravity is a mere manner of using the equations. The equations for the parallel and perpendicular components are:.

10 Forces and the Laws of MotionSection 4 Practice problems A 25 kg chair initially at rest on a horizontal floor require a 365 N horizontal force to set it in motion. Once the chair is in motion, a 327 N horizontal force keeps it moving at a constant velocity. Find the coefficient of static and kinetic friction between the chair and the floor.

11 Forces and the Laws of MotionSection 4 Practice problems A box of book weighting 325 N moves with a constant velocity across the floor when it is pushed with a force of 425 N exerted downward at an angle of 35.2° below the horizontal. Find the μ k between the box and the floor.

12 Forces and the Laws of MotionSection 4 Practice problems A student moves a box of books down the hall by pulling on a rope attached to the box. The student pulls with a force of 185 N at an angle of 25° above the horizontal. The box has a mass of 35 kg, and μ k between the box and the floor is Find the acceleration of the box.

13 Forces and the Laws of MotionSection 4 Practice problems The student in the above item moves the box up a ramp inclined at 12° with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25° with respect to the incline and with the same 185 N force, what is the acceleration up the ramp? μk =.27

14 Forces and the Laws of MotionSection 4 Practice problems A 75 kg box slides down a 25° ramp with an acceleration of 3.6 m/s 2. Find μ k What acceleration would a 175 kg box have on the same ramp?


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