APPLICATION OF NEWTON’S SECOND LAW To solve a problem using Newton’s 2nd Law of Motion: • Draw a free body diagram for the object. • Use the free body diagram to write a net force equation for each direction (x and y). Recall that forces are vectors, so forces in the positive directions (up or to the right) should be added, and forces in the negative directions (down or to the left) should be subtracted. Set the equation equal to zero if there is no net force/acceleration in that direction or mass times acceleration if there is a non-zero net force/acceleration in that direction. Remember you are creating a separate equation for each direction so the up and down forces should not be in the same equation as the left and right forces. See the examples for more clarification. At this level, we will not be dealing with any forces at angles in these types of problems, so all of your forces will be up, down, left, or right. Additionally, the net force on each object will not be at an angle, therefore at least one of the net force equations will always be set equal to zero. • Plug in individual force equations (ex. Fg = mg) where appropriate. • Use algebra to rearrange the net force equation(s) to solve for the variable you are looking for.

FRICTION FORCE Friction force is caused by the interactions between two surfaces. There are two types of friction, static and kinetic. Kinetic friction refers to friction between two surfaces that are moving relative to each other. Kinetic friction can be calculated using the equation Ff,k = μkFN where FN is the normal force exerted by the second surface, and μk is the coefficient of kinetic friction. The coefficient of friction is a unitless value that depends on the materials of the two surfaces. The larger the coefficient of friction, the larger the friction force will be. Static friction refers to friction between two surfaces that are not moving relative to each other. The maximum static friction force between two surfaces can be calculated using the equation Ff,s ≤ μsFN where once again, FN is the normal force exerted by the second surface, and μs is the coefficient of static friction. This equation only gives the maximum possible static friction force, so it is important to pay at- tention to the other forces acting on the object to determine what the actual static friction force is acting on the object. Static friction will increase or decrease to balance other forces acting along the same axis.

HOOKE’S LAW m=50kg μk =0.4 ΣFy =FN −Fg =0N FN =Fg =mg Ff,k = μkFN = μkmg = 0.4(50kg)(9.81m/s2) Ff,k = 196.2N Hooke’s Law details the relationship between the force exerted by a spring and the displacement (how far the spring has been stretched or compressed) of the spring. Fs = k∆x In the equation, k is the spring constant. The spring constant is a value that corresponds to the individual spring and indicates how much force is needed to stretch or compress the spring a distance of one meter. Spring constant is expressed in units of newtons per meter (N/m). Spring force is a restorative force. This means that the force acts in a direction opposite the displacement. A stretched spring will want to contract and a compressed spring will want to decompress.

NEWTON’S THIRD LAW OF MOTION Newton’s Third Law of Motion - When an object exerts a force on a second object, the second object will simultaneously exert a force equal in magnitude and in the opposite direction on the first object. When applying Newton’s third law, it is important to be able to differentiate between third law pairs and balanced forces. Forces that make up a third law pair must act simultaneously and act on different objects. Balanced forces do not have to occur simultaneously and act on the same object.

NEWTON’S LAW OF UNIVERSAL GRAVITATION Newton’s Law of Universal Gravitation states that all objects that have mass are gravitationally attracted to all other objects that have mass. This gravitational force can be calculated using the equation Fg =Gm1m2 r2 where G is the gravitational constant (6.67 × 10−11 m3 ), m and m are the masses of the objects, and r kgs2 1 2 is the distance between the objects’ centers of mass.