Free-body diagrams Pg. 15 in NB A free-body diagram is a vital tool for applying Newton's laws. It shows a single object isolated from its environment, with all interactions replaced by forces. Students learn the rules for drawing these diagrams. Weight is shown acting at the center of mass of the object. The normal force acts at the point of contact with a supporting surface. When the total, or net, force acting on an object is zero, the object is in equilibrium. Anything not moving (static) is in equilibrium. In this lesson students learn the rules fro drawing free-body diagrams and use them to calculate forces on objects in equilibrium.

Objectives Physics terms Identify normal forces. Draw a free-body diagram with forces in vertical and/or horizontal directions. Calculate the net force on an object acted on by forces in vertical and/or horizontal directions. net force normal force free-body diagram The lesson objectives describe what the student should know or be able to do upon completion of the lesson.

Free-body diagrams If you know the forces acting on an object, you can predict its motion. Free-body diagrams are valuable tools for figuring out the magnitudes and directions of the forces that act on an object. FN1 FN2 Fw Students sometimes think that free-body diagrams are an end unto themselves. Stress the fact that learning to draw these diagrams is a valuable skill. These diagrams are calculation tools as well as a way of visually presenting key relationships.

Free-body diagrams A free-body diagram is a sketch of an object isolated from its surroundings. ONLY the forces exerted ON the object are included in the diagram. Forces are drawn as arrows. FN1 FN2 Fw Key point: the free-body diagram includes all forces acting ON an object, but NOT the forces that this object exerts on its surroundings. Point out that the two surfaces are applying forces to the ball that hold it in place. In the free-body diagram the surfaces disappear but the forces they exert are shown with force vector arrows. Reassure students that will learn about the force subscripts for FN later in this lesson.

Free-body diagrams Real object Free-body diagram Step 1 Start a free-body diagram by drawing an outline of the object. This program will use the outline method for drawing free-body diagrams. This is the first step.

Free-body diagrams Real object Free-body diagram Step 1 Start a free-body diagram by drawing an outline of the object. This program will use the outline method for drawing free-body diagrams. This is the first step.

Weight Step 2 Next, draw the forces acting ON the object, starting with weight. mg The weight vector is drawn from the center of mass of the object, and points straight down. Students should start a free-body diagram by drawing the weight vector because weight ALWAYS points straight down and ALWAYS equals mg. The values/ of other forces may depend on the weight. Center of mass is explained on the next slide.

Center of mass The center of mass is the “balance point” around which all of an object’s mass is equally distributed. It is at the center of symmetrical shapes. Demonstrate the concept of center of mass by balancing a pair of scissors on your finger. The place where it balances is the place where the weight vector can be drawn.

Weight acts at center of mass Draw the weight force on a free-body diagram approximately at the center of mass of the object. 10 kg 10 kg Assure the students that these diagrams do not need to be exact. They will be used to guide thinking and calculations.

Weight acts at center of mass Draw the weight force on a free-body diagram approximately at the center of mass of the object. Assure the students that these diagrams do not need to be exact. They will be used to guide thinking and calculations.

Applied forces F This spring pulls upward on the object. mg Step 3 Add applied forces to the diagram. Applied forces are drawn at the point where they act, and in the correct direction.

Normal or support forces FN FN mg Surfaces that contact the object exert a normal or support force, FN. There is no easy equation for the magnitude of the normal force. It must be figured out based on Newton’s laws and depends on the other forces acting on the object. Students CAN immediately know the direction of the normal force.

Direction of the normal force FN FN mg Surfaces always push, NEVER pull. The table pushes up on the barbell, so the normal forces point up. Notice that with this style of free-body diagram, forces that push on the object point toward it, while forces that pull on the object point away from it.

Direction of the normal force FN FN mg Normal means perpendicular so that means normal forces are ALWAYS perpendicular to the surface the object is touching. Students may need to be reminded that the word perpendicular means that there is a right angle (90°), as shown. Mention that the normal force ALWAYS acts perpendicular to the surface, even if the surface is tilted. it can be helpful to demonstrate this with a pencil (representing the vector) held perpendicular to a book (the surface), even as you tilt the book.

Free-body diagrams This is a complete free-body diagram. It contains ALL the forces that act ON the object. Every force is identified with a label and direction. It does not have too much detail—a rough sketch is all you need. mg FN FN Some teachers may feel very strongly about drawing free-body diagrams to scale, so that the length of the vector arrows reflects the magnitude of the vectors. At this point in the process, these magnitudes have not yet been calculated. The object at this stage of a problem-solving process is simply to identify all forces acting on the object, and correctly show their direction.

Identify all the forces On a free-body diagram, include every force that acts ON the object: weight, normal forces, and applied forces from springs, ropes, and other sources. The isolated object acts exactly as it did before being “removed” from contact with the environment. FN FN mg

More on the normal force Every contact with a surface creates a normal force. Normal forces may be vertical, horizontal, or act at an angle. Remember: normal forces are always perpendicular to the surfaces that applied them. Examples of normal forces Be sure to assign different names to different normal forces! Point out the direction of the normal force is always perpendicular to the surface and always pushes away from the surface toward the object.

More on the normal force Every contact with a surface creates a normal force. Normal forces may be vertical, horizontal, or act at an angle. Remember: normal forces are always perpendicular to the surfaces that applied them. Examples of normal forces Be sure to assign different names to different normal forces! Point out the direction of the normal force is always perpendicular to the surface and always pushes away from the surface toward the object.

