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Chapter 5. Force and Motion I. Classical Mechanics. Describes the relationship between the motion of objects in our everyday world and the forces acting on them Conditions when Classical Mechanics does not apply very tiny objects (< atomic sizes) objects moving near the speed of light.
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Chapter 5 Force and Motion I
Classical Mechanics • Describes the relationship between the motion of objects in our everyday world and the forces acting on them • Conditions when Classical Mechanics does not apply • very tiny objects (< atomic sizes) • objects moving near the speed of light
Forces • Usually think of a force as a push or pull • Vector quantity • May be contact or field force
Fundamental Forces • Types • Strong nuclear force • Electromagnetic force • Weak nuclear force • Gravity • Characteristics • All field forces • Listed in order of decreasing strength • Only gravity and electromagnetic in mechanics
Measuring the Strength of a Force • Use the deformation of a spring to determine the strength of F. • The spring elongates under the action of F1 and the pointer on the scale indicates the magnitude of F1. • Calibration is necessary
Newton’s First Law • If no forces act on an object, it continues in its original state of motion; that is, unless something exerts an external force on it, an object at rest remains at rest and an object moving with some velocity continues with that same velocity.
Newton’s First Law, cont. • External force • any force that results from the interaction between the object and its environment • Alternative statement of Newton’s First Law • When there are no external forces acting on an object, the acceleration of the object is zero.
Inertia • Is the tendency of an object to continue in its original motion
Mass • A measure of the resistance of an object to changes in its motion due to a force • Scalar quantity • SI units are kg
Newton’s Second Law • The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. • F and a are both vectors • Can also be applied three-dimensionally
Units of Force • SI unit of force is a Newton (N) • US Customary unit of force is a pound (lb) N. B. (1 N = 0.225 lb)
A teacher performs the following demonstration: A sphere and a block are connected by a string placed over a pulley. The two objects are at rest at the same level, as shown in the Figure. The block is now pulled down and held while in its new position, as shown in Figure 2. The teacher asks the class to predict what will happen in this real situation when the block is released, and to give a reason supporting the prediction.
A teacher performs the following demonstration: • The block will move down because the gravitational force on the block will be greater at this lower level. • 2. The block will remain stationary because there will be no resultant force on the block. • 3. The block will move up and return to its original position because this will conserve potential energy. • 4. The block will move up and return to its original position because that was its equilibrium position.
More Question: The adjacent figure shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If the force magnitudes are chosen properly, in which situations is it possible that the block is (a) stationary and (b) moving with a constant velocity?
Gravitational Force • Mutual force of attraction between any two objects • Expressed by Newton’s Law of Universal Gravitation:
Weight • The magnitude of the gravitational force acting on an object of mass m near the Earth’s surface is called the weight w of the object • w = m g is a special case of Newton’s Second Law • g can also be found from the Law of Universal Gravitation
More about weight • Weight is not an inherent property of an object • mass is an inherent property • Weight depends upon location
Newton’s Third Law • If two objects interact, the force F12 exerted by object 1 on object 2 is equal in magnitude but opposite in direction to the force F21 exerted by object 2 on object 1. • Equivalent to saying a single isolated force cannot exist
Newton’s Third Law cont. • F12 may be called the action force and F21 the reaction force • Actually, either force can be the action or the reaction force • The action and reaction forces act on different objects
Some Action-Reaction Pairs • n and n’ • n is the normal force, the force the table exerts on the TV • n is always perpendicular to the surface • n’ is the reaction – the TV on the table • n = - n’
More Action-Reaction pairs • Fg and Fg’ • Fg is the force the Earth exerts on the object • Fg’ is the force the object exerts on the earth • Fg = -Fg’
Forces Acting on an Object • Newton’s Law uses the forces acting on an object • n and Fg are acting on the object • n’ and Fg’ are acting on other objects
Forces Acting on an Object Ex.: The adjacent Figure shows four choices for the direction of a force of magnitude F to be applied to a block on an inclined plane. The directions are either horizontal or vertical. (For choices a and b, the force is not enough to lift the block off the plane.) Rank the choices according to the magnitude of the normal force on the block from the plane, greatest first.
Applying Newton’s Laws • Assumptions • Objects behave as particles • can ignore rotational motion (for now) • Masses of strings or ropes are negligible • Interested only in the forces acting on the object
Free Body Diagram • Must identify all the forces acting on the object of interest • Choose an appropriate coordinate system • If the free body diagram is incorrect, the solution will likely be incorrect
Equilibrium • An object either at rest or moving with a constant velocity is said to be in equilibrium • The net force acting on the object is zero
Equilibrium cont. • Easier to work with the equation in terms of its components:
Solving Equilibrium Problems • Make a sketch of the situation described in the problem • Draw a free body diagram for the isolated object under consideration and label all the forces acting on it • Resolve the forces into x- and y-components, using a convenient coordinate system • Apply equations, keeping track of signs • Solve the resulting equations
Newton’s Second Law Problems • Similar to equilibrium except • Use components • ax or ay may be zero
Solving Newton’s Second LawProblems • Make a sketch of the situation described in the problem • Draw a free body diagram for the isolated object under consideration and label all the forces acting on it • If more than one object is present, draw free body diagram for each object • Resolve the forces into x- and y-components, using a convenient coordinate system • Apply equations, keeping track of signs • Solve the resulting equations
Inclined Planes • Choose the coordinate system with x along the incline and y perpendicular to the incline • Replace the force of gravity with its components
Example: • A car of mass m is on a icy driveway inclined at angle q as shown in the adjacent Figure. Find the acceleration of the car, assuming No Friction.
Solution: • Chose of x- and y-axes. • Free Body diagram. • Newton’s 2nd Law: mg + n = ma Projection along the x-axis: mgsin(q) = max Projection along the y-axis: mgcos(q) + n = 0 ax = gsin(q)
More Examples: • Two Blocks of masses m1 and m2, with m1 > m2, are placed in contact with each other on a frictionless horizontal surface. A constant force F is applied to m1 as shown. a) Find the magnitude of the acceleration of the system? b) Determine the magnitude of the contact force between the two blocks?
Solution: • Both Blocks must experience the same acceleration since they are in contact. F is the only force acting horizontally: F = (m1+m2)ax
y x Solution Cont’d: • Free body diagram of m1. Along the x-axis, we have: F – P21 = m1ax
Solution: • Note that one can get P12 from m2: Along the x-axis: P12 = m2ax We see that |P12|= |P21|
Example, The Atwood Machine: • Two object of unequal masses (m2>m1) are hung vertically over a frictionless pulley of negligible mass. Determine the magnitude of the acceleration of the two objects and the tension in the lightweight cord.
a a Solution Atwood Machine: • m2 will move down and m1 will move up with the same a. • N.B. The tensions in the cord from both side are equal since mass of pulley is negligible and there is no Friction.
a Projection along the direction of motion (upwards for m1) -m1g + T = m1a T = m1g + m1a Projection along the direction of motion (downwards for m2) m2g - T = m2a T = m2g – m2a Atwood Cont’d: • Drawing the free body diagram and applying Newton’s second Law.
(m2 – m1)g = Net Force N.B. (m2 + m1) = Total mass Atwood Cont’d: • T = T m1g + m1a = m2g – m2a