Presentation on theme: "Newton’s Second Law The net force on a body is equal to the product of the body’s mass and its acceleration."— Presentation transcript:
1Newton’s Second LawThe net force on a body is equal to the product of the body’s mass and its acceleration.
2UnitsProblem 20: A car traveling at 53 km/h hits a bridge abutment. A passenger in the car moves forward a distance of 65 cm (with respect to the road) while being brought to rest by an inflated air bag. What magnitude of force (assumed constant) acts on the passenger’s upper torso, which has a mass of 41 kg?
3Gravitational Force and Weight Gravitational force is the force that the Earth exerts on any object. It is directed toward the center of the Earth.The magnitude of the gravitational force is equal to the product of mass and acceleration due to gravity.The weight W of a body is the magnitude of the net force required to prevent the body from falling freely, as measured by someone on the ground.The weight W of a body is equal to the magnitude of the gravitational force on the body. A body’s weight is related to the body’s mass by,
4Contact Forces: As the name implies, these forces act between two objects that are in contact. The contact forces have two components: one that is acting along the normal to the contact surface (normal force) and a second component that is acting parallel to the contact surface (frictional force).Normal Force: When a body presses against a surface, the surface deforms and pushes on the body with a normal force perpendicular to the contact surface. An example is shown in the picture to the left. A block of mass m rests on a table.Note: In this case FN = mg. This is not always the case.Friction: If we slide or attempt to slide an object over a surface, the motion is resisted by a bonding between the object and the surface. This force is known as “friction.” More on friction in Chapter 6.(5-7)
5Tension: This is the force exerted by a rope or a cable attached to an object. Tension has the following characteristics:It is always directed along the rope.It is always pulling the object.3. It has the same value along the rope (for example, between points A and B).The following assumptions are made:The rope has negligible mass compared to the mass of the object it pulls.The rope does not stretch.If a pulley is used as in fig.(b) and fig.(c), we assume that the pulley is massless and frictionless.AB
6Newton’s Third Law:When two bodies interact by exerting forces on each other, the forces are equal in magnitude and opposite in direction.
7Recipe for the Application of Newton’s Laws of Motion 1. Choose the system to be studied.2. Make a simple sketch of the system.3. Choose a convenient coordinate system.4. Identify all the forces that act on the system. Label them on the diagram.5. Apply Newton’s laws of motion to the system.P 17, page 109: In the figure , let the mass of the block be 8.5 kg and the angle θ be 30°. Find (a) the tension in the cord and (b) the normal force acting on the block. (c) If the cord is cut, find the magnitude of the resulting acceleration of the block.
8Sample ProblemFigure 5-12 shows a block S (the sliding block) with mass M = 3.3 kg. The block is free to move along a horizontal frictionless surface and connected, by a cord that wraps over a frictionless pulley, to a second block H (the hanging block), with mass m = 2.1 kg. The cord and pulley have negligible masses compared to the blocks (they are “massless”). The hanging block H falls as the sliding block S accelerates to the right. Find (a) the acceleration of block S, (b) the acceleration of block H, and (c) the tension in the cord.