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Miscellaneous Forces. Weight. Weight is the force of gravity upon an object. This force is represented by the symbol F g (units of Newtons) Acceleration of gravity on the earth is always -9.81 m/s/s, so weight can be calculated by multiplying the mass and -9.81 m/s/s. Normal Force.
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Weight • Weight is the force of gravity upon an object. This force is represented by the symbol Fg (units of Newtons) • Acceleration of gravity on the earth is always -9.81 m/s/s, so weight can be calculated by multiplying the mass and -9.81 m/s/s
Normal Force • Normal Force is the force of a surface on an object, and acts against the force of gravity (weight). It is always perpendicular to the contact surface. It is represented by Fn
Force of Friction • Friction is a force that opposes applied force • Static friction is the force that is exerted on a motionless object, and is always equal but opposite to an applied force. It is represented by Fs. • There is a limit where the static friction can no longer keep the object motionless. This limit is called the Fs,max
Force of Friction • Kinetic friction is the force that opposes motion of a moving object. It is a fixed amount that depends on the surfaces involved. It is represented by the symbol Fk
Force of Friction • The coefficient of friction describes the relative amount of friction between two surfaces. It is represented by μs or μk • It can be calculated by: μs= Fs,max / Fn • μk = Fk / Fn
6.3 Equilibrium and Hooke's Law When the net force acting on an object is zero, the forces on the object are balanced. We call this condition equilibrium.
6.3 Equilibrium and Hooke's Law • A moving object continues to move with the same speed and direction. • Newton’s second law states that for an object to be in equilibrium, the net force, or the sum of the forces, has to be zero.
6.3 Equilibrium and Hooke's Law Accelerationresults from a net force that is not equal to zero.
Calculating the net force from four forces • You are asked for acceleration. • You are given mass and force. • Use a = F ÷ m. • First add the forces to find the net force. • F = - 75N - 25N + 45N + 55N = 0 N, so a = 0 Four people are pulling on the same 200 kg box with the forces shown. Calculate the acceleration of the box.
6.3 Free-body diagrams • To keep track of the number and direction of all the forces in a system, it is useful to draw a free-body diagram. • A free-body diagram makes it possible to focus on all forces and where they act
6.3 Free-body diagrams • Forces due to weight or acceleration may be assumed to act directly on an object, often at its center. • A reaction force is usually present at any point an object is in contact with another object or the floor. • If a force comes out negative, it means it opposes another force.
6.3 Applications of equilibrium • If an object is not moving, then you know it is in equilibrium and the net force must be zero. • You know the total upward force from the cables must equal the downward force of the sign’s weight because the sign is in equilibrium. What is the upward force in each cable?
Using equilibrium to find anunknown force • You are asked for the force on one chain. • You are given 2 forces and the mass. • Use: net force = zero, Fw= mg and g = 9.8 N/kg. • Substitute values: Fw = mg = (150 kg)(9.8 N/kg) = 1,470 N. • Let F be the force in the other chain, equilibrium requires: • F + (600 N) = 1,470 N F = 1,470 N – 600 N • F = 870 N. Two chains are used to lift a small boat. One of the chains has a force of 600 newtons. Find the force on the other chain if the mass of the boat is 150 kilograms.
6.3 Applications of equilibrium • Real objects can move in three directions: up-down, right-left, and front-back. • The three directions are called three dimensions and usually given the names x, y, and z. • When an object is in equilibrium, forces must balance separatelyin each of the x, y, and z dimensions.
6.3 The force from a spring A spring is a device designed to expand or contract, and thereby make forces in a controlled way. Springs are used in many devices to create force. There are springs holding up the wheels in a car, springs to close doors, and a spring in a toaster that pops up the toast.
6.3 The force from a spring The most common type of spring is a coil of metal or plastic that creates a force when it is extended(stretched) or compressed (squeezed).
6.3 The force from a spring The force from a spring has two important characteristics: The force always acts in a direction that tries to return the spring to its unstretched shape. The strength of the force is proportional to the amount of extension or compression in the spring.
6.3 Restoring force and Hooke’s Law • The force created by an extended or compressed spring is called a “restoring force” because it always acts in a direction to restore the spring to its natural length. • The change a natural, unstretched length from extension or compression is called deformation. • The relationship between the restoring force and deformation of a spring is given by the spring constant (k).
6.3 Restoring force and Hooke’s Law • The relationship between force, spring constant, and deformation is called Hooke’s law. • The spring constant has units of newtons per meter, abbreviated N/m.
6.3 Hooke's Law • The negative sign indicates that positive deformation, or extension, creates a restoring force in the opposite direction. Deformation (m) F = - k x Force (N) Spring constant N/m
Calculate the force from a spring • You are asked for force. • You are given k and x. • Use F = - kx • Substitute values: F = - (250 N/m)(0.01 m) F = - 2.5 N A spring with k = 250 N/m is extended by one centimeter. How much force does the spring exert?
6.3 More about action-reaction and normal forces The restoring force from a wall is always exactly equal and opposite to the force you apply, because it is caused by the deformationresulting from the force you apply.
Calculate the restoring force The spring constant for a piece of solid wood is 1 × 108 N/m. Use Hooke’s law to calculate the deformation when a force of 500 N (112 lbs) is applied. • You are asked for the deformation, x. • You are given force, F and spring constant, k. • Use F = - kx, so x = - F ÷ k • Substitute values: x = - (500 N/m) ÷ (1 × 108 N/m) • x = - 5 × 10-6 meters (a very small deformation)