# Newtons First Law Chapter 4 section 2. Newtons First Law of Motion An object at rest remains at rest, and an object in motion continues in motion with.

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Newtons First Law Chapter 4 section 2

Newtons First Law of Motion An object at rest remains at rest, and an object in motion continues in motion with a constant velocity (that is, constant speed in a straight line) unless it experiences a net external force. An object at rest remains at rest, and an object in motion continues in motion with a constant velocity (that is, constant speed in a straight line) unless it experiences a net external force.

Inertia Newtons First Law is sometimes referred to as the, Law of Inertia Newtons First Law is sometimes referred to as the, Law of Inertia Inertia – The tendency of an object to maintain its state of motion and resist change. Inertia – The tendency of an object to maintain its state of motion and resist change.

How to find Change in Motion

Net External Force Net External Force – The total force resulting from a combination of external forces on an object. Net External Force – The total force resulting from a combination of external forces on an object. –Sometimes called the Resultant Force The net external force is found by summing all the forces acting on an object and can be seen in a free-body diagram. The net external force is found by summing all the forces acting on an object and can be seen in a free-body diagram.

Labeling a Free-Body Diagram Variables used to describe forces: Variables used to describe forces: –W or mg or F g = Gravitational Force –T or F T = Tension Force –F push or F p = Compression Force –n or N or F n = Normal Force –F f = Friction Force

Normal Force Normal Force – A contact force exerted by an object on another object in a direction perpendicular to the surface of contact. Normal Force – A contact force exerted by an object on another object in a direction perpendicular to the surface of contact. –Also known as the Support Force

Forces as a Vector When calculating the net external force acting on an object, vector addition can be used to find the sum of all the forces. When calculating the net external force acting on an object, vector addition can be used to find the sum of all the forces. It may be necessary to break the forces into their x- and y- components to find the sum of the forces. It may be necessary to break the forces into their x- and y- components to find the sum of the forces. Take a look at sample problem 4A on page 132 Take a look at sample problem 4A on page 132

The x-y Coordinate System When talking about the x-y coordinate system, the x- axis is always horizontal and the y- axis as always vertical. When talking about the x-y coordinate system, the x- axis is always horizontal and the y- axis as always vertical. –This is true in math class, but in physics we can rotate the axis to fit the specific problem. Usually the x- axis is parallel to the surface the object is resting on and the y- axis is perpendicular to the surface. Usually the x- axis is parallel to the surface the object is resting on and the y- axis is perpendicular to the surface.

Solving for Force Components When a vector lies at some angle (θ), then the x- and y-components must be found. When a vector lies at some angle (θ), then the x- and y-components must be found. –The vector that lies opposite to the angle (θ) uses the sine function to find the magnitude of the vector. –The vector that lies adjacent to the angle (θ) uses the cosine function to find the magnitude of the vector.

Example Problem #1 A boy is pulling on a wagon with a force of 50.0 N directed at an angle of 25.0 degrees to the horizontal. What is the x component of this force? What is the y component of this force? A boy is pulling on a wagon with a force of 50.0 N directed at an angle of 25.0 degrees to the horizontal. What is the x component of this force? What is the y component of this force?

Example Problem #1 Answer Fx = 45.32 N Fx = 45.32 N Fy = 21.13 N Fy = 21.13 N

Example Problem #2 A box is pulled to the East with a force of 190N, to the West with a force of 120N, to the North with 465N and South with a force of 230N. What is the magnitude and direction of the Net Force acting on the object? A box is pulled to the East with a force of 190N, to the West with a force of 120N, to the North with 465N and South with a force of 230N. What is the magnitude and direction of the Net Force acting on the object?

Example Problem #2 Answer Fnet = 245.20N @ 73.41 degrees Fnet = 245.20N @ 73.41 degrees

Mass vs. Inertia Mass is directly proportional to inertia. Mass is directly proportional to inertia. –If the mass increases, the inertia of the object increases. –If the mass decreases, the inertia of the object decreases.

Tendency to Maintain its Motion Inertia is the tendency of an object to maintain its state of motion unless acted on by an unbalanced net force. Inertia is the tendency of an object to maintain its state of motion unless acted on by an unbalanced net force. –Object at rest remains at rest –Object in motion stays in motion in a stright line.

Equilibrium Equilibrium – The state of a body in which there is no change in its motion Equilibrium – The state of a body in which there is no change in its motion –If the Sum of the forces acting on the object are equal to zero then there is no change in motion. ΣF = 0 ΣF = 0 Object at rest Object at rest Object traveling at a constant velocity Object traveling at a constant velocity

Equilibrium in the x and y direction To determine if an object is in equilibrium it is best to break all the forces into their x- and y- components, then find the vector sum of the forces. To determine if an object is in equilibrium it is best to break all the forces into their x- and y- components, then find the vector sum of the forces. –If the sum of the forces in the x-direction is equal to zero (ΣFx = 0) and the sum of the forces in the y- direction is equal to zero (ΣFy = 0), then the object is in equilibrium.

Object not in Equilibrium Example: Example: –Object in free fall dropped off the edge of a cliff. ΣFx = 0 ΣFx = 0 ΣFy 0 ΣFy 0 –Therefore, there will be an acceleration in the y- direction, but not in the x- direction.

Example Problem #3 A person with a weight of 600N is at the gym and lifting weights above their head. If the person is holding 1250N above their head in a stable position, what is the force normal force exerted on the person? What is the force exerted on each of the persons feet? A person with a weight of 600N is at the gym and lifting weights above their head. If the person is holding 1250N above their head in a stable position, what is the force normal force exerted on the person? What is the force exerted on each of the persons feet?

Example Problem #3 Answer ΣF = 0 ΣF = 0 F N = 1850 N F N = 1850 N F feet = 925 N F feet = 925 N

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