A PHYSICS EXAM Don’t be Static!

Solving problems when  F = 0

Total force,  F sum of the forces, must equal zero To be in static or translational equilibrium the Net force, Total force,  F sum of the forces, must equal zero about any point. Therefore the object may be: Stationary Moving with a constant velocity

How to approach an equilibrium problem: Draw a free-body diagram Choose a coordinate axis and resolve all forces into components. Set the sum of the force components each equal to 0. Solve the resulting equations for the unknowns.

Free Body Diagrams are a very important things for an engineer to know how to draw and use. What ? - It is a drawing that shows all external forces acting on the particle. Why ? - It helps you write the equations of equilibrium used to solve for unknowns (usually forces or angles).

Free Body Diagrams

When a particle is in equilibrium, the sum of forces acting on it equals ___ . a constant a positive number zero a negative number an integer. Answers: 1. C 2. B

Free-Body Diagram: A sketch showing all the forces acting on the body. If the weight of the lumber is 736 N find the tension in cables TAB and TAC

Find the tension in the cables, A and B, if the mass of the engine is 250 kg, neglect the mass of the cables themselves. Explain the following terms : Coplanar force system – The forces lie on a sheet of paper (plane). A particle – It has a mass, but a size that can be neglected. Equilibrium – A condition of rest or constant velocity. m = 250 kg

If the car is towed at constant speed and angle  = 25° 1 m If the car is towed at constant speed and angle  = 25° Find: The forces in the ropes AB and AC.

Find T1 and T2 if the weight of the signal, w, is 100 Newtons.

 Fx = FAC cos 30° – FAB cos 25° = 0 600 N Find FAC and FAB if the object located at A is in equilibrium A 25° 30° FAC FAB  Fx = FAC cos 30° – FAB cos 25° = 0  Fy = -FAC sin 30° – FAB sin 25° + 600 = 0 Solving the above equations, we get; FAB = 634 N FAC = 664 N

Using this FBD of Point O, the sum of forces in the x-direction ( FX ) is ___ . F2 sin 50° – 20 = 0 F2 cos 50° – 20 = 0 F2 sin 50° – F1 = 0 F2 cos 50° + 20 = 0 20 N 50° O F1 Answers: 2. B

TBy TB TA TAy A B TAx TBx q f q f M W W = Mg Hypothetical Example: mass hanging from two ropes as shown with free body diagram and component diagram.

Example: Determine the tension in each rope. 10kg 60o A B Example: Determine the tension in each rope. Draw a free-body diagram) Choose a convenient set of coordinate axis and resolve all forces into components. Watch carefully for appropriate use of +/- signs. Set the sum of the force components along each axis equal to 0. Solve the resulting equations for the unknown quantity or quantities. Substitute numerical values of the known quantities to find the answer.

A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with a constant velocity of 1 m/s, what must the tension in the rope be?

Choose two of the largest students in the class to pull on the opposite ends of a strong 10 m long rope. Ask the students if the rope appears to be straight. Have the smallest student attempt to push down on the middle of the rope with one finger. The small student will easily be able to push the rope down. Material: 10 m rope.

To demonstrate the non-perpendicular components of vectors, tie two loops in the ends of a strong cord that is about 1 m long. Have a student hold a loop in each hand as a classmate carefully hangs a 1 kg mass on the cord. He/she should start with his/her hands together and slowly move them apart. It is not possible to keep the cord straight.

To quantitatively demonstrate vector addition of forces, attach large dial type spring balances to the top of the chalkboard. Hang a known weight at various points on a string between them. The vectors then can easily be placed on the chalkboard and the equilibrium conditions worked out.