SECTION Madison Evans, Mitchell Peters, Ryder Briggs

5.1 - EQUILIBRIUM If an object is in equilibrium, Newton’s Second Law is used to find the forces that keep it in equilibrium. Two types: -Static: an object at rest -Dynamic: moving constant speed along a straight line In both types of equilibrium, Newton’s Second Law shows that there is no net force acting on the object meaning the sum of all the forces acting upon an object is zero. Component form is putting the equation in terms of x and y (shown later in presentation). This equation is used to solve for missing forces.

5.1-EQUILIBRIUM PROBLEMS Equilibrium refers to Newton’s 2 nd law of motion. 1.Check that the object is in equilibrium: does a=0 -an object at rest is in static equilibrium -an object moving at a constant velocity is in dynamic equilibrium Identify all forces acting on the object. Determine which forces you know and need to solve. 2. Solve- an object in equilibrium must follow Newton's Second Law - ∑ F x = ma x = 0 and ∑ F y = ma y = 0 - you can find force components directly from free-body diagram. Solve for unknown 3. Make sure have correct units, is reasonable, and answers the question

5.2 – DYNAMICS & NEWTON’S SECOND LAW Newton’s Second Law - a = F net /m - object’s motion can be determined by using kinematics equations Component equations - ∑ F x = ma x and ∑ F y = ma y Dynamics problems are solved using these equations and the kinematics equations -one can start from know to find acceleration -or one can start from kinematics to find a

1.Sketch visual overview - list known quantities and what problem wants you to find - force identification diagram for all forces - free-body diagram * When using kinematics have motion diagrams for acceleration and pictorial representation 2. Solve - writing Newton 2 nd law in component form depending on problem - solve for acceleration then use kinematics velocity and positions - use kinematics to determine acceleration and then solve for forces 3. Make sure have correct units, is reasonable, and answers the question 5.2 – DYNAMICS PROBLEMS SOLVING STRATEGY

5.3-MASS AND WEIGHT Mass describes an object’s inertia. It describes the amount of matter within an object. Weight is a force based. Its is determined by gravity’s force on an object. Weight is a vector and not scalar. W=mg The magnitude of the weight force is directly proportional to the mass. Weight varies based on gravity but mass is always consistent. -1lb=4.45N -M=W/g=4.45/9.8m/s 2 -1m=3.28ft Example: W gymnast =90lb×(4.45/1lb)=400N

5.3-APPARENT WEIGHT AND WEIGHTLESSNESS Apparent weight=W app =magnitude of supporting contact forces. -weight and apparent weight are generally the same in equilibrium A person in freefall has no apparent weight -weightless means no apparent weight Example of apparent weight is on an elevator, you feel light going down, but going up you feel heavier and becomes harder to jump or get off the ground.

VIDEO A lift of mass 500 kg is lowered or raised by means of metal cable attached to it top. The lift contains passengers whose total mass is 300 kg. The lift starts from rest and accelerates at constant rate reaching a speed of 3m/s after moving a distance of 5 m. Find A. acceleration of the lift B. tension in the cable if the lift is moving vertically downwards C. tension in the cable if the lift is moving vertically upwards