Styles of free-body diagrams There are two different styles you may see for drawing free-body diagrams. A block of mass m sits on a floor partially suspended by two springs. The outline method has the advantage of getting students to decide if a force is pushing or pulling on an object. The point-mass method makes sense when an object is being treated as a particle, but cannot be used when considering torques and rotation later in the program.

Force is a vector To solve force problems, you have to choose which directions will be positive and which will be negative. For beginning students it can be helpful to be consistent with this choice. As the students advance, they should learn to be flexible and choose to positive direction based on whatever makes the problem easy to solve.

Force is a vector This choice is arbitrary. Choose the positive direction that makes the problem easiest to solve. Always make a diagram to remind yourself which direction is positive!

The net force In most situations there are many forces acting at once. Objects respond to the net force. In physics “net” means total, taking directions into account. FN FN mg What is the net force acting on the dumbbell? 𝐹 𝑛𝑒𝑡 =𝐹+ 𝐹 𝑁 + 𝐹 𝑁 − 𝐹 𝑤 The net force is the vector sum of the forces, not the arithmetic sum. Fnet = F + 2FN - Fw

Equilibrium Equilibrium exists when the net force is zero. In equilibrium there is no change in motion. An object at rest stays at rest. F Fnet = 0 FN FN mg The dumbbell is at rest so the net force on it must be zero: Fnet = F + 2FN - Fw= 0 This is really a partial statement of Newton’s first law, which will be introduced formally in the next lesson.

Find the normal force The box shown is at rest, so Fnet = 0. What is FN in these examples? Pulled up with a force of 4 N. Pressed down with a 4 N force Pressed against the ceiling with a 15 N force F = 4 N. F = 4 N mg = 10 N mg = 10 N mg = 10 N F = 15 N.

Find the normal force Notice: there is no formula for calculating the normal force. Its magnitude depends on the situation. Pulled up with a force of 4 N. Pressed down with a 4 N force Pressed against the ceiling with a 15 N force FN = 5 N. F = 4 N. F = 4 N mg = 10 N mg = 10 N mg = 10 N F = 15 N. FN = 6 N. FN = 14 N.

Equilibrium problems An object at rest is in equilibrium. Therefore the net force is zero. What is the tension in the ropes that support the gymnast? Fnet = 0 Point out the the tension forces point up and the weight points down. The tensions must equal the weight.

Equilibrium problems 𝐹+𝐹−𝑚𝑔=0 𝐹= 637 2 =318 𝑁 2𝐹−637=0 The gymnast is centered, so we can assume the forces from the ropes are equal. Therefore we give them the same name, F. 𝐹+𝐹−𝑚𝑔=0 𝐹= 637 2 =318 𝑁 2𝐹−637=0 Tell the students they can make this argument by symmetry. The weight down equals 637 N, so the total tension up must also equal 637 N. When an object is in equilibrium, forces up must cancel forces down. Also true: forces left must cancel forces right.

Solving equilibrium problems Here are the EQUILIBRIUM problem-solving steps. Draw the free-body diagram of the object. Name all the forces. Set the net force to zero, taking account of +/- directions. Solve for the unknown force. Students will use these steps to solve problems on the student assignment.

Test your knowledge Two masses are at rest, connected by a rope that passes over two frictionless pulleys. The string tension T equals m2g. Which of these free body diagrams best represents the forces acting on mass m1? This problem is a little bit advanced, but students can get to the right answer by elimination, even if they do not fully understand all aspects of the situation. Because the masses are at rest, the forces on the hanging mass must equal zero. Therefore, the force of tension in the rope must be equal to the weight of the hanging mass. The pulleys do not effect the magnitude of the tension. They only serve to change the direction of the tension force. the tension is an ideal rope is constant throughout its length, and always pulls at both ends.

Test your knowledge Diagram A is best. Which of these free body diagrams best represents the forces acting on mass m1? Diagram A is best. Diagram (B) omits the normal force from the floor. Diagram (C) incorrectly shows the string tension acting downward.

Assessment A 10 kg dumbbell resting on a table is partly supported by a spring that pulls upward with a force of 50 N. Draw the free-body diagram for the dumbbell. What is the magnitude of the net force acting on the dumbbell? What force does the table exert on the dumbbell to hold it up? This assessment is keyed to the objectives. Students should be able to: Identify normal forces, draw a free-body diagram with forces in vertical and/or horizontal directions, and calculate the net force on an object acted on by forces in vertical and/or horizontal directions. The answers appear on the following slides.

Assessment Draw the free-body diagram for the dumbbell. FN FN mg 10 kg FN FN mg

Assessment F = 50 N A 10 kg dumbbell resting on a table is partly supported by a spring that pulls upward with a force of 50 N. 2. What is the magnitude of the net force acting on the dumbbell? 10 kg FN FN mg Zero. It is at rest (in equilibrium).

Assessment F = 50 N A 10 kg dumbbell resting on a table is partly supported by a spring that pulls upward with a force of 50 N. 3. What force does the table exert on the dumbbell to hold it up? 10 kg FN FN mg Fw = mg = (10 kg)(9.8 N/kg) = 98 N Fnet = 50 N + 2FN - 98 N = 0 2FN = 98 N – 50 N = 48 N (TOTAL) 24N on EACH END of the dumbbell Each normal force equals 24 N